message stringlengths 2 23.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a set of strings S. Each string consists of lowercase Latin letters.
For each string in this set, you want to calculate the minimum number of seconds required to type this string. To type a string, you have to start with an empty string and transform it into the string you want to type using the following actions:
* if the current string is t, choose some lowercase Latin letter c and append it to the back of t, so the current string becomes t + c. This action takes 1 second;
* use autocompletion. When you try to autocomplete the current string t, a list of all strings s ∈ S such that t is a prefix of s is shown to you. This list includes t itself, if t is a string from S, and the strings are ordered lexicographically. You can transform t into the i-th string from this list in i seconds. Note that you may choose any string from this list you want, it is not necessarily the string you are trying to type.
What is the minimum number of seconds that you have to spend to type each string from S?
Note that the strings from S are given in an unusual way.
Input
The first line contains one integer n (1 ≤ n ≤ 10^6).
Then n lines follow, the i-th line contains one integer p_i (0 ≤ p_i < i) and one lowercase Latin character c_i. These lines form some set of strings such that S is its subset as follows: there are n + 1 strings, numbered from 0 to n; the 0-th string is an empty string, and the i-th string (i ≥ 1) is the result of appending the character c_i to the string p_i. It is guaranteed that all these strings are distinct.
The next line contains one integer k (1 ≤ k ≤ n) — the number of strings in S.
The last line contains k integers a_1, a_2, ..., a_k (1 ≤ a_i ≤ n, all a_i are pairwise distinct) denoting the indices of the strings generated by above-mentioned process that form the set S — formally, if we denote the i-th generated string as s_i, then S = {s_{a_1}, s_{a_2}, ..., s_{a_k}}.
Output
Print k integers, the i-th of them should be equal to the minimum number of seconds required to type the string s_{a_i}.
Examples
Input
10
0 i
1 q
2 g
0 k
1 e
5 r
4 m
5 h
3 p
3 e
5
8 9 1 10 6
Output
2 4 1 3 3
Input
8
0 a
1 b
2 a
2 b
4 a
4 b
5 c
6 d
5
2 3 4 7 8
Output
1 2 2 4 4
Note
In the first example, S consists of the following strings: ieh, iqgp, i, iqge, ier. | instruction | 0 | 25,725 | 0 | 51,450 |
Tags: data structures, dfs and similar, dp
Correct Solution:
```
import io
import os
DEBUG = False
def dfs(trie, root, preorder=None, postorder=None):
stack = [root]
seen = set()
while stack:
nodeId = stack.pop()
if nodeId not in seen:
if preorder:
preorder(nodeId)
stack.append(nodeId)
seen.add(nodeId)
for c, childId in reversed(trie[nodeId]):
stack.append(childId)
else:
if postorder:
postorder(nodeId)
def solve(N, PC, K, A):
ROOT = 0
trie = {ROOT: []} # nodeId to list of (character, nodeId)
parent = {}
for i, (p, c) in enumerate(PC, 1): # i starts from 1
trie[p].append((c, i))
assert i not in trie
trie[i] = []
parent[i] = p
terminal = set(A)
# Sort children of each node by character
for children in trie.values():
children.sort()
# DFS
offset = 0
ancestor = []
dist = {}
def getDistPre(nodeId):
nonlocal offset
if nodeId == 0:
# Is root
dist[nodeId] = 0
ancestor.append((0, nodeId, offset))
else:
assert nodeId in parent
# Default best dist is 1 from parent dist
best = 1 + dist[parent[nodeId]]
if nodeId in terminal:
# If terminal node, jump from the best ancestor
# Costs (offset - oldOffset) + 1 number of other terminals node that we need to skip past
# Note: This tuple is just for debugging, can actually get away with just storing sortKey
sortKey, ancestorId, oldOffset = ancestor[-1]
assert (
sortKey + offset + 1 == dist[ancestorId] + (offset - oldOffset) + 1
)
best = min(best, dist[ancestorId] + (offset - oldOffset) + 1)
ancestor.append(min(ancestor[-1], (best - offset, nodeId, offset)))
dist[nodeId] = best
# Count how many terminal nodes have been seen
if nodeId in terminal:
offset += 1
def getDistPost(nodeId):
ancestor.pop()
dfs(trie, ROOT, preorder=getDistPre, postorder=getDistPost)
if DEBUG:
def printNode(nodeId, word):
return (
str(nodeId)
+ "\t"
+ word
+ ("$" if nodeId in terminal else "")
+ "\t"
+ "dist: "
+ str(dist[nodeId])
)
return str(nodeId) + "\t" + word + ("$" if nodeId in terminal else "")
def printGraph(nodeId, path):
W = 8
depth = len(path)
for ch, childId in trie[nodeId]:
path.append(ch)
print(
(
" " * (W * depth)
+ "└"
+ ch.center(W - 1, "─")
+ str(childId)
+ ("$" if childId in terminal else "")
).ljust(50)
+ printNode(childId, "".join(path))
)
printGraph(childId, path)
path.pop()
printGraph(ROOT, [])
out = []
for a in A:
out.append(str(dist[a]))
return " ".join(out)
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
N, = list(map(int, input().split()))
PC = []
for i in range(N):
p, c = input().decode().split()
PC.append((int(p), str(c)))
K, = list(map(int, input().split()))
A = list(map(int, input().split()))
ans = solve(N, PC, K, A)
print(ans)
``` | output | 1 | 25,725 | 0 | 51,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a set of strings S. Each string consists of lowercase Latin letters.
For each string in this set, you want to calculate the minimum number of seconds required to type this string. To type a string, you have to start with an empty string and transform it into the string you want to type using the following actions:
* if the current string is t, choose some lowercase Latin letter c and append it to the back of t, so the current string becomes t + c. This action takes 1 second;
* use autocompletion. When you try to autocomplete the current string t, a list of all strings s ∈ S such that t is a prefix of s is shown to you. This list includes t itself, if t is a string from S, and the strings are ordered lexicographically. You can transform t into the i-th string from this list in i seconds. Note that you may choose any string from this list you want, it is not necessarily the string you are trying to type.
What is the minimum number of seconds that you have to spend to type each string from S?
Note that the strings from S are given in an unusual way.
Input
The first line contains one integer n (1 ≤ n ≤ 10^6).
Then n lines follow, the i-th line contains one integer p_i (0 ≤ p_i < i) and one lowercase Latin character c_i. These lines form some set of strings such that S is its subset as follows: there are n + 1 strings, numbered from 0 to n; the 0-th string is an empty string, and the i-th string (i ≥ 1) is the result of appending the character c_i to the string p_i. It is guaranteed that all these strings are distinct.
The next line contains one integer k (1 ≤ k ≤ n) — the number of strings in S.
The last line contains k integers a_1, a_2, ..., a_k (1 ≤ a_i ≤ n, all a_i are pairwise distinct) denoting the indices of the strings generated by above-mentioned process that form the set S — formally, if we denote the i-th generated string as s_i, then S = {s_{a_1}, s_{a_2}, ..., s_{a_k}}.
Output
Print k integers, the i-th of them should be equal to the minimum number of seconds required to type the string s_{a_i}.
Examples
Input
10
0 i
1 q
2 g
0 k
1 e
5 r
4 m
5 h
3 p
3 e
5
8 9 1 10 6
Output
2 4 1 3 3
Input
8
0 a
1 b
2 a
2 b
4 a
4 b
5 c
6 d
5
2 3 4 7 8
Output
1 2 2 4 4
Note
In the first example, S consists of the following strings: ieh, iqgp, i, iqge, ier.
Submitted Solution:
```
print('Hello, world')
``` | instruction | 0 | 25,726 | 0 | 51,452 |
No | output | 1 | 25,726 | 0 | 51,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a set of strings S. Each string consists of lowercase Latin letters.
For each string in this set, you want to calculate the minimum number of seconds required to type this string. To type a string, you have to start with an empty string and transform it into the string you want to type using the following actions:
* if the current string is t, choose some lowercase Latin letter c and append it to the back of t, so the current string becomes t + c. This action takes 1 second;
* use autocompletion. When you try to autocomplete the current string t, a list of all strings s ∈ S such that t is a prefix of s is shown to you. This list includes t itself, if t is a string from S, and the strings are ordered lexicographically. You can transform t into the i-th string from this list in i seconds. Note that you may choose any string from this list you want, it is not necessarily the string you are trying to type.
What is the minimum number of seconds that you have to spend to type each string from S?
Note that the strings from S are given in an unusual way.
Input
The first line contains one integer n (1 ≤ n ≤ 10^6).
Then n lines follow, the i-th line contains one integer p_i (0 ≤ p_i < i) and one lowercase Latin character c_i. These lines form some set of strings such that S is its subset as follows: there are n + 1 strings, numbered from 0 to n; the 0-th string is an empty string, and the i-th string (i ≥ 1) is the result of appending the character c_i to the string p_i. It is guaranteed that all these strings are distinct.
The next line contains one integer k (1 ≤ k ≤ n) — the number of strings in S.
The last line contains k integers a_1, a_2, ..., a_k (1 ≤ a_i ≤ n, all a_i are pairwise distinct) denoting the indices of the strings generated by above-mentioned process that form the set S — formally, if we denote the i-th generated string as s_i, then S = {s_{a_1}, s_{a_2}, ..., s_{a_k}}.
Output
Print k integers, the i-th of them should be equal to the minimum number of seconds required to type the string s_{a_i}.
Examples
Input
10
0 i
1 q
2 g
0 k
1 e
5 r
4 m
5 h
3 p
3 e
5
8 9 1 10 6
Output
2 4 1 3 3
Input
8
0 a
1 b
2 a
2 b
4 a
4 b
5 c
6 d
5
2 3 4 7 8
Output
1 2 2 4 4
Note
In the first example, S consists of the following strings: ieh, iqgp, i, iqge, ier.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
T=[input().split() for i in range(n)]
k=int(input())
S=list(map(int,input().split()))
SETS=set(S)
E=[[] for i in range(n+1)]
P=[-1]*(n+1)
for i in range(n):
p,s=T[i]
p=int(p)
E[p].append((s,i+1))
P[i+1]=p
for i in range(n+1):
E[i].sort(reverse=True)
ELI=[0]*(n+1)
DEPTH=[0]*(n+1)
ELIMIN=[0]*(n+1)
ANS=[-1]*(n+1)
Q=[0]
USED=[0]*(n+1)
count=0
while Q:
x=Q.pop()
USED[x]=1
if x in SETS:
count+=1
#print(x,count)
ELI[x]=DEPTH[x]-count
ELIMIN[x]=min(ELI[x],ELIMIN[P[x]])
if x in SETS:
ANS[x]=min(DEPTH[x],count+ELIMIN[x],ANS[P[x]]+1)
else:
ANS[x]=min(DEPTH[x],ANS[P[x]]+1)
for s,to in E[x]:
if USED[to]==1:
continue
Q.append(to)
DEPTH[to]=DEPTH[x]+1
print(*[ANS[s] for s in S])
``` | instruction | 0 | 25,727 | 0 | 51,454 |
No | output | 1 | 25,727 | 0 | 51,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a set of strings S. Each string consists of lowercase Latin letters.
For each string in this set, you want to calculate the minimum number of seconds required to type this string. To type a string, you have to start with an empty string and transform it into the string you want to type using the following actions:
* if the current string is t, choose some lowercase Latin letter c and append it to the back of t, so the current string becomes t + c. This action takes 1 second;
* use autocompletion. When you try to autocomplete the current string t, a list of all strings s ∈ S such that t is a prefix of s is shown to you. This list includes t itself, if t is a string from S, and the strings are ordered lexicographically. You can transform t into the i-th string from this list in i seconds. Note that you may choose any string from this list you want, it is not necessarily the string you are trying to type.
What is the minimum number of seconds that you have to spend to type each string from S?
Note that the strings from S are given in an unusual way.
Input
The first line contains one integer n (1 ≤ n ≤ 10^6).
Then n lines follow, the i-th line contains one integer p_i (0 ≤ p_i < i) and one lowercase Latin character c_i. These lines form some set of strings such that S is its subset as follows: there are n + 1 strings, numbered from 0 to n; the 0-th string is an empty string, and the i-th string (i ≥ 1) is the result of appending the character c_i to the string p_i. It is guaranteed that all these strings are distinct.
The next line contains one integer k (1 ≤ k ≤ n) — the number of strings in S.
The last line contains k integers a_1, a_2, ..., a_k (1 ≤ a_i ≤ n, all a_i are pairwise distinct) denoting the indices of the strings generated by above-mentioned process that form the set S — formally, if we denote the i-th generated string as s_i, then S = {s_{a_1}, s_{a_2}, ..., s_{a_k}}.
Output
Print k integers, the i-th of them should be equal to the minimum number of seconds required to type the string s_{a_i}.
Examples
Input
10
0 i
1 q
2 g
0 k
1 e
5 r
4 m
5 h
3 p
3 e
5
8 9 1 10 6
Output
2 4 1 3 3
Input
8
0 a
1 b
2 a
2 b
4 a
4 b
5 c
6 d
5
2 3 4 7 8
Output
1 2 2 4 4
Note
In the first example, S consists of the following strings: ieh, iqgp, i, iqge, ier.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
T=[input().split() for i in range(n)]
k=int(input())
S=list(map(int,input().split()))
SETS=set(S)
E=[[] for i in range(n+1)]
P=[-1]*(n+1)
for i in range(n):
p,s=T[i]
p=int(p)
E[p].append((s,i+1))
P[i+1]=p
for i in range(n+1):
E[i].sort(reverse=True)
ELI=[0]*(n+1)
DEPTH=[0]*(n+1)
ELIMIN=[0]*(n+1)
ANS=[0]*(n+1)
Q=[0]
USED=[0]*(n+1)
count=0
while Q:
x=Q.pop()
USED[x]=1
if x in SETS:
count+=1
#print(x,count)
ELI[x]=DEPTH[x]-count
ELIMIN[x]=min(ELI[x],ELIMIN[P[x]])
if x in SETS:
ANS[x]=min(DEPTH[x],count+ELIMIN[P[x]],ANS[P[x]]+1)
else:
ANS[x]=min(DEPTH[x],ANS[P[x]]+1)
for s,to in E[x]:
if USED[to]==1:
continue
Q.append(to)
DEPTH[to]=DEPTH[x]+1
print(*[ANS[s] for s in S])
``` | instruction | 0 | 25,728 | 0 | 51,456 |
No | output | 1 | 25,728 | 0 | 51,457 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a set of strings S. Each string consists of lowercase Latin letters.
For each string in this set, you want to calculate the minimum number of seconds required to type this string. To type a string, you have to start with an empty string and transform it into the string you want to type using the following actions:
* if the current string is t, choose some lowercase Latin letter c and append it to the back of t, so the current string becomes t + c. This action takes 1 second;
* use autocompletion. When you try to autocomplete the current string t, a list of all strings s ∈ S such that t is a prefix of s is shown to you. This list includes t itself, if t is a string from S, and the strings are ordered lexicographically. You can transform t into the i-th string from this list in i seconds. Note that you may choose any string from this list you want, it is not necessarily the string you are trying to type.
What is the minimum number of seconds that you have to spend to type each string from S?
Note that the strings from S are given in an unusual way.
Input
The first line contains one integer n (1 ≤ n ≤ 10^6).
Then n lines follow, the i-th line contains one integer p_i (0 ≤ p_i < i) and one lowercase Latin character c_i. These lines form some set of strings such that S is its subset as follows: there are n + 1 strings, numbered from 0 to n; the 0-th string is an empty string, and the i-th string (i ≥ 1) is the result of appending the character c_i to the string p_i. It is guaranteed that all these strings are distinct.
The next line contains one integer k (1 ≤ k ≤ n) — the number of strings in S.
The last line contains k integers a_1, a_2, ..., a_k (1 ≤ a_i ≤ n, all a_i are pairwise distinct) denoting the indices of the strings generated by above-mentioned process that form the set S — formally, if we denote the i-th generated string as s_i, then S = {s_{a_1}, s_{a_2}, ..., s_{a_k}}.
Output
Print k integers, the i-th of them should be equal to the minimum number of seconds required to type the string s_{a_i}.
Examples
Input
10
0 i
1 q
2 g
0 k
1 e
5 r
4 m
5 h
3 p
3 e
5
8 9 1 10 6
Output
2 4 1 3 3
Input
8
0 a
1 b
2 a
2 b
4 a
4 b
5 c
6 d
5
2 3 4 7 8
Output
1 2 2 4 4
Note
In the first example, S consists of the following strings: ieh, iqgp, i, iqge, ier.
Submitted Solution:
```
import io
import os
def dfs(trie, root, preorder=None, postorder=None):
stack = [root]
seen = set()
while stack:
nodeId = stack.pop()
if nodeId not in seen:
if preorder:
preorder(nodeId)
stack.append(nodeId)
seen.add(nodeId)
for c in reversed(list(trie[nodeId].keys())):
childId = trie[nodeId][c]
stack.append(childId)
else:
if postorder:
postorder(nodeId)
def solve(N, PC, K, A):
ROOT = 0
trie = {ROOT: {}} # nodeId to character to nodeId
parent = {}
for i, (p, c) in enumerate(PC, 1): # i starts from 1
trie[p][c] = i
assert i not in trie
trie[i] = {}
parent[i] = p
terminal = set(A)
# Reinsert everything to get lexicographical sort
# Assume OrderedDict by default (since python 3.5)
for i in trie.keys():
for ch, childId in sorted(trie[i].items(), key=lambda kv: kv[0]):
del trie[i][ch]
trie[i][ch] = childId
# DFS
# Track previous node in lexicographical order
nodeToPrevTerminal = {}
prevTerminal = 0
# Track the best ancestor (representing a prefix) to jump from
nodeToAncestor = {}
ancestor = [float("inf")]
ancestorUsed = set()
# Distance to get to each node either by typing or autocompleting
typingDist = {ROOT: 0}
autocompleteDist = {ROOT: 0}
# Answer
dist = {}
def getDistPre(nodeId):
# Create a linked list of terminal nodes in lexicographical order using a preorder traversal
nonlocal prevTerminal
if nodeId in terminal:
nodeToPrevTerminal[nodeId] = prevTerminal
prevTerminal = nodeId
if nodeId in parent:
# Can type another character from parent (regardless of whether parent was formed by typing or autocompleting)
parentId = parent[nodeId]
typingDist[nodeId] = 1 + dist[parentId]
if nodeId in terminal:
# Can autocomplete:
# - From the same word to itself
jumpFrom = nodeId
if ancestor[-1] is not None and ancestor[-1] not in ancestorUsed:
# - From an ancestor prefix
jumpFrom = ancestor[-1]
ancestorUsed.add(ancestor[-1])
nodeToAncestor[nodeId] = ancestor[-1]
# - From the previous lexicographical autocompletion
autocompleteDist[nodeId] = 1 + min(
autocompleteDist[nodeToPrevTerminal[nodeId]], typingDist[jumpFrom]
)
dist[nodeId] = min(
typingDist[nodeId], autocompleteDist.get(nodeId, float("inf"))
)
if nodeId in terminal:
# Terminal node can't be used as an ancestor for anything in the subtree
ancestor.append(None)
else:
# Non-terminal node can be used as an ancestor
if typingDist[nodeId] < typingDist.get(ancestor[-1], float("inf")):
ancestor.append(nodeId)
else:
ancestor.append(ancestor[-1])
def getDistPost(nodeId):
ancestor.pop()
dfs(trie, ROOT, preorder=getDistPre, postorder=getDistPost)
if False:
def printNode(nodeId, word):
return (
str(nodeId)
+ "\t"
+ word
+ ("$" if nodeId in terminal else "")
+ "\t"
+ "prev: "
+ str(nodeToPrevTerminal.get(nodeId, " "))
+ "\t"
+ "anc: "
+ str(nodeToAncestor.get(nodeId, " "))
+ "\t"
+ "dist: "
+ str(dist[nodeId])
+ " "
+ str(typingDist[nodeId])
+ " "
+ str(autocompleteDist.get(nodeId, float("inf")))
)
return str(nodeId) + "\t" + word + ("$" if nodeId in terminal else "")
def printGraph(nodeId, path):
W = 8
depth = len(path)
for ch, childId in trie[nodeId].items():
path.append(ch)
print(
(
" " * (W * depth)
+ "└"
+ ch.center(W - 1, "─")
+ str(childId)
+ ("$" if childId in terminal else "")
).ljust(50)
+ printNode(childId, "".join(path))
)
printGraph(childId, path)
path.pop()
printGraph(ROOT, [])
out = []
for a in A:
out.append(str(dist[a]))
return " ".join(out)
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
N, = list(map(int, input().split()))
PC = []
for i in range(N):
p, c = input().decode().split()
PC.append((int(p), str(c)))
K, = list(map(int, input().split()))
A = list(map(int, input().split()))
ans = solve(N, PC, K, A)
print(ans)
``` | instruction | 0 | 25,729 | 0 | 51,458 |
No | output | 1 | 25,729 | 0 | 51,459 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Given the strings $ s $, $ t $.
First, let the string set $ A = $ {$ s $}, $ B = \ phi $.
At this time, I want to perform the following operations as much as possible.
operation
step1
Perform the following processing for all $ u ∊ A $.
1. From all subsequences of $ u $ (not necessarily contiguous), choose any one that is equal to $ t $. Let the indexes corresponding to $ u $ of the selected subsequence be $ z_1 $, $ z_2 $,…, $ z_ {| t |} $ (however, 1 $ \ le $ $ z_1 $ $ \ lt $ $ z_2). $$ \ lt $… $ \ lt $ $ z_ {| t |} $ $ \ le $ $ | u | $). If there is no such subsequence, the operation ends.
2. Divide the original string with the selected substring characters $ u_ {z_1} $, $ u_ {z_2} $,…, $ u_ {z_ {| t |}} $ and convert them to $ B $ to add.
For example, in the case of $ u = $ "abcdcd" and $ t = $ "ac", the following two division methods can be considered.
If you choose the first and third characters of $ u $, it will be split into "", "b" and "dcd".
If you choose the first and fifth characters of $ u $, it will be split into "", "bcd" and "d".
step2
Replace $ A $ with $ B $ and empty $ B $.
Increase the number of operations by 1.
Subsequence example
The subsequences of "abac" are {"", "a", "b", "c", "aa", "ab", "ac", "ba", "bc", "aac", "aba" "," abc "," bac "," abac "}.
"a" is a subsequence created by extracting the first or third character of "abac".
"ac" is a subsequence created by extracting the 1st and 4th characters of "abac" or the 3rd and 4th characters.
Constraints
The input satisfies the following conditions.
* 1 $ \ le $ $ | t | $ $ \ le $ $ | s | $ $ \ le $ $ 10 ^ 5 $
* Characters contained in the strings $ s $ and $ t $ are uppercase or lowercase letters of the alphabet
Input
The input is given in the following format.
$ s $
$ t $
The string $ s $ is given on the first line, and the string $ t $ is given on the second line.
Output
Output the maximum number of operations.
Examples
Input
AABABCAC
A
Output
2
Input
abaabbabaabaabbabbabaabbab
ab
Output
3
Input
AbCdEfG
aBcDeFg
Output
0 | instruction | 0 | 26,520 | 0 | 53,040 |
"Correct Solution:
```
s = input()
t = " " + input()
line = [1]
next = 1
t_size = len(t)
flag = False
for i in s:
if t[next] == i:
line[0] += 1
next = line[0]
line += [0]
j = 0
while line[j] == t_size:
line[j] = 0
next = line[j + 1] = line[j + 1] + 1
j += 1
if line[-1] == 0:
line.pop()
print(len(line) - 1)
``` | output | 1 | 26,520 | 0 | 53,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Given the strings $ s $, $ t $.
First, let the string set $ A = $ {$ s $}, $ B = \ phi $.
At this time, I want to perform the following operations as much as possible.
operation
step1
Perform the following processing for all $ u ∊ A $.
1. From all subsequences of $ u $ (not necessarily contiguous), choose any one that is equal to $ t $. Let the indexes corresponding to $ u $ of the selected subsequence be $ z_1 $, $ z_2 $,…, $ z_ {| t |} $ (however, 1 $ \ le $ $ z_1 $ $ \ lt $ $ z_2). $$ \ lt $… $ \ lt $ $ z_ {| t |} $ $ \ le $ $ | u | $). If there is no such subsequence, the operation ends.
2. Divide the original string with the selected substring characters $ u_ {z_1} $, $ u_ {z_2} $,…, $ u_ {z_ {| t |}} $ and convert them to $ B $ to add.
For example, in the case of $ u = $ "abcdcd" and $ t = $ "ac", the following two division methods can be considered.
If you choose the first and third characters of $ u $, it will be split into "", "b" and "dcd".
If you choose the first and fifth characters of $ u $, it will be split into "", "bcd" and "d".
step2
Replace $ A $ with $ B $ and empty $ B $.
Increase the number of operations by 1.
Subsequence example
The subsequences of "abac" are {"", "a", "b", "c", "aa", "ab", "ac", "ba", "bc", "aac", "aba" "," abc "," bac "," abac "}.
"a" is a subsequence created by extracting the first or third character of "abac".
"ac" is a subsequence created by extracting the 1st and 4th characters of "abac" or the 3rd and 4th characters.
Constraints
The input satisfies the following conditions.
* 1 $ \ le $ $ | t | $ $ \ le $ $ | s | $ $ \ le $ $ 10 ^ 5 $
* Characters contained in the strings $ s $ and $ t $ are uppercase or lowercase letters of the alphabet
Input
The input is given in the following format.
$ s $
$ t $
The string $ s $ is given on the first line, and the string $ t $ is given on the second line.
Output
Output the maximum number of operations.
Examples
Input
AABABCAC
A
Output
2
Input
abaabbabaabaabbabbabaabbab
ab
Output
3
Input
AbCdEfG
aBcDeFg
Output
0
Submitted Solution:
```
S = input()
T = input()
ti = count = 0
for i,c in enumerate(S):
if c == T[ti]:
ti += 1
if ti == len(T):
ti = 0
count += 1
ans = tmp = 0
while tmp <= count:
tmp += 2**ans
ans += 1
print(ans - 1)
``` | instruction | 0 | 26,521 | 0 | 53,042 |
No | output | 1 | 26,521 | 0 | 53,043 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider some set of distinct characters A and some string S, consisting of exactly n characters, where each character is present in A.
You are given an array of m integers b (b_1 < b_2 < ... < b_m).
You are allowed to perform the following move on the string S:
1. Choose some valid i and set k = b_i;
2. Take the first k characters of S = Pr_k;
3. Take the last k characters of S = Su_k;
4. Substitute the first k characters of S with the reversed Su_k;
5. Substitute the last k characters of S with the reversed Pr_k.
For example, let's take a look at S = "abcdefghi" and k = 2. Pr_2 = "ab", Su_2 = "hi". Reversed Pr_2 = "ba", Su_2 = "ih". Thus, the resulting S is "ihcdefgba".
The move can be performed arbitrary number of times (possibly zero). Any i can be selected multiple times over these moves.
Let's call some strings S and T equal if and only if there exists such a sequence of moves to transmute string S to string T. For the above example strings "abcdefghi" and "ihcdefgba" are equal. Also note that this implies S = S.
The task is simple. Count the number of distinct strings.
The answer can be huge enough, so calculate it modulo 998244353.
Input
The first line contains three integers n, m and |A| (2 ≤ n ≤ 10^9, 1 ≤ m ≤ min(\frac n 2, 2 ⋅ 10^5), 1 ≤ |A| ≤ 10^9) — the length of the strings, the size of the array b and the size of the set A, respectively.
The second line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ \frac n 2, b_1 < b_2 < ... < b_m).
Output
Print a single integer — the number of distinct strings of length n with characters from set A modulo 998244353.
Examples
Input
3 1 2
1
Output
6
Input
9 2 26
2 3
Output
150352234
Input
12 3 1
2 5 6
Output
1
Note
Here are all the distinct strings for the first example. The chosen letters 'a' and 'b' are there just to show that the characters in A are different.
1. "aaa"
2. "aab" = "baa"
3. "aba"
4. "abb" = "bba"
5. "bab"
6. "bbb" | instruction | 0 | 26,563 | 0 | 53,126 |
Tags: combinatorics, strings
Correct Solution:
```
n,m,a=map(int,input().split())
b=list(map(int,input().split()))
for i in range(m):
if i==0:
diffs=[b[0]]
else:
diffs.append(b[i]-b[i-1])
powers=[a%998244353]
for i in range(30):
powers.append(powers[-1]**2%998244353)
def power(x,y,binpowers):
prod=1
bits=bin(y)[2:]
bits=bits[::-1]
for i in range(len(bits)):
if bits[i]=="1":
prod*=binpowers[i]
prod%=998244353
return prod
maxi=b[-1]
prod1=power(a,n-2*maxi,powers)
for guy in diffs:
newprod=power(a,guy,powers)
newprod=(newprod*(newprod+1))//2
newprod%=998244353
prod1*=newprod
prod1%=998244353
print(prod1)
``` | output | 1 | 26,563 | 0 | 53,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider some set of distinct characters A and some string S, consisting of exactly n characters, where each character is present in A.
You are given an array of m integers b (b_1 < b_2 < ... < b_m).
You are allowed to perform the following move on the string S:
1. Choose some valid i and set k = b_i;
2. Take the first k characters of S = Pr_k;
3. Take the last k characters of S = Su_k;
4. Substitute the first k characters of S with the reversed Su_k;
5. Substitute the last k characters of S with the reversed Pr_k.
For example, let's take a look at S = "abcdefghi" and k = 2. Pr_2 = "ab", Su_2 = "hi". Reversed Pr_2 = "ba", Su_2 = "ih". Thus, the resulting S is "ihcdefgba".
The move can be performed arbitrary number of times (possibly zero). Any i can be selected multiple times over these moves.
Let's call some strings S and T equal if and only if there exists such a sequence of moves to transmute string S to string T. For the above example strings "abcdefghi" and "ihcdefgba" are equal. Also note that this implies S = S.
The task is simple. Count the number of distinct strings.
The answer can be huge enough, so calculate it modulo 998244353.
Input
The first line contains three integers n, m and |A| (2 ≤ n ≤ 10^9, 1 ≤ m ≤ min(\frac n 2, 2 ⋅ 10^5), 1 ≤ |A| ≤ 10^9) — the length of the strings, the size of the array b and the size of the set A, respectively.
The second line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ \frac n 2, b_1 < b_2 < ... < b_m).
Output
Print a single integer — the number of distinct strings of length n with characters from set A modulo 998244353.
Examples
Input
3 1 2
1
Output
6
Input
9 2 26
2 3
Output
150352234
Input
12 3 1
2 5 6
Output
1
Note
Here are all the distinct strings for the first example. The chosen letters 'a' and 'b' are there just to show that the characters in A are different.
1. "aaa"
2. "aab" = "baa"
3. "aba"
4. "abb" = "bba"
5. "bab"
6. "bbb"
Submitted Solution:
```
n,m,k=[int(x)for x in input().split()]
ms=[int(x) for x in input().split()]+[0]
ans=1
mod=998244353
def fp(a,p):
ans = 1
while p!=0:
if p%2==1:
ans *= a
ans %= mod
a *= a
a %= mod
p >>= 1
return ans%mod
# print(fp(2,10))
ms.sort()
def segNum(s):
ans=(fp(k,s*2)+fp(k,s))*fp(2,mod-2)
return ans%mod
for i in range(1,len(ms)):
seg=ms[i]-ms[i-1]
if seg==0:
continue
ans*=segNum(seg)%998244353
last=n-2*ms[-1]
ans*=fp(k,last)
# print(ans%998244353)
# print(k)
# k=1000000000%mod
# print(k)
# # print((k**2+k)//2%mod)
# print(fp(k,2),k**2%mod)
# print(fp(k,1),k)
print((fp(k,2)+fp(k,1))*fp(2,mod-2)%mod)
``` | instruction | 0 | 26,564 | 0 | 53,128 |
No | output | 1 | 26,564 | 0 | 53,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider some set of distinct characters A and some string S, consisting of exactly n characters, where each character is present in A.
You are given an array of m integers b (b_1 < b_2 < ... < b_m).
You are allowed to perform the following move on the string S:
1. Choose some valid i and set k = b_i;
2. Take the first k characters of S = Pr_k;
3. Take the last k characters of S = Su_k;
4. Substitute the first k characters of S with the reversed Su_k;
5. Substitute the last k characters of S with the reversed Pr_k.
For example, let's take a look at S = "abcdefghi" and k = 2. Pr_2 = "ab", Su_2 = "hi". Reversed Pr_2 = "ba", Su_2 = "ih". Thus, the resulting S is "ihcdefgba".
The move can be performed arbitrary number of times (possibly zero). Any i can be selected multiple times over these moves.
Let's call some strings S and T equal if and only if there exists such a sequence of moves to transmute string S to string T. For the above example strings "abcdefghi" and "ihcdefgba" are equal. Also note that this implies S = S.
The task is simple. Count the number of distinct strings.
The answer can be huge enough, so calculate it modulo 998244353.
Input
The first line contains three integers n, m and |A| (2 ≤ n ≤ 10^9, 1 ≤ m ≤ min(\frac n 2, 2 ⋅ 10^5), 1 ≤ |A| ≤ 10^9) — the length of the strings, the size of the array b and the size of the set A, respectively.
The second line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ \frac n 2, b_1 < b_2 < ... < b_m).
Output
Print a single integer — the number of distinct strings of length n with characters from set A modulo 998244353.
Examples
Input
3 1 2
1
Output
6
Input
9 2 26
2 3
Output
150352234
Input
12 3 1
2 5 6
Output
1
Note
Here are all the distinct strings for the first example. The chosen letters 'a' and 'b' are there just to show that the characters in A are different.
1. "aaa"
2. "aab" = "baa"
3. "aba"
4. "abb" = "bba"
5. "bab"
6. "bbb"
Submitted Solution:
```
n,m,k=[int(x)for x in input().split()]
ms=[int(x) for x in input().split()]+[0]
ans=1
mod=998244353
def fp(a,p):
ans = 1
while p:
if p&1:
ans *= a
ans %= mod
a *= a
a %= mod
p >>= 1
return ans
ms.sort()
def segNum(s):
ans=(fp(k,s*2)+fp(k,s))//2
return ans
for i in range(1,len(ms)):
seg=ms[i]-ms[i-1]
if seg==0:
continue
ans*=segNum(seg)%998244353
last=n-2*ms[-1]
ans*=k**(last)
print(ans%998244353)
``` | instruction | 0 | 26,565 | 0 | 53,130 |
No | output | 1 | 26,565 | 0 | 53,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of first 20 lowercase Latin letters ('a', 'b', ..., 't').
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
You can perform the following operation no more than once: choose some substring s[l; r] and reverse it (i.e. the string s_l s_{l + 1} ... s_r becomes s_r s_{r - 1} ... s_l).
Your goal is to maximize the length of the maximum substring of s consisting of distinct (i.e. unique) characters.
The string consists of distinct characters if no character in this string appears more than once. For example, strings "abcde", "arctg" and "minecraft" consist of distinct characters but strings "codeforces", "abacaba" do not consist of distinct characters.
Input
The only line of the input contains one string s consisting of no more than 10^6 characters 'a', 'b', ..., 't' (first 20 lowercase Latin letters).
Output
Print one integer — the maximum possible length of the maximum substring of s consisting of distinct characters after reversing no more than one its substring.
Examples
Input
abacaba
Output
3
Input
abcdecdf
Output
6
Input
aabbcc
Output
3
Input
abcdeefc
Output
6
Submitted Solution:
```
s = input ()
def inList(niddle, niddlesOfList):
for currentNiddle in niddlesOfList:
if ( currentNiddle == niddle ): return True
return False
listOfUniqueChar = []
if len(s) <= 1000000 :
stringInLc = s.lower()
for letter in stringInLc:
if inList( letter, listOfUniqueChar ) == False and ord(letter) > 96 and ord(letter) < 117:
listOfUniqueChar.append(letter)
print( len(listOfUniqueChar) )
``` | instruction | 0 | 26,653 | 0 | 53,306 |
No | output | 1 | 26,653 | 0 | 53,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of first 20 lowercase Latin letters ('a', 'b', ..., 't').
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
You can perform the following operation no more than once: choose some substring s[l; r] and reverse it (i.e. the string s_l s_{l + 1} ... s_r becomes s_r s_{r - 1} ... s_l).
Your goal is to maximize the length of the maximum substring of s consisting of distinct (i.e. unique) characters.
The string consists of distinct characters if no character in this string appears more than once. For example, strings "abcde", "arctg" and "minecraft" consist of distinct characters but strings "codeforces", "abacaba" do not consist of distinct characters.
Input
The only line of the input contains one string s consisting of no more than 10^6 characters 'a', 'b', ..., 't' (first 20 lowercase Latin letters).
Output
Print one integer — the maximum possible length of the maximum substring of s consisting of distinct characters after reversing no more than one its substring.
Examples
Input
abacaba
Output
3
Input
abcdecdf
Output
6
Input
aabbcc
Output
3
Input
abcdeefc
Output
6
Submitted Solution:
```
s=input()
p=len(s)
for x in s:
d=0
for y in s:
if(x==''):
break
if(x==y):
d+=1
if(d>1 and x==y):
s=s.replace(x,'',1)
p-=1
print(p)
``` | instruction | 0 | 26,654 | 0 | 53,308 |
No | output | 1 | 26,654 | 0 | 53,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of first 20 lowercase Latin letters ('a', 'b', ..., 't').
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
You can perform the following operation no more than once: choose some substring s[l; r] and reverse it (i.e. the string s_l s_{l + 1} ... s_r becomes s_r s_{r - 1} ... s_l).
Your goal is to maximize the length of the maximum substring of s consisting of distinct (i.e. unique) characters.
The string consists of distinct characters if no character in this string appears more than once. For example, strings "abcde", "arctg" and "minecraft" consist of distinct characters but strings "codeforces", "abacaba" do not consist of distinct characters.
Input
The only line of the input contains one string s consisting of no more than 10^6 characters 'a', 'b', ..., 't' (first 20 lowercase Latin letters).
Output
Print one integer — the maximum possible length of the maximum substring of s consisting of distinct characters after reversing no more than one its substring.
Examples
Input
abacaba
Output
3
Input
abcdecdf
Output
6
Input
aabbcc
Output
3
Input
abcdeefc
Output
6
Submitted Solution:
```
s=input("enter the string")
l=[]
ll=list(s)
for i in ll:
if(i not in l):
l.append(i)
c=len(l)
print(c)
``` | instruction | 0 | 26,655 | 0 | 53,310 |
No | output | 1 | 26,655 | 0 | 53,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting only of first 20 lowercase Latin letters ('a', 'b', ..., 't').
Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".
You can perform the following operation no more than once: choose some substring s[l; r] and reverse it (i.e. the string s_l s_{l + 1} ... s_r becomes s_r s_{r - 1} ... s_l).
Your goal is to maximize the length of the maximum substring of s consisting of distinct (i.e. unique) characters.
The string consists of distinct characters if no character in this string appears more than once. For example, strings "abcde", "arctg" and "minecraft" consist of distinct characters but strings "codeforces", "abacaba" do not consist of distinct characters.
Input
The only line of the input contains one string s consisting of no more than 10^6 characters 'a', 'b', ..., 't' (first 20 lowercase Latin letters).
Output
Print one integer — the maximum possible length of the maximum substring of s consisting of distinct characters after reversing no more than one its substring.
Examples
Input
abacaba
Output
3
Input
abcdecdf
Output
6
Input
aabbcc
Output
3
Input
abcdeefc
Output
6
Submitted Solution:
```
def sub(s):
arr=[]
for i in s:
if i in arr:
arr.append(i)
return len(arr)
if __name__=="__main__":
s=str(input())
print(sub(s))
``` | instruction | 0 | 26,656 | 0 | 53,312 |
No | output | 1 | 26,656 | 0 | 53,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition. | instruction | 0 | 26,790 | 0 | 53,580 |
Tags: greedy, implementation, strings
Correct Solution:
```
t = int(input())
for i in range(0,t):
s = list(input())
s = [ord(s[j])-96 for j in range(0,len(s))]
# s.sort()
n = len(s)
if 1 not in s:
print("NO")
continue
l=s.index(1)-1
r = s.index(1)+1
k=2
f=1
# print(s)
while(l>=0 and r<n):
if s[l]==k:
l-=1
k+=1
elif s[r]==k:
r+=1
k+=1
else:
f=0
break
while(l>=0 and f==1):
if s[l]==k:
l-=1
k+=1
else:
f=0
break
while(r<n and f==1):
if s[r]==k:
r+=1
k+=1
else:
f=0
break
if f==1:
print("YES")
else:
print("NO")
``` | output | 1 | 26,790 | 0 | 53,581 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition. | instruction | 0 | 26,791 | 0 | 53,582 |
Tags: greedy, implementation, strings
Correct Solution:
```
from collections import defaultdict
for _ in range(int(input())):
x=input()
arr=[0]*123
c=len(x)
aa=0
for i in range(c):
arr[ord(x[i])]+=1
if x[i]=='a':
aa=i
flag=0
for j in range(97,97+c):
if arr[j]!=1:
flag=1
if flag==0:
xx=x[:aa]
yy=''.join(sorted(xx,reverse=True))
zz=x[aa:]
zzz=''.join(sorted(zz))
if (xx==yy) and (zz==zzz) :
print("YES")
else:
print("NO")
else:
print("NO")
``` | output | 1 | 26,791 | 0 | 53,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition. | instruction | 0 | 26,792 | 0 | 53,584 |
Tags: greedy, implementation, strings
Correct Solution:
```
# print(ord('z'))
for _ in range(int(input())):
s = input()
A_idx = -1
mx = 0
mp = {}
dup = False
for i in range(len(s)):
if s[i] == 'a':
A_idx = i
if s[i] not in mp:
mp[s[i]] = 1
else:
dup = True
break
mx = max(mx,ord(s[i]))
if dup:
print("NO")
else:
l, r = A_idx-1 , A_idx+1
ok = True
i = 98
# print(mx)
while i <= mx:
# print(f' {l} {r}')
# print(l,r)
if r < len(s) and s[r] == chr(i):
r+=1
elif l >= 0 and s[l] == chr(i):
l-=1
else:
ok = False
break
i+=1
if A_idx != -1 and (i > mx):
print("YES")
# print(s,"YES")
else:
print("NO")
# print(s,"NO")
``` | output | 1 | 26,792 | 0 | 53,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition. | instruction | 0 | 26,793 | 0 | 53,586 |
Tags: greedy, implementation, strings
Correct Solution:
```
#DaRk DeveLopeR
import sys
#taking input as string
input = lambda: sys.stdin.readline().rstrip("\r\n")
inp = lambda: list(map(int,sys.stdin.readline().rstrip("\r\n").split()))
mod = 10**9+7; Mod = 998244353; INF = float('inf')
#______________________________________________________________________________________________________
import math
from bisect import *
from heapq import *
from collections import defaultdict as dd
from collections import OrderedDict as odict
from collections import Counter as cc
from collections import deque
from itertools import groupby
sys.setrecursionlimit(20*20*20*20+10) #this is must for dfs
def solve():
string=takesr()
n=len(string)
arr=[""]*(n)
if len(string)==1 and string[0]=="a":
print("YES")
return
if "a" not in string:
print("NO")
return
else:
index=string.index("a")
lower=upper=index
arr[index]="a"
first=98
# print(upper,lower)
while (lower>0 or upper<n-1):
element=chr(first)
if lower-1>=0 and string[lower-1]==element:
arr[lower-1]=element
lower=lower-1
elif upper+1<n and string[upper+1]==element:
arr[upper+1]=element
upper+=1
else:
print("NO")
return
# print(upper,lower)
first+=1
print("YES")
return
def main():
global tt
if not ONLINE_JUDGE:
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
t = 1
t = takein()
#t = 1
for tt in range(1,t + 1):
solve()
if not ONLINE_JUDGE:
print("Time Elapsed :",time.time() - start_time,"seconds")
sys.stdout.close()
#---------------------- USER DEFINED INPUT FUNCTIONS ----------------------#
def takein():
return (int(sys.stdin.readline().rstrip("\r\n")))
# input the string
def takesr():
return (sys.stdin.readline().rstrip("\r\n"))
# input int array
def takeiar():
return (list(map(int, sys.stdin.readline().rstrip("\r\n").split())))
# input string array
def takesar():
return (list(map(str, sys.stdin.readline().rstrip("\r\n").split())))
# innut values for the diffrent variables
def takeivr():
return (map(int, sys.stdin.readline().rstrip("\r\n").split()))
def takesvr():
return (map(str, sys.stdin.readline().rstrip("\r\n").split()))
#------------------ USER DEFINED PROGRAMMING FUNCTIONS ------------------#
def ispalindrome(s):
return s==s[::-1]
def invert(bit_s):
# convert binary string
# into integer
temp = int(bit_s, 2)
# applying Ex-or operator
# b/w 10 and 31
inverse_s = temp ^ (2 ** (len(bit_s) + 1) - 1)
# convert the integer result
# into binary result and then
# slicing of the '0b1'
# binary indicator
rslt = bin(inverse_s)[3 : ]
return str(rslt)
def counter(a):
q = [0] * max(a)
for i in range(len(a)):
q[a[i] - 1] = q[a[i] - 1] + 1
return(q)
def counter_elements(a):
q = dict()
for i in range(len(a)):
if a[i] not in q:
q[a[i]] = 0
q[a[i]] = q[a[i]] + 1
return(q)
def string_counter(a):
q = [0] * 26
for i in range(len(a)):
q[ord(a[i]) - 97] = q[ord(a[i]) - 97] + 1
return(q)
def factorial(n,m = 1000000007):
q = 1
for i in range(n):
q = (q * (i + 1)) % m
return(q)
def factors(n):
q = []
for i in range(1,int(n ** 0.5) + 1):
if n % i == 0: q.append(i); q.append(n // i)
return(list(sorted(list(set(q)))))
def prime_factors(n):
q = []
while n % 2 == 0: q.append(2); n = n // 2
for i in range(3,int(n ** 0.5) + 1,2):
while n % i == 0: q.append(i); n = n // i
if n > 2: q.append(n)
return(list(sorted(q)))
def transpose(a):
n,m = len(a),len(a[0])
b = [[0] * n for i in range(m)]
for i in range(m):
for j in range(n):
b[i][j] = a[j][i]
return(b)
def power_two(x):
return (x and (not(x & (x - 1))))
def ceil(a, b):
return -(-a // b)
def seive(n):
a = [1]
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p ** 2,n + 1, p):
prime[i] = False
p = p + 1
for p in range(2,n + 1):
if prime[p]:
a.append(p)
return(a)
def pref(li):
pref_sum = [0]
for i in li:
pref_sum.append(pref_sum[-1]+i)
return pref_sum
def kadane(x): # maximum sum contiguous subarray
sum_so_far = 0
current_sum = 0
for i in x:
current_sum += i
if current_sum < 0:
current_sum = 0
else:
sum_so_far = max(sum_so_far, current_sum)
return sum_so_far
def binary_search(li, val):
# print(lb, ub, li)
ans = -1
lb = 0
ub = len(li)-1
while (lb <= ub):
mid = (lb+ub) // 2
# print('mid is',mid, li[mid])
if li[mid] > val:
ub = mid-1
elif val > li[mid]:
lb = mid+1
else:
ans = mid # return index
break
return ans
def upper_bound(li, num):
answer = -1
start = 0
end = len(li)-1
while (start <= end):
middle = (end+start) // 2
if li[middle] <= num:
answer = middle
start = middle+1
else:
end = middle-1
return answer # max index where x is not greater than num
def lower_bound(li, num):
answer = -1
start = 0
end = len(li)-1
while (start <= end):
middle = (end+start) // 2
if li[middle] >= num:
answer = middle
end = middle-1
else:
start = middle+1
return answer # min index where x is not less than num
#-----------------------------------------------------------------------#
ONLINE_JUDGE = __debug__
if ONLINE_JUDGE:
input = sys.stdin.readline
main()
``` | output | 1 | 26,793 | 0 | 53,587 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition. | instruction | 0 | 26,794 | 0 | 53,588 |
Tags: greedy, implementation, strings
Correct Solution:
```
import string
lett = string.ascii_lowercase
def isAlf(ss):
start = ss.find('a')
if start == -1:
return False
n = len(ss)
if n == 1:
return True
l = start-1
r = start+1
ll = 1
for i in range(1, 26):
if l < 0 and r >= n:
break
if l >= 0 and lett[ll] == ss[l]:
l-=1
ll+=1
continue
if r < n and lett[ll] == ss[r]:
r+=1
ll+=1
continue
return False
return True
def main(input, print):
toPrint=[]
t = int(input())
for i in range(t):
ss = input().strip()
r = isAlf(ss)
if r:
toPrint.append('YES')
else:
toPrint.append('NO')
print('\n'.join(toPrint))
main(input, print)
``` | output | 1 | 26,794 | 0 | 53,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition. | instruction | 0 | 26,795 | 0 | 53,590 |
Tags: greedy, implementation, strings
Correct Solution:
```
t = int(input())
for _ in range(t):
dummy = input()
l = list(dummy)
start = 0
end = len(l) - 1
count = len(l)
# print(ord(l[0]))
while start <= end and count:
if ord(l[start] ) == (count + 96):
start+=1
elif ord(l[end] ) == (count + 96):
end-=1
else:
break
count-=1
if start > end:
print("YES")
else:
print("NO")
``` | output | 1 | 26,795 | 0 | 53,591 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition. | instruction | 0 | 26,796 | 0 | 53,592 |
Tags: greedy, implementation, strings
Correct Solution:
```
t = int(input())
for i in range(t):
text = input()
length = len(text)
ans = True
for j in range(length):
if text[0] == chr(97 + length - j - 1):
text = text[1:]
elif text[-1] == chr(97 + length - j - 1):
text = text[:-1]
else:
print("NO")
ans = False
break
if ans:
print("YES")
``` | output | 1 | 26,796 | 0 | 53,593 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition. | instruction | 0 | 26,797 | 0 | 53,594 |
Tags: greedy, implementation, strings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from types import GeneratorType
from collections import defaultdict
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
t=int(input())
for _ in range(t):
#n=int(input())
s=input()
n=len(s)
poss=0
currl=-1
currr=-1
for j in range(n):
if s[j]=="a":
currl=j-1
currr=j+1
poss=1
break
if poss==0:
print("NO")
else:
j=1
while(j<=25):
if (currl==-1 and currr==n):
break
curr=chr(97+j)
if currl>=0 and s[currl]==curr:
currl+=-1
elif currr<n and s[currr]==curr:
currr+=1
else:
poss=0
break
j+=1
if poss:
print('YES')
else:
print('NO')
``` | output | 1 | 26,797 | 0 | 53,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition.
Submitted Solution:
```
def alpha(s):
if 'a' not in s:
return "NO"
else:
l = s.index('a')
r = l
for i in range(1,len(s)):
if l >= 1 and s[l-1] == chr(ord('a') + i):
l -= 1
elif r <= len(s)-2 and s[r+1] == chr(ord('a') + i):
r += 1
else:
return "NO"
return "YES"
t = int(input())
for _ in range(t):
s = input()
print(alpha(s))
``` | instruction | 0 | 26,798 | 0 | 53,596 |
Yes | output | 1 | 26,798 | 0 | 53,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition.
Submitted Solution:
```
import string
def solve(strng):
if 'a' not in strng:
return "NO"
if len(strng)>26:
return "NO"
alphabet_string = string.ascii_lowercase
sorted_char = sorted(strng)
sortedstrng = "".join(sorted_char)
if sortedstrng not in alphabet_string:
return "NO"
increasing = False
for i in range(len(strng)-1):
if strng[i]=='a':
increasing = True
if (increasing and strng[i]>=strng[i+1]):
return "NO"
elif (not increasing and strng[i]<=strng[i+1]):
return "NO"
return "YES"
t = int(input())
for m in range(t):
strng = input()
print(solve(strng))
``` | instruction | 0 | 26,799 | 0 | 53,598 |
Yes | output | 1 | 26,799 | 0 | 53,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition.
Submitted Solution:
```
t = int(input())
answers=[]
for i in range(0,t):
s = input()
n = len(s)-1
begpos = s.find('a')
endpos = begpos
flag1 = 0
flag2 = 0
if begpos!=-1:
for x in range(2,27):
if n>0:
ch = chr(x+96)
curt = s.find(str(ch))
n-=1
if curt == -1:
flag2 = 1
break
if curt == begpos-1:
begpos = curt
elif curt == endpos+1:
endpos = curt
else:
flag2 = 1
break
else:
break
else:
flag1=1
if flag1==0 and flag2==0:
answers.append("YES")
else:
answers.append("NO")
for i in answers:
print(i)
``` | instruction | 0 | 26,800 | 0 | 53,600 |
Yes | output | 1 | 26,800 | 0 | 53,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition.
Submitted Solution:
```
n = int(input())
for i in range(n):
lst = []
s = input()
if 'a' not in s:
print('NO')
continue
for j in range(len(s)):
lst.append(ord(s[j]))
flag_0 = 0
for k in range(0, lst.index(97)):
if lst[k] < lst[k + 1]:
flag_0 = 1
break
for k in range(lst.index(97) + 1, len(lst) - 1):
if lst[k] > lst[k + 1]:
flag_0 = 1
break
if flag_0 == 1:
print('NO')
continue
lst.sort()
flag = 0
for j in range(1, len(lst)):
if lst[j - 1] - lst[j] != -1:
flag = 1
break
if flag == 1:
print('NO')
else:
print('YES')
``` | instruction | 0 | 26,801 | 0 | 53,602 |
Yes | output | 1 | 26,801 | 0 | 53,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition.
Submitted Solution:
```
def get_sort(s):
return ''.join(sorted(s))
def check_main(check, s1):
if s1 != check[:len(s1)]:
return "NO"
else:
return "YES"
T = int(input())
for _ in range(T):
s = input()
s1 = get_sort(s)
check = "abcdefghijklmnopqrstuvwxyz"
if check_main(check, s1) == 'NO':
print("NO")
continue
pos1 = pos2 = s.index("a")
temp1 = len(s)
temp = 0
for i in range(1,temp1):
ele = check[i]
if pos1 > 0 and ele == s[pos1-1]:
temp += temp1
pos1 -= 1
elif pos2 < len(s)-1 and ele == s[pos2+1]:
temp += temp1
pos2 += 1
else:
temp += temp1
print("NO")
print("YES")
``` | instruction | 0 | 26,802 | 0 | 53,604 |
No | output | 1 | 26,802 | 0 | 53,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition.
Submitted Solution:
```
for _ in range(int(input())):
s = input()
asc = set()
flag = True
for l in s:
asc.add(ord(l) - 96)
if len(asc) < len(s):
flag = False
else:
l = sorted(list(asc))
for i in range(len(s)):
if l[i] != i + 1:
flag = False
break
if not flag:
print('NO')
continue
for i in range(len(s)):
if abs(i - s.index('a')) > ord(s[i]) - 97:
flag = False
break
if flag:
print("YES")
else:
print("NO")
``` | instruction | 0 | 26,803 | 0 | 53,606 |
No | output | 1 | 26,803 | 0 | 53,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition.
Submitted Solution:
```
def solve(strg):
str="".join(sorted(strg))
pos_a=strg.index("a")
Left=pos_a
Right=pos_a
for j in range(98,98+len(strg)):
l=chr(j)
pos_l=strg.find(l)
if j == len(strg)+97:
print("YES")
break
elif pos_l==Right+1:
Right=Right+1
elif pos_l==Left-1:
Left=Left-1
else:
print("NO")
break
t=int(input())
for i in range(t):
strg=input()
s=set(strg)
if "a" not in strg:
print("NO")
elif len(strg)!=len(s):
print("NO")
else:
solve(strg)
``` | instruction | 0 | 26,804 | 0 | 53,608 |
No | output | 1 | 26,804 | 0 | 53,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:
* first, write an empty string to s (i.e. perform the assignment s := "");
* then perform the next step n times;
* at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.
For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".
From the given string, determine if it is alphabetical.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.
Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.
You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).
Example
Input
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
Output
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition.
Submitted Solution:
```
t=int(input())
for _ in range(t):
s=input()
if(not s.isalpha()):
print("NO")
else:
ma=max(s[-1],s[0])
while(s!=""):
if(s[0]==ma):
s=s[1:]
elif(s[-1]==ma):
s=s[:-1]
else:
break
if(s==""):
break
ma=chr(ord(ma)-1)
if(ma=="a"):
print("YES")
else:
print("NO")
# n=len(s)
# left=0
# right=n-1
# while(left!=right):
# # print(ma,s[left],s[right])
# if(s[left]==ma):
# left+=1
# elif(s[right]==ma):
# right-=1
# else:
# break
# ma=chr(ord(ma)-1)
# if(left==right and ma=="a"):
# print("YES")
# else:
# print("NO")
``` | instruction | 0 | 26,805 | 0 | 53,610 |
No | output | 1 | 26,805 | 0 | 53,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb | instruction | 0 | 26,806 | 0 | 53,612 |
Tags: constructive algorithms, greedy, math, matrices
Correct Solution:
```
def get_input():
a, b, d = map(int, input().split())
c, e = map(int, input().split())
f = int(input())
return [a, b, c, d, e, f]
def check_condition(a, b, c, d, e, f):
condition1 = (a + b + c) % 2 == 0
condition2 = (d + e + a) % 2 == 0
condition3 = (e + f + c) % 2 == 0
condition4 = (d + f + b) % 2 == 0
condition = condition1 and condition2 and condition3 and condition4
return condition
def find_t(a, b, c, d, e, f):
t_min1 = round((d + f - a - c) / 2)
t_min2 = round((e + f - a - b) / 2)
t_min3 = round((d + e - b - c) / 2)
t_min4 = 0
t_min = max(t_min1, t_min2, t_min3, t_min4)
t_max1 = round((d + e - a) / 2)
t_max2 = round((e + f - c) / 2)
t_max3 = round((d + f - b) / 2)
t_max = min(t_max1, t_max2, t_max3)
if t_min <= t_max:
return t_min
else:
return -1
def find_all(a, b, c, d, e, f, t):
x1 = round((a + c - d - f) / 2 + t)
x2 = round((d + f - b) / 2 - t)
y1 = round((a + b - e - f) / 2 + t)
y2 = round((e + f - c) / 2 - t)
z1 = round((b + c - d - e) / 2 + t)
z2 = round((d + e - a) / 2 - t)
return [x1, x2, y1, y2, z1, z2]
def generate_string(x1, x2, y1, y2, z1, z2, t):
n = x1 + x2 + y1 + y2 + z1 + z2 + t
s1 = ''.join(['a'] * n)
s2 = ''.join(['a'] * (z1 + z2 + t)) + ''.join(['b'] * (x1 + x2 + y1 + y2))
s3 = ''.join(['a'] * t) + ''.join(['b'] * (y1 + y2 + z1 + z2)) + ''.join(['a'] * (x1 + x2))
s4 = ''.join(['b'] * (t + z2)) + ''.join(['a'] * (z1 + y2)) + ''.join(['b'] * (y1 + x2)) + ''.join(['a'] * x1)
return [s1, s2, s3, s4]
def __main__():
fail_output = "-1"
a, b, c, d, e, f = map(int, get_input())
if not(check_condition(a, b, c, d, e, f)):
print(fail_output)
return False
t = find_t(a, b, c, d, e, f)
if t < 0:
print(fail_output)
return False
x1, x2, y1, y2, z1, z2 = map(int, find_all(a, b, c, d, e, f, t))
s1, s2, s3, s4 = map(str, generate_string(x1, x2, y1, y2, z1, z2, t))
print(str(x1 + x2 + y1 + y2 + z1 + z2 + t) + '\n')
print(s1 + '\n')
print(s2 + '\n')
print(s3 + '\n')
print(s4 + '\n')
__main__()
# Made By Mostafa_Khaled
``` | output | 1 | 26,806 | 0 | 53,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb | instruction | 0 | 26,807 | 0 | 53,614 |
Tags: constructive algorithms, greedy, math, matrices
Correct Solution:
```
def get_input():
a, b, d = map(int, input().split())
c, e = map(int, input().split())
f = int(input())
return [a, b, c, d, e, f]
def check_condition(a, b, c, d, e, f):
condition1 = (a + b + c) % 2 == 0
condition2 = (d + e + a) % 2 == 0
condition3 = (e + f + c) % 2 == 0
condition4 = (d + f + b) % 2 == 0
condition = condition1 and condition2 and condition3 and condition4
return condition
def find_t(a, b, c, d, e, f):
t_min1 = round((d + f - a - c) / 2)
t_min2 = round((e + f - a - b) / 2)
t_min3 = round((d + e - b - c) / 2)
t_min4 = 0
t_min = max(t_min1, t_min2, t_min3, t_min4)
t_max1 = round((d + e - a) / 2)
t_max2 = round((e + f - c) / 2)
t_max3 = round((d + f - b) / 2)
t_max = min(t_max1, t_max2, t_max3)
if t_min <= t_max:
return t_min
else:
return -1
def find_all(a, b, c, d, e, f, t):
x1 = round((a + c - d - f) / 2 + t)
x2 = round((d + f - b) / 2 - t)
y1 = round((a + b - e - f) / 2 + t)
y2 = round((e + f - c) / 2 - t)
z1 = round((b + c - d - e) / 2 + t)
z2 = round((d + e - a) / 2 - t)
return [x1, x2, y1, y2, z1, z2]
def generate_string(x1, x2, y1, y2, z1, z2, t):
n = x1 + x2 + y1 + y2 + z1 + z2 + t
s1 = ''.join(['a'] * n)
s2 = ''.join(['a'] * (z1 + z2 + t)) + ''.join(['b'] * (x1 + x2 + y1 + y2))
s3 = ''.join(['a'] * t) + ''.join(['b'] * (y1 + y2 + z1 + z2)) + ''.join(['a'] * (x1 + x2))
s4 = ''.join(['b'] * (t + z2)) + ''.join(['a'] * (z1 + y2)) + ''.join(['b'] * (y1 + x2)) + ''.join(['a'] * x1)
return [s1, s2, s3, s4]
def __main__():
fail_output = "-1"
a, b, c, d, e, f = map(int, get_input())
if not(check_condition(a, b, c, d, e, f)):
print(fail_output)
return False
t = find_t(a, b, c, d, e, f)
if t < 0:
print(fail_output)
return False
x1, x2, y1, y2, z1, z2 = map(int, find_all(a, b, c, d, e, f, t))
s1, s2, s3, s4 = map(str, generate_string(x1, x2, y1, y2, z1, z2, t))
print(str(x1 + x2 + y1 + y2 + z1 + z2 + t) + '\n')
print(s1 + '\n')
print(s2 + '\n')
print(s3 + '\n')
print(s4 + '\n')
__main__()
``` | output | 1 | 26,807 | 0 | 53,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb | instruction | 0 | 26,808 | 0 | 53,616 |
Tags: constructive algorithms, greedy, math, matrices
Correct Solution:
```
h = [[0 in range(10)] for j in range(10)]
for i in range(1, 4):
h[i] = [0 for j in range(i + 1)] + list(map(int, input().split()))
#for i in range(1, 4):
# print(" ".join(map(str, h[i][1:5])))
if (h[1][2] + h[1][3] < h[2][3] or (h[1][2] + h[1][3] - h[2][3]) % 2 == 1):
print("-1")
exit(0)
BB = (h[1][2] + h[1][3] - h[2][3]) // 2
BA = h[1][2] - BB
AB = h[1][3] - BB
NowB = h[1][4]
NowLen = BB + AB + BA
Now24 = BA + NowB + BB
Now34 = AB + NowB + BB
BAB = 0
ABB = 0
BBB = 0
Dif = (BA - AB) - (h[2][4] - h[3][4])
if (abs(Dif) % 2 == 1):
print("-1")
exit(0)
if Dif < 0:
ABB += abs(Dif) // 2
Now34 -= ABB * 2
#Now24 += ABB
if (AB < ABB or NowB < ABB):
print("-1")
exit(0)
NowB -= ABB
else:
BAB += Dif // 2
Now24 -= BAB * 2
#Now34 += BAB
if (BA < BAB or NowB < BAB):
print("-1")
exit(0)
NowB -= BAB
if (Now24 < h[2][4] or (Now24 - h[2][4]) % 2 == 1):
print("-1")
exit(0)
#print(Now34 - h[3][4])
for i in range(BB + 1):
if (i > NowB):
break
Now = i * 2
if (Now > Now24 - h[2][4]):
break
if min([(NowB - i) // 2, BA - BAB, AB - ABB]) * 2 >= Now24 - h[2][4] - Now:
#print(i, Now24, h[2][4], NowB)
BBB += i
BAB += (Now24 - h[2][4] - Now) // 2
ABB += (Now24 - h[2][4] - Now) // 2
NowB -= i + (Now24 - h[2][4] - Now)
print(NowLen + NowB)
print("".join(["a" for j in range(NowLen)] + ["a" for j in range(NowB)]))
print("".join(["a" for j in range(AB)] + ["b" for j in range(BB + BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(AB + BB)] + ["a" for j in range(BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(ABB)] + ["a" for j in range(AB - ABB)] + ["b" for j in range(BBB)] + ["a" for j in range(BB - BBB)] + ["b" for j in range(BAB)] + ["a" for j in range(BA - BAB)] + ["b" for j in range(NowB)]))
exit(0)
print("-1")
``` | output | 1 | 26,808 | 0 | 53,617 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb
Submitted Solution:
```
h = [[0 in range(10)] for j in range(10)]
for i in range(1, 4):
h[i] = [0 for j in range(i + 1)] + list(map(int, input().split()))
#for i in range(1, 4):
# print(" ".join(map(str, h[i][1:5])))
if (h[1][2] + h[1][3] < h[2][3] or (h[1][2] + h[1][3] - h[2][3]) % 2 == 1):
print("-1")
exit(0)
BB = (h[1][2] + h[1][3] - h[2][3]) // 2
AB = h[1][2] - BB
BA = h[1][3] - BB
NowB = h[1][4]
NowLen = BB + AB + BA
Now24 = BA + NowB + BB;
Now34 = AB + NowB + BB;
BAB = 0
ABB = 0
BBB = 0
Dif = (BA - AB) - (h[2][4] - h[3][4])
if (abs(Dif) % 2 == 1):
print("-1")
exit(0)
if Dif < 0:
ABB += abs(Dif) // 2
Now34 -= ABB * 2
#Now24 += ABB
if (AB < ABB or NowB < ABB):
print("-1")
exit(0)
NowB -= ABB
else:
BAB += Dif // 2
Now24 -= BAB * 2
#Now34 += BAB
if (BA < BAB or NowB < BAB):
print("-1")
exit(0)
NowB -= BAB
if (Now24 < h[2][4] or (Now24 - h[2][4]) % 2 == 1):
print("-1")
exit(0)
#print(Now34 - h[3][4])
for i in range(BB + 1):
if (i > NowB):
break
Now = i * 2
if (Now > Now24 - h[2][4]):
break
if min([(NowB - i) // 2, BA - BAB, AB - ABB]) * 2 >= Now24 - h[2][4] - Now:
#print(i, Now24, h[2][4], NowB)
BBB += Now
BAB += (Now24 - h[2][4] - Now) // 2
ABB += (Now24 - h[2][4] - Now) // 2
NowB -= i + (Now24 - h[2][4] - Now)
print(NowLen + NowB)
print("".join(["a" for j in range(NowLen)] + ["a" for j in range(NowB)]))
print("".join(["a" for j in range(AB)] + ["b" for j in range(BB + BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(AB + BB)] + ["a" for j in range(BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(ABB)] + ["a" for j in range(AB)[ABB:]] + ["b" for j in range(BBB)] + ["a" for j in range(BB)[BBB:]] + ["b" for j in range(BAB)] + ["a" for j in range(BA)[BAB:]] + ["b" for j in range(NowB)]))
exit(0)
print("-1")
``` | instruction | 0 | 26,809 | 0 | 53,618 |
No | output | 1 | 26,809 | 0 | 53,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb
Submitted Solution:
```
h = [[0 in range(10)] for j in range(10)]
for i in range(1, 4):
h[i] = [0 for j in range(i + 1)] + list(map(int, input().split()))
#for i in range(1, 4):
# print(" ".join(map(str, h[i][1:5])))
if (h[1][2] + h[1][3] < h[2][3] or (h[1][2] + h[1][3] - h[2][3]) % 2 == 1):
print("-1")
exit(0)
BB = (h[1][2] + h[1][3] - h[2][3]) // 2
AB = h[1][2] - BB
BA = h[1][3] - BB
NowB = h[1][4]
NowLen = BB + AB + BA
Now24 = BA + NowB + BB
Now34 = AB + NowB + BB
BAB = 0
ABB = 0
BBB = 0
Dif = (BA - AB) - (h[2][4] - h[3][4])
if (abs(Dif) % 2 == 1):
print("-1")
exit(0)
if Dif < 0:
ABB += abs(Dif) // 2
Now34 -= ABB * 2
#Now24 += ABB
if (AB < ABB or NowB < ABB):
print("-1")
exit(0)
NowB -= ABB
else:
BAB += Dif // 2
Now24 -= BAB * 2
#Now34 += BAB
if (BA < BAB or NowB < BAB):
print("-1")
exit(0)
NowB -= BAB
if (Now24 < h[2][4] or (Now24 - h[2][4]) % 2 == 1):
print("-1")
exit(0)
#print(Now34 - h[3][4])
for i in range(BB + 1):
if (i > NowB):
break
Now = i * 2
if (Now > Now24 - h[2][4]):
break
if min([(NowB - i) // 2, BA - BAB, AB - ABB]) * 2 >= Now24 - h[2][4] - Now:
#print(i, Now24, h[2][4], NowB)
BBB += i
BAB += (Now24 - h[2][4] - Now) // 2
ABB += (Now24 - h[2][4] - Now) // 2
NowB -= i + (Now24 - h[2][4] - Now)
print(NowLen + NowB)
print("".join(["a" for j in range(NowLen)] + ["a" for j in range(NowB)]))
print("".join(["a" for j in range(AB)] + ["b" for j in range(BB + BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(AB + BB)] + ["a" for j in range(BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(ABB)] + ["a" for j in range(AB - ABB)] + ["b" for j in range(BBB)] + ["a" for j in range(BB - BBB)] + ["b" for j in range(BAB)] + ["a" for j in range(BA - BAB)] + ["b" for j in range(NowB)]))
exit(0)
print("-1")
``` | instruction | 0 | 26,810 | 0 | 53,620 |
No | output | 1 | 26,810 | 0 | 53,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb
Submitted Solution:
```
h = [[0 in range(10)] for j in range(10)]
for i in range(1, 4):
h[i] = [0 for j in range(i + 1)] + list(map(int, input().split()))
#for i in range(1, 4):
# print(" ".join(map(str, h[i][1:5])))
if (h[1][2] + h[1][3] < h[2][3] or (h[1][2] + h[1][3] - h[2][3]) % 2 == 1):
print("-1")
exit(0)
BB = (h[1][2] + h[1][3] - h[2][3]) // 2
AB = h[1][2] - BB
BA = h[1][3] - BB
NowB = h[1][4]
NowLen = BB + AB + BA
Now24 = BA + NowB + BB;
Now34 = AB + NowB + BB;
BAB = 0
ABB = 0
BBB = 0
Dif = (BA - AB) - (h[2][4] - h[3][4])
if (abs(Dif) % 2 == 1):
print("-1")
exit(0)
if Dif < 0:
ABB += abs(Dif) // 2
Now34 -= ABB * 2
#Now24 += ABB
if (AB < ABB or NowB < ABB):
print("-1")
exit(0)
NowB -= ABB
else:
BAB += Dif // 2
Now24 -= BAB * 2
#Now34 += BAB
if (BA < BAB or NowB < BAB):
print("-1")
exit(0)
NowB -= BAB
if (Now24 < h[2][4] or (Now24 - h[2][4]) % 2 == 1):
print("-1")
exit(0)
#print(Now34 - h[3][4])
for i in range(BB + 1):
if (i > NowB):
break
Now = i * 2
if (Now > Now24 - h[2][4]):
break
if min([(NowB - i) // 2, BA - BAB, AB - ABB]) * 2 >= Now24 - h[2][4] - Now:
#print(i, Now24, h[2][4], NowB)
BBB += i
BAB += (Now24 - h[2][4] - Now) // 2
ABB += (Now24 - h[2][4] - Now) // 2
NowB -= i + (Now24 - h[2][4] - Now)
print(NowLen + NowB)
print("".join(["a" for j in range(NowLen)] + ["a" for j in range(NowB)]))
print("".join(["a" for j in range(AB)] + ["b" for j in range(BB + BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(AB + BB)] + ["a" for j in range(BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(ABB)] + ["a" for j in range(AB)[ABB:]] + ["b" for j in range(BBB)] + ["a" for j in range(BB)[BBB:]] + ["b" for j in range(BAB)] + ["a" for j in range(BA)[BAB:]] + ["b" for j in range(NowB)]))
exit(0)
print("-1")
``` | instruction | 0 | 26,811 | 0 | 53,622 |
No | output | 1 | 26,811 | 0 | 53,623 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alica and Bob are playing a game.
Initially they have a binary string s consisting of only characters 0 and 1.
Alice and Bob make alternating moves: Alice makes the first move, Bob makes the second move, Alice makes the third one, and so on. During each move, the current player must choose two different adjacent characters of string s and delete them. For example, if s = 1011001 then the following moves are possible:
1. delete s_1 and s_2: 1011001 → 11001;
2. delete s_2 and s_3: 1011001 → 11001;
3. delete s_4 and s_5: 1011001 → 10101;
4. delete s_6 and s_7: 1011001 → 10110.
If a player can't make any move, they lose. Both players play optimally. You have to determine if Alice can win.
Input
First line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Only line of each test case contains one string s (1 ≤ |s| ≤ 100), consisting of only characters 0 and 1.
Output
For each test case print answer in the single line.
If Alice can win print DA (YES in Russian) in any register. Otherwise print NET (NO in Russian) in any register.
Example
Input
3
01
1111
0011
Output
DA
NET
NET
Note
In the first test case after Alice's move string s become empty and Bob can not make any move.
In the second test case Alice can not make any move initially.
In the third test case after Alice's move string s turn into 01. Then, after Bob's move string s become empty and Alice can not make any move. | instruction | 0 | 27,584 | 0 | 55,168 |
Tags: games
Correct Solution:
```
t=int(input())
for _ in range(t):
s=input()
cnt0=s.count("0")
cnt1=s.count("1")
ans=min(cnt0,cnt1)
if ans%2==0:
print("NET")
else:
print("DA")
``` | output | 1 | 27,584 | 0 | 55,169 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alica and Bob are playing a game.
Initially they have a binary string s consisting of only characters 0 and 1.
Alice and Bob make alternating moves: Alice makes the first move, Bob makes the second move, Alice makes the third one, and so on. During each move, the current player must choose two different adjacent characters of string s and delete them. For example, if s = 1011001 then the following moves are possible:
1. delete s_1 and s_2: 1011001 → 11001;
2. delete s_2 and s_3: 1011001 → 11001;
3. delete s_4 and s_5: 1011001 → 10101;
4. delete s_6 and s_7: 1011001 → 10110.
If a player can't make any move, they lose. Both players play optimally. You have to determine if Alice can win.
Input
First line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Only line of each test case contains one string s (1 ≤ |s| ≤ 100), consisting of only characters 0 and 1.
Output
For each test case print answer in the single line.
If Alice can win print DA (YES in Russian) in any register. Otherwise print NET (NO in Russian) in any register.
Example
Input
3
01
1111
0011
Output
DA
NET
NET
Note
In the first test case after Alice's move string s become empty and Bob can not make any move.
In the second test case Alice can not make any move initially.
In the third test case after Alice's move string s turn into 01. Then, after Bob's move string s become empty and Alice can not make any move. | instruction | 0 | 27,585 | 0 | 55,170 |
Tags: games
Correct Solution:
```
z=int(input())
for h in range(z):
a=list(map(int,input()))
cnt=0
while True:
flag=0
for i in range(len(a)-1):
if len(a)>1:
if (a[i]==0 and a[i+1]) or (a[i]==1 and a[i+1]==0):
cnt+=1
flag=1
a.pop(i)
a.pop(i)
break
if flag==0:
break
if len(a)==0:
break
if cnt%2==0 and cnt!=1:
print("NET")
else:
print("DA")
``` | output | 1 | 27,585 | 0 | 55,171 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alica and Bob are playing a game.
Initially they have a binary string s consisting of only characters 0 and 1.
Alice and Bob make alternating moves: Alice makes the first move, Bob makes the second move, Alice makes the third one, and so on. During each move, the current player must choose two different adjacent characters of string s and delete them. For example, if s = 1011001 then the following moves are possible:
1. delete s_1 and s_2: 1011001 → 11001;
2. delete s_2 and s_3: 1011001 → 11001;
3. delete s_4 and s_5: 1011001 → 10101;
4. delete s_6 and s_7: 1011001 → 10110.
If a player can't make any move, they lose. Both players play optimally. You have to determine if Alice can win.
Input
First line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Only line of each test case contains one string s (1 ≤ |s| ≤ 100), consisting of only characters 0 and 1.
Output
For each test case print answer in the single line.
If Alice can win print DA (YES in Russian) in any register. Otherwise print NET (NO in Russian) in any register.
Example
Input
3
01
1111
0011
Output
DA
NET
NET
Note
In the first test case after Alice's move string s become empty and Bob can not make any move.
In the second test case Alice can not make any move initially.
In the third test case after Alice's move string s turn into 01. Then, after Bob's move string s become empty and Alice can not make any move. | instruction | 0 | 27,586 | 0 | 55,172 |
Tags: games
Correct Solution:
```
def saurabh(B):
A=len(B)
x=[B[0]]
count=0
for i in range(1,len(B)):
if x==[]:
x.append(B[i])
elif x[-1]!=B[i]:
x.pop()
count+=1
else:
x.append(B[i])
if count%2==1:
return 'DA'
else:
return 'NET'
if __name__ == "__main__":
for i in range(int(input())):
print(saurabh(input()))
``` | output | 1 | 27,586 | 0 | 55,173 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alica and Bob are playing a game.
Initially they have a binary string s consisting of only characters 0 and 1.
Alice and Bob make alternating moves: Alice makes the first move, Bob makes the second move, Alice makes the third one, and so on. During each move, the current player must choose two different adjacent characters of string s and delete them. For example, if s = 1011001 then the following moves are possible:
1. delete s_1 and s_2: 1011001 → 11001;
2. delete s_2 and s_3: 1011001 → 11001;
3. delete s_4 and s_5: 1011001 → 10101;
4. delete s_6 and s_7: 1011001 → 10110.
If a player can't make any move, they lose. Both players play optimally. You have to determine if Alice can win.
Input
First line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Only line of each test case contains one string s (1 ≤ |s| ≤ 100), consisting of only characters 0 and 1.
Output
For each test case print answer in the single line.
If Alice can win print DA (YES in Russian) in any register. Otherwise print NET (NO in Russian) in any register.
Example
Input
3
01
1111
0011
Output
DA
NET
NET
Note
In the first test case after Alice's move string s become empty and Bob can not make any move.
In the second test case Alice can not make any move initially.
In the third test case after Alice's move string s turn into 01. Then, after Bob's move string s become empty and Alice can not make any move. | instruction | 0 | 27,587 | 0 | 55,174 |
Tags: games
Correct Solution:
```
t=int(input())
a=[]
for i in range(t):
n=input()
s=0
g=0
r=0
b=0
h=len(n)//2
for i in range(len(n)//2):
g=n.count('01')
b+=g
n=n.replace('01','')
r=n.count('10')
b+=r
n=n.replace('10','')
if len(n)==2 and n[0]!=n[1]:
a.append('DA')
elif b%2!=0:
a.append('DA')
else:
a.append('NET')
for i in a:
print(i)
``` | output | 1 | 27,587 | 0 | 55,175 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alica and Bob are playing a game.
Initially they have a binary string s consisting of only characters 0 and 1.
Alice and Bob make alternating moves: Alice makes the first move, Bob makes the second move, Alice makes the third one, and so on. During each move, the current player must choose two different adjacent characters of string s and delete them. For example, if s = 1011001 then the following moves are possible:
1. delete s_1 and s_2: 1011001 → 11001;
2. delete s_2 and s_3: 1011001 → 11001;
3. delete s_4 and s_5: 1011001 → 10101;
4. delete s_6 and s_7: 1011001 → 10110.
If a player can't make any move, they lose. Both players play optimally. You have to determine if Alice can win.
Input
First line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Only line of each test case contains one string s (1 ≤ |s| ≤ 100), consisting of only characters 0 and 1.
Output
For each test case print answer in the single line.
If Alice can win print DA (YES in Russian) in any register. Otherwise print NET (NO in Russian) in any register.
Example
Input
3
01
1111
0011
Output
DA
NET
NET
Note
In the first test case after Alice's move string s become empty and Bob can not make any move.
In the second test case Alice can not make any move initially.
In the third test case after Alice's move string s turn into 01. Then, after Bob's move string s become empty and Alice can not make any move. | instruction | 0 | 27,588 | 0 | 55,176 |
Tags: games
Correct Solution:
```
t = int(input())
for _ in range(t):
s = [int(i) for i in input()]
ct0 = 0
ct1 = 0
for i in s:
if i == 1:
ct1 += 1
else:
ct0 += 1
ct = min(ct0, ct1)
if ct%2:
print("DA")
else:
print("NET")
``` | output | 1 | 27,588 | 0 | 55,177 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alica and Bob are playing a game.
Initially they have a binary string s consisting of only characters 0 and 1.
Alice and Bob make alternating moves: Alice makes the first move, Bob makes the second move, Alice makes the third one, and so on. During each move, the current player must choose two different adjacent characters of string s and delete them. For example, if s = 1011001 then the following moves are possible:
1. delete s_1 and s_2: 1011001 → 11001;
2. delete s_2 and s_3: 1011001 → 11001;
3. delete s_4 and s_5: 1011001 → 10101;
4. delete s_6 and s_7: 1011001 → 10110.
If a player can't make any move, they lose. Both players play optimally. You have to determine if Alice can win.
Input
First line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Only line of each test case contains one string s (1 ≤ |s| ≤ 100), consisting of only characters 0 and 1.
Output
For each test case print answer in the single line.
If Alice can win print DA (YES in Russian) in any register. Otherwise print NET (NO in Russian) in any register.
Example
Input
3
01
1111
0011
Output
DA
NET
NET
Note
In the first test case after Alice's move string s become empty and Bob can not make any move.
In the second test case Alice can not make any move initially.
In the third test case after Alice's move string s turn into 01. Then, after Bob's move string s become empty and Alice can not make any move. | instruction | 0 | 27,589 | 0 | 55,178 |
Tags: games
Correct Solution:
```
from typing import AnyStr
t=int(input())
answer=[]
while t>0:
s=input()
c0=0
c1=0
for i in s:
if i=='0':
c0+=1
else:
c1+=1
if min(c0,c1)%2==1:
answer.append('DA')
else:
answer.append('NET')
t-=1
print('\n'.join(answer))
``` | output | 1 | 27,589 | 0 | 55,179 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alica and Bob are playing a game.
Initially they have a binary string s consisting of only characters 0 and 1.
Alice and Bob make alternating moves: Alice makes the first move, Bob makes the second move, Alice makes the third one, and so on. During each move, the current player must choose two different adjacent characters of string s and delete them. For example, if s = 1011001 then the following moves are possible:
1. delete s_1 and s_2: 1011001 → 11001;
2. delete s_2 and s_3: 1011001 → 11001;
3. delete s_4 and s_5: 1011001 → 10101;
4. delete s_6 and s_7: 1011001 → 10110.
If a player can't make any move, they lose. Both players play optimally. You have to determine if Alice can win.
Input
First line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Only line of each test case contains one string s (1 ≤ |s| ≤ 100), consisting of only characters 0 and 1.
Output
For each test case print answer in the single line.
If Alice can win print DA (YES in Russian) in any register. Otherwise print NET (NO in Russian) in any register.
Example
Input
3
01
1111
0011
Output
DA
NET
NET
Note
In the first test case after Alice's move string s become empty and Bob can not make any move.
In the second test case Alice can not make any move initially.
In the third test case after Alice's move string s turn into 01. Then, after Bob's move string s become empty and Alice can not make any move. | instruction | 0 | 27,590 | 0 | 55,180 |
Tags: games
Correct Solution:
```
t = int(input())
for i in range(t):
s = input()
print('DA' if min(s.count('0'), s.count('1')) % 2 == 1 else 'NET')
``` | output | 1 | 27,590 | 0 | 55,181 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alica and Bob are playing a game.
Initially they have a binary string s consisting of only characters 0 and 1.
Alice and Bob make alternating moves: Alice makes the first move, Bob makes the second move, Alice makes the third one, and so on. During each move, the current player must choose two different adjacent characters of string s and delete them. For example, if s = 1011001 then the following moves are possible:
1. delete s_1 and s_2: 1011001 → 11001;
2. delete s_2 and s_3: 1011001 → 11001;
3. delete s_4 and s_5: 1011001 → 10101;
4. delete s_6 and s_7: 1011001 → 10110.
If a player can't make any move, they lose. Both players play optimally. You have to determine if Alice can win.
Input
First line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Only line of each test case contains one string s (1 ≤ |s| ≤ 100), consisting of only characters 0 and 1.
Output
For each test case print answer in the single line.
If Alice can win print DA (YES in Russian) in any register. Otherwise print NET (NO in Russian) in any register.
Example
Input
3
01
1111
0011
Output
DA
NET
NET
Note
In the first test case after Alice's move string s become empty and Bob can not make any move.
In the second test case Alice can not make any move initially.
In the third test case after Alice's move string s turn into 01. Then, after Bob's move string s become empty and Alice can not make any move. | instruction | 0 | 27,591 | 0 | 55,182 |
Tags: games
Correct Solution:
```
t=int(input())
def alice(st):
j=1
while True:
if "10" in st:
a=st.index("10")
st=st[:a]+st[a+2:]
j+=1
elif "01" in st:
a=st.index("01")
st=st[:a]+st[a+2:]
j+=1
else:
break
if j%2==0:
return "DA"
else:
return "NET"
li=[]
for i in range(t):
li=li+[input()]
for i in li:
print(alice(i))
``` | output | 1 | 27,591 | 0 | 55,183 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Alica and Bob are playing a game.
Initially they have a binary string s consisting of only characters 0 and 1.
Alice and Bob make alternating moves: Alice makes the first move, Bob makes the second move, Alice makes the third one, and so on. During each move, the current player must choose two different adjacent characters of string s and delete them. For example, if s = 1011001 then the following moves are possible:
1. delete s_1 and s_2: 1011001 → 11001;
2. delete s_2 and s_3: 1011001 → 11001;
3. delete s_4 and s_5: 1011001 → 10101;
4. delete s_6 and s_7: 1011001 → 10110.
If a player can't make any move, they lose. Both players play optimally. You have to determine if Alice can win.
Input
First line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Only line of each test case contains one string s (1 ≤ |s| ≤ 100), consisting of only characters 0 and 1.
Output
For each test case print answer in the single line.
If Alice can win print DA (YES in Russian) in any register. Otherwise print NET (NO in Russian) in any register.
Example
Input
3
01
1111
0011
Output
DA
NET
NET
Note
In the first test case after Alice's move string s become empty and Bob can not make any move.
In the second test case Alice can not make any move initially.
In the third test case after Alice's move string s turn into 01. Then, after Bob's move string s become empty and Alice can not make any move. | instruction | 0 | 27,592 | 0 | 55,184 |
Tags: games
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return map(int,raw_input().split())
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
for t in range(ni()):
s=raw_input().strip()
c1=s.count('0')
c2=s.count('1')
c=min(c1,c2)
if c%2:
pr('DA\n')
else:
pr('NET\n')
``` | output | 1 | 27,592 | 0 | 55,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alica and Bob are playing a game.
Initially they have a binary string s consisting of only characters 0 and 1.
Alice and Bob make alternating moves: Alice makes the first move, Bob makes the second move, Alice makes the third one, and so on. During each move, the current player must choose two different adjacent characters of string s and delete them. For example, if s = 1011001 then the following moves are possible:
1. delete s_1 and s_2: 1011001 → 11001;
2. delete s_2 and s_3: 1011001 → 11001;
3. delete s_4 and s_5: 1011001 → 10101;
4. delete s_6 and s_7: 1011001 → 10110.
If a player can't make any move, they lose. Both players play optimally. You have to determine if Alice can win.
Input
First line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
Only line of each test case contains one string s (1 ≤ |s| ≤ 100), consisting of only characters 0 and 1.
Output
For each test case print answer in the single line.
If Alice can win print DA (YES in Russian) in any register. Otherwise print NET (NO in Russian) in any register.
Example
Input
3
01
1111
0011
Output
DA
NET
NET
Note
In the first test case after Alice's move string s become empty and Bob can not make any move.
In the second test case Alice can not make any move initially.
In the third test case after Alice's move string s turn into 01. Then, after Bob's move string s become empty and Alice can not make any move.
Submitted Solution:
```
t = int(input())
for i in range(t):
s = input()
list1 = list(s)
flag = 0
c = 0
d = 0
for j in range(0,len(list1)):
if(list1[j]=="1"):
c=c+1
else:
d=d+1
chal = min(c,d)
print(chal)
if(c%2!=0):
print("DA")
else:
print("NET")
``` | instruction | 0 | 27,598 | 0 | 55,196 |
No | output | 1 | 27,598 | 0 | 55,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of n lowercase Latin letters.
Let's denote k-substring of s as a string subsk = sksk + 1..sn + 1 - k. Obviously, subs1 = s, and there are exactly <image> such substrings.
Let's call some string t an odd proper suprefix of a string T iff the following conditions are met:
* |T| > |t|;
* |t| is an odd number;
* t is simultaneously a prefix and a suffix of T.
For evey k-substring (<image>) of s you have to calculate the maximum length of its odd proper suprefix.
Input
The first line contains one integer n (2 ≤ n ≤ 106) — the length s.
The second line contains the string s consisting of n lowercase Latin letters.
Output
Print <image> integers. i-th of them should be equal to maximum length of an odd proper suprefix of i-substring of s (or - 1, if there is no such string that is an odd proper suprefix of i-substring).
Examples
Input
15
bcabcabcabcabca
Output
9 7 5 3 1 -1 -1 -1
Input
24
abaaabaaaabaaabaaaabaaab
Output
15 13 11 9 7 5 3 1 1 -1 -1 1
Input
19
cabcabbcabcabbcabca
Output
5 3 1 -1 -1 1 1 -1 -1 -1
Note
The answer for first sample test is folowing:
* 1-substring: bcabcabcabcabca
* 2-substring: cabcabcabcabc
* 3-substring: abcabcabcab
* 4-substring: bcabcabca
* 5-substring: cabcabc
* 6-substring: abcab
* 7-substring: bca
* 8-substring: c
Submitted Solution:
```
print("9 7 5 3 1 -1 -1 -1")
``` | instruction | 0 | 28,015 | 0 | 56,030 |
No | output | 1 | 28,015 | 0 | 56,031 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string $t$ and a set $S$ of $N$ different strings. You need to separate $t$ such that each part is included in $S$.
For example, the following 4 separation methods satisfy the condition when $t = abab$ and $S = \\{a, ab, b\\}$.
* $a,b,a,b$
* $a,b,ab$
* $ab,a,b$
* $ab,ab$
Your task is to count the number of ways to separate $t$. Because the result can be large, you should output the remainder divided by $1,000,000,007$.
Input
The input consists of a single test case formatted as follows.
$N$
$s_1$
:
$s_N$
$t$
The first line consists of an integer $N$ ($1 \leq N \leq 100,000$) which is the number of the elements of $S$. The following $N$ lines consist of $N$ distinct strings separated by line breaks. The $i$-th string $s_i$ represents the $i$-th element of $S$. $s_i$ consists of lowercase letters and the length is between $1$ and $100,000$, inclusive. The summation of length of $s_i$ ($1 \leq i \leq N$) is at most $200,000$. The next line consists of a string $t$ which consists of lowercase letters and represents the string to be separated and the length is between $1$ and $100,000$, inclusive.
Output
Calculate the number of ways to separate $t$ and print the remainder divided by $1,000,000,007$.
Examples
Input
3
a
b
ab
abab
Output
4
Input
3
a
b
c
xyz
Output
0
Input
7
abc
ab
bc
a
b
c
aa
aaabcbccababbc
Output
160
Input
10
a
aa
aaa
aaaa
aaaaa
aaaaaa
aaaaaaa
aaaaaaaa
aaaaaaaaa
aaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
Output
461695029 | instruction | 0 | 28,219 | 0 | 56,438 |
"Correct Solution:
```
from collections import defaultdict
import sys
def solve():
readline = sys.stdin.readline
write = sys.stdout.write
mod = 10**9 + 9
base = 37
ca = ord('a')
N = int(readline())
SS = [readline().strip() for i in range(N)]
SS.sort(key = len)
T = readline().strip()
L = len(T)
L0 = max(max(map(len, SS)), L)
r = 1
pw = [1]*(L0+1)
for i in range(L0):
pw[i+1] = r = r * base % mod
SM = defaultdict(dict)
E = [None]*N
for i in range(N):
s = SS[i][::-1]
h = 0
p = -1
for j, c in enumerate(s):
h = (h + pw[j] * (ord(c) - ca)) % mod
if j+1 in SM and h in SM[j+1]:
p = SM[j+1][h]
l = len(s)
E[i] = (E[p] + [l]) if p != -1 else [l]
SM[l][h] = i
*SI, = SM.items()
SI.sort()
H = [0]*(L+1)
r = 0
for i, c in enumerate(T):
H[i+1] = r = (base * r + (ord(c) - ca)) % mod
MOD = 10**9 + 7
dp = [0]*(L+1)
dp[0] = 1
SI.append((L+10, set()))
it = iter(SI).__next__
ln, sn = it()
SL = []
for i in range(1, L+1):
if i == ln:
SL.append((ln, sn, pw[ln]))
ln, sn = it()
hi = H[i]
for l, hs, w in reversed(SL):
v = (hi - H[i-l]*w) % mod
if v in hs:
dp[i] = sum(dp[i-e] for e in E[hs[v]]) % MOD
break
write("%d\n" % dp[L])
solve()
``` | output | 1 | 28,219 | 0 | 56,439 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, the Fair Nut has written k strings of length n, consisting of letters "a" and "b". He calculated c — the number of strings that are prefixes of at least one of the written strings. Every string was counted only one time.
Then, he lost his sheet with strings. He remembers that all written strings were lexicographically not smaller than string s and not bigger than string t. He is interested: what is the maximum value of c that he could get.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
The first line contains two integers n and k (1 ≤ n ≤ 5 ⋅ 10^5, 1 ≤ k ≤ 10^9).
The second line contains a string s (|s| = n) — the string consisting of letters "a" and "b.
The third line contains a string t (|t| = n) — the string consisting of letters "a" and "b.
It is guaranteed that string s is lexicographically not bigger than t.
Output
Print one number — maximal value of c.
Examples
Input
2 4
aa
bb
Output
6
Input
3 3
aba
bba
Output
8
Input
4 5
abbb
baaa
Output
8
Note
In the first example, Nut could write strings "aa", "ab", "ba", "bb". These 4 strings are prefixes of at least one of the written strings, as well as "a" and "b". Totally, 6 strings.
In the second example, Nut could write strings "aba", "baa", "bba".
In the third example, there are only two different strings that Nut could write. If both of them are written, c=8. | instruction | 0 | 28,294 | 0 | 56,588 |
Tags: greedy, strings
Correct Solution:
```
n,k=map(int,input().split())
s=input()
t=input()
res,tp=0,1
for i in range(n):
tp*=2
if s[i]=='b':
tp-=1
if t[i]=='a':
tp-=1
res+=min(tp,k)
tp=min(tp,1e9)
print(res)
``` | output | 1 | 28,294 | 0 | 56,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently, the Fair Nut has written k strings of length n, consisting of letters "a" and "b". He calculated c — the number of strings that are prefixes of at least one of the written strings. Every string was counted only one time.
Then, he lost his sheet with strings. He remembers that all written strings were lexicographically not smaller than string s and not bigger than string t. He is interested: what is the maximum value of c that he could get.
A string a is lexicographically smaller than a string b if and only if one of the following holds:
* a is a prefix of b, but a ≠ b;
* in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.
Input
The first line contains two integers n and k (1 ≤ n ≤ 5 ⋅ 10^5, 1 ≤ k ≤ 10^9).
The second line contains a string s (|s| = n) — the string consisting of letters "a" and "b.
The third line contains a string t (|t| = n) — the string consisting of letters "a" and "b.
It is guaranteed that string s is lexicographically not bigger than t.
Output
Print one number — maximal value of c.
Examples
Input
2 4
aa
bb
Output
6
Input
3 3
aba
bba
Output
8
Input
4 5
abbb
baaa
Output
8
Note
In the first example, Nut could write strings "aa", "ab", "ba", "bb". These 4 strings are prefixes of at least one of the written strings, as well as "a" and "b". Totally, 6 strings.
In the second example, Nut could write strings "aba", "baa", "bba".
In the third example, there are only two different strings that Nut could write. If both of them are written, c=8. | instruction | 0 | 28,295 | 0 | 56,590 |
Tags: greedy, strings
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
n,k = map(int,input().split())
s = input().strip()
t = input().strip()
diff,ans = 0,0
for i in range(n):
if diff+1 < k:
diff <<= 1
if s[i] == 'b':
diff -= 1
if t[i] == 'b':
diff += 1
ans += min(k,diff+1)
print(ans)
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | output | 1 | 28,295 | 0 | 56,591 |
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