question_id stringlengths 7 10 | question stringlengths 42 12.9k | starter_code stringclasses 37
values | response stringlengths 1.25k 174k | code stringlengths 14 11k | verified bool 1
class | is_passed bool 1
class | normalized_code stringlengths 14 5.01k |
|---|---|---|---|---|---|---|---|
seed_10096 | Given a Directed Graph, find a Mother Vertex in the Graph (if present).
A Mother Vertex is a vertex through which we can reach all the other vertices of the Graph.
Example 1:
Input:
Output: 0
Explanation: According to the given edges, all
nodes can be reaced from nodes from 0, 1 and 2.
But, since 0 is minimum amo... |
class Solution:
#Function to find a Mother Vertex in the Graph.
def findMotherVertex(self, V, adj):
#Code here | Okay, I need to solve this problem where I have to find a mother vertex in a directed graph. A mother vertex is one from which all other vertices can be reached. If there are multiple, pick the smallest one. If none exists, return -1. Hmm, how do I approach this?
Let me think. The straightforward way is for each verte... | from collections import deque
class Solution:
def findMotherVertex(self, V, adj):
visited = [False] * V
finish_order = []
# Step 1: Find the last finished node using DFS
for i in range(V):
if not visited[i]:
stack = [(i, False)]
w... | true | true | from collections import deque
class Solution:
def findMotherVertex(self, V, adj):
visited = [False] * V
finish_order = []
for i in range(V):
if not visited[i]:
stack = [(i, False)]
while stack:
node, processed = stack.pop()
if processed:
finish_order.append(node)
continue
if visited[node]:
continue
visited[node] = True... |
seed_101 | "Night gathers, and now my watch begins. It shall not end until my death. I shall take no wife, hold no lands, father no children. I shall wear no crowns and win no glory. I shall live and die at my post. I am the sword in the darkness. I am the watcher on the walls. I am the shield that guards the realms of men. I ple... | Okay, let's tackle this problem. So the task is to find out how many stewards Jon will support. The condition is that a steward is supported only if there's at least one steward with a strictly lower strength and another with a strictly higher strength than him.
Hmm, right. So first, I need to figure out for each ste... | n = int(input())
a = list(map(int, input().split()))
if n < 2:
print(0)
else:
min_val = min(a)
max_val = max(a)
if min_val == max_val:
print(0)
else:
count = 0
for num in a:
if num > min_val and num < max_val:
count += 1
print(count) | true | true | n = int(input())
a = list(map(int, input().split()))
if n < 2:
print(0)
else:
min_val = min(a)
max_val = max(a)
if min_val == max_val:
print(0)
else:
count = 0
for num in a:
if num > min_val and num < max_val:
count += 1
print(count) | |
seed_10138 | Given a [tree] containing N nodes.
Each node can be coloured using exactly one of the C colours. Find the number of colourings of the tree such that:
All nodes in a path of length at most two have distinct colours.
As the answer can be huge, print it modulo (10 ^ 9 + 7).
Note:
Two colourings of a tree are different ... | Okay, I need to solve this problem where I have to count the number of valid colorings of a tree such that all nodes in any path of length at most two have distinct colors. Hmm, let's see.
First, the constraints are big, with N and C up to 1e6. So I need an efficient algorithm, probably O(N) time. Let's think of the ... | import sys
from sys import stdin
from collections import deque
MOD = 10**9 + 7
def main():
sys.setrecursionlimit(1 << 25)
N, C = map(int, stdin.readline().split())
if N == 1:
print(C % MOD)
return
# Build adjacency list
adj = [[] for _ in range(N+1)]
for _ in range(N-1):
... | true | true | import sys
from sys import stdin
from collections import deque
MOD = 10**9 + 7
def main():
sys.setrecursionlimit(1 << 25)
N, C = map(int, stdin.readline().split())
if N == 1:
print(C % MOD)
return
adj = [[] for _ in range(N+1)]
for _ in range(N-1):
u, v = map(int, stdin.readline().split())
adj[u].append(v)
adj[v].appen... | |
seed_10164 | Given a binary number (of $N$ bits) $X$. Find the highest power of 2 that divides this number.
Note: If the binary number is "100" then the highest power of 2 that divides it is 2 (as $2^{2}$ = 4)
------ Input: ------
The first line contains N the number of bits in the number
The next line contains a binary number o... | Okay, I need to solve this problem where I have to find the highest power of 2 that divides a given binary number. Let's think about how to approach this.
Hmm, the key here is understanding what the highest power of 2 dividing the number means. Like, for example, if the binary is "100", which is 4 in decimal, the hig... | n = int(input())
s = input().strip()
count = 0
for c in reversed(s):
if c == '0':
count += 1
else:
break
print(count) | true | true | n = int(input())
s = input().strip()
count = 0
for c in reversed(s):
if c == '0':
count += 1
else:
break
print(count) | |
seed_10170 | "Given a binary string $S$ consisting of only 1’s and 0’s where 1 represents a Square and 0 repr(...TRUNCATED) | "Okay, let's tackle this problem. Hmm, the task is to find the maximum number of circles we can insc(...TRUNCATED) | "import sys\n\ndef main():\n N = int(sys.stdin.readline())\n for _ in range(N):\n S = s(...TRUNCATED) | true | true | "import sys\ndef main():\nN = int(sys.stdin.readline())\nfor _ in range(N):\nS = sys.stdin.readline((...TRUNCATED) | |
seed_10278 | "Given a complete binary tree with the height of H, we index the nodes respectively top-down and lef(...TRUNCATED) | "Okay, I need to solve this programming problem. Let's read the question again carefully to make sur(...TRUNCATED) | "import sys\n\nMOD = 10**9 + 7\n\ndef main():\n for line in sys.stdin:\n H = int(line.stri(...TRUNCATED) | true | true | "import sys\nMOD = 10**9 + 7\ndef main():\nfor line in sys.stdin:\nH = int(line.strip())\nif H == 0:(...TRUNCATED) | |
seed_10298 | "Given a directed acyclic graph(DAG) with n nodes labeled from 0 to n-1. Given edges, s and d ,count(...TRUNCATED) | "#User function Template for python3\n\n\n\nclass Solution:\n\n def possible_paths(self, edges, n(...TRUNCATED) | "Okay, I need to solve this problem where I have to find the number of paths from node s to d in a d(...TRUNCATED) | "class Solution:\n def possible_paths(self, edges, n, s, d):\n # Build adjacency list\n (...TRUNCATED) | true | true | "class Solution:\ndef possible_paths(self, edges, n, s, d):\nadj = [[] for _ in range(n)]\nfor u, v (...TRUNCATED) |
seed_103 | "\"Pizza At\", a fast food chain, offers three kinds of pizza: \"A-pizza\", \"B-pizza\" and \"AB-piz(...TRUNCATED) | "Okay, so I have this programming problem to solve. Let me read through it again carefully.\n\nThe p(...TRUNCATED) | "def main():\n import sys\n A, B, C, X, Y = map(int, sys.stdin.readline().split())\n max_ne(...TRUNCATED) | true | true | "def main():\nimport sys\nA, B, C, X, Y = map(int, sys.stdin.readline().split())\nmax_needed = max(X(...TRUNCATED) | |
seed_10302 | "Given a directed graph, find out if a vertex j is reachable from another vertex i for all vertex pa(...TRUNCATED) | "#User function Template for python3\n\nclass Solution:\n def transitiveClosure(self, N, graph):\(...TRUNCATED) | "Okay, I need to solve this problem where I have to find the transitive closure of a directed graph.(...TRUNCATED) | "class Solution:\n def transitiveClosure(self, N, graph):\n # Initialize the reachability (...TRUNCATED) | true | true | "class Solution:\ndef transitiveClosure(self, N, graph):\nreach = [[graph[i][j] for j in range(N)] f(...TRUNCATED) |
seed_10303 | "Given a directed graph. The task is to do Breadth First Traversal of this graph starting from 0.\nN(...TRUNCATED) | "#User function Template for python3\n\n\n\nfrom typing import List\n\nfrom queue import Queue\n\ncl(...TRUNCATED) | "Okay, I need to solve this BFS traversal problem for a directed graph. Let's see. The task is to pe(...TRUNCATED) | "from typing import List\nfrom collections import deque\n\nclass Solution:\n def bfsOfGraph(self,(...TRUNCATED) | true | true | "from typing import List\nfrom collections import deque\nclass Solution:\ndef bfsOfGraph(self, V: in(...TRUNCATED) |
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