Muennighoff/xP3x-sample huggingface dataset:
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1328/F:
You are given the array $a$ consisting of $n$ elements and the integer $k \le n$.

You want to obtain at least $k$ equal elements in the array $a$. In one move, you can make one of the following two operations:

  Take one of the minimum elements of the array and increase its value by one (more formally, if the minimum value of $a$ is $mn$ then you choose such index $i$ that $a_i = mn$ and set $a_i := a_i + 1$);  take one of the maximum elements of the array and decrease its value by one (more formally, if the maximum value of $a$ is $mx$ then you choose such index $i$ that $a_i = mx$ and set $a_i := a_i - 1$). 

Your task is to calculate the minimum number of moves required to obtain at least $k$ equal elements in the array.


-----Input-----

The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of elements in $a$ and the required number of equal elements.

The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the $i$-th element of $a$.


-----Output-----

Print one integer — the minimum number of moves required to obtain at least $k$ equal elements in the array.


-----Examples-----
Input
6 5
1 2 2 4 2 3

Output
3

Input
7 5
3 3 2 1 1 1 3

Output
4

I tried it in Python, but could not do it. Can you solve it?
targets: #!usr/bin/env python3
from collections import defaultdict, deque
from heapq import heappush, heappop
from itertools import permutations, accumulate
import sys
import math
import bisect
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
    res = list(sys.stdin.readline())
    if res[-1] == "\n":
        return res[:-1]
    return res
def IR(n):
    return [I() for i in range(n)]
def LIR(n):
    return [LI() for i in range(n)]
def SR(n):
    return [S() for i in range(n)]
def LSR(n):
    return [LS() for i in range(n)]

sys.setrecursionlimit(1000000)
mod = 1000000007

def solve():
    n,k = LI()
    a = LI()
    a.sort()
    d = defaultdict(lambda : 0)
    c = defaultdict(lambda : 0)
    s = [0]
    for i in a:
        d[i] += i
        c[i] += 1
        s.append(s[-1]+i)
    ans = float("inf")
    p = -1
    for i in a:
        if i == p:
            continue
        if k <= c[i]:
            ans = 0
            break
        l,r = bisect.bisect_left(a,i),bisect.bisect_right(a,i)
        m = r
        if m >= k:
            ns = l*(i-1)-s[l]
            su = ns+k-c[i]
            if su < ans:
                ans = su
        m = n-l
        if m >= k:
            ns = s[n]-s[r]-(n-r)*(i+1)
            su = ns+k-c[i]
            if su < ans:
                ans = su
        ns = s[n]-s[r]-(n-r)*(i+1)+l*(i-1)-s[l]
        su = ns+k-c[i]
        if su < ans:
            ans = su
        p = i
    print(ans)
    return

#Solve
def __starting_point():
    solve()

__starting_point()
inputs: Solve in Python:
Let's denote a function 

$d(x, y) = \left\{\begin{array}{ll}{y - x,} & {\text{if}|x - y|> 1} \\{0,} & {\text{if}|x - y|\leq 1} \end{array} \right.$

You are given an array a consisting of n integers. You have to calculate the sum of d(a_{i}, a_{j}) over all pairs (i, j) such that 1 ≤ i ≤ j ≤ n.


-----Input-----

The first line contains one integer n (1 ≤ n ≤ 200000) — the number of elements in a.

The second line contains n integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ 10^9) — elements of the array. 


-----Output-----

Print one integer — the sum of d(a_{i}, a_{j}) over all pairs (i, j) such that 1 ≤ i ≤ j ≤ n.


-----Examples-----
Input
5
1 2 3 1 3

Output
4

Input
4
6 6 5 5

Output
0

Input
4
6 6 4 4

Output
-8



-----Note-----

In the first example:

  d(a_1, a_2) = 0;  d(a_1, a_3) = 2;  d(a_1, a_4) = 0;  d(a_1, a_5) = 2;  d(a_2, a_3) = 0;  d(a_2, a_4) = 0;  d(a_2, a_5) = 0;  d(a_3, a_4) =  - 2;  d(a_3, a_5) = 0;  d(a_4, a_5) = 2.
targets: N = 2 * 10**5 + 3
n = int(input())
A, cnt = list(map(int,input().split(' '))), {}

s, a = 0, 0

for i in range(n-1,-1,-1):
    a += s
    if (A[i]-1) in cnt:
        a += cnt[A[i]-1]
    if (A[i]+1) in cnt:
        a -= cnt[A[i]+1]
    a -= (n-(i+1))*A[i]
    s += A[i]
    if A[i] not in cnt:
        cnt[A[i]]=0
    cnt[A[i]] += 1

print(a)
inputs: Solve in Python:
*** No Loops Allowed ***

You will be given an array (a) and a value (x). All you need to do is check whether the provided array contains the value, without using a loop.

Array can contain numbers or strings. X can be either. Return true if the array contains the value, false if not. With strings you will need to account for case.

Looking for more, loop-restrained fun? Check out the other kata in the series:

 https://www.codewars.com/kata/no-loops-1-small-enough
 https://www.codewars.com/kata/no-loops-3-copy-within
targets: def check(a, x): 
    test = a.count(x)
    if test >= 1:
        return True
    else:
        return False
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/327/A:
Iahub got bored, so he invented a game to be played on paper. 

He writes n integers a_1, a_2, ..., a_{n}. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values a_{k} for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.

The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.


-----Input-----

The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a_1, a_2, ..., a_{n}. It is guaranteed that each of those n values is either 0 or 1.


-----Output-----

Print an integer — the maximal number of 1s that can be obtained after exactly one move. 


-----Examples-----
Input
5
1 0 0 1 0

Output
4

Input
4
1 0 0 1

Output
4



-----Note-----

In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].

In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1.

I tried it in Python, but could not do it. Can you solve it?
targets: n = int(input())
I = list(map(int, input().split()))
d = []
for val in I:
    d.append(val)
res = 0;
for i in range (0, n):
    for j in range (i, n):
        cnt = 0;
        for k in range (0, n):
            if k >= i and k <= j:
                cnt = cnt + 1 - d[k];
            else:
                cnt = cnt + d[k];
        if cnt > res:
            res = cnt;
print (res)
inputs: Solve in Python:
On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't wait to destroy the toy.

The toy consists of n parts and m ropes. Each rope links two parts, but every pair of parts is linked by at most one rope. To split the toy, the child must remove all its parts. The child can remove a single part at a time, and each remove consume an energy. Let's define an energy value of part i as v_{i}. The child spend v_{f}_1 + v_{f}_2 + ... + v_{f}_{k} energy for removing part i where f_1, f_2, ..., f_{k} are the parts that are directly connected to the i-th and haven't been removed.

Help the child to find out, what is the minimum total energy he should spend to remove all n parts.


-----Input-----

The first line contains two integers n and m (1 ≤ n ≤ 1000; 0 ≤ m ≤ 2000). The second line contains n integers: v_1, v_2, ..., v_{n} (0 ≤ v_{i} ≤ 10^5). Then followed m lines, each line contains two integers x_{i} and y_{i}, representing a rope from part x_{i} to part y_{i} (1 ≤ x_{i}, y_{i} ≤ n; x_{i} ≠ y_{i}).

Consider all the parts are numbered from 1 to n.


-----Output-----

Output the minimum total energy the child should spend to remove all n parts of the toy.


-----Examples-----
Input
4 3
10 20 30 40
1 4
1 2
2 3

Output
40

Input
4 4
100 100 100 100
1 2
2 3
2 4
3 4

Output
400

Input
7 10
40 10 20 10 20 80 40
1 5
4 7
4 5
5 2
5 7
6 4
1 6
1 3
4 3
1 4

Output
160



-----Note-----

One of the optimal sequence of actions in the first sample is:  First, remove part 3, cost of the action is 20.  Then, remove part 2, cost of the action is 10.  Next, remove part 4, cost of the action is 10.  At last, remove part 1, cost of the action is 0. 

So the total energy the child paid is 20 + 10 + 10 + 0 = 40, which is the minimum.

In the second sample, the child will spend 400 no matter in what order he will remove the parts.
targets: (n, m), f, s = list(map(int, input().split())), [int(x) for x in input().split()], 0

for i in range(m):
    (a, b) = list(map(int, input().split()))
    s = s + min(f[a - 1], f[b - 1])

print( s )
inputs: I found an interesting problem on https://www.codewars.com/kata/559e5b717dd758a3eb00005a:
DropCaps means that the first letter of the starting word of the paragraph should be in caps and the remaining lowercase, just like you see in the newspaper. 

But for a change, let's do that for each and every word of the given String. Your task is to capitalize every word that has length greater than 2, leaving smaller words as they are.

*should work also on Leading and Trailing Spaces and caps.

```python
drop_cap('apple') => "Apple"
drop_cap('apple of banana'); => "Apple of Banana"
drop_cap('one   space'); => "One   Space" 
drop_cap('   space WALK   '); => "   Space Walk   " 
```

**Note:** you will be provided atleast one word and should take string as input and return string as output.

I tried it in Python, but could not do it. Can you solve it?
targets: import re

def drop_cap(s):
    return re.sub(r'\S{3,}', lambda m: m.group(0).title(), s)
inputs: Solve in Python:
#It's show time!
Archers have gathered from all around the world to participate in the Arrow Function Faire. But the faire will only start if there are archers signed and if they all have enough arrows in their quivers - at least 5 is the requirement! Are all the archers ready?

#Reference
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions/Arrow_functions

#Argument
`archers` is an array of integers, in which each element corresponds to the number of arrows each archer has.

#Return
Your function must return `true` if the requirements are met or `false` otherwise.

#Examples
`archersReady([1, 2, 3, 4, 5, 6, 7, 8])` returns `false` because there are archers with not enough arrows.

`archersReady([5, 6, 7, 8])` returns `true`.
targets: def archers_ready(archers):
    if not archers:
        return False
    for arrows in archers:
        if arrows < 5:
            return False
    return True
inputs: I found an interesting problem on https://www.codewars.com/kata/5750699bcac40b3ed80001ca:
It is 2050 and romance has long gone, relationships exist solely for practicality.

MatchMyHusband is a website that matches busy working women with perfect house husbands. You have been employed by MatchMyHusband to write a function that determines who matches!!

The rules are... a match occurs providing the husband's "usefulness" rating is greater than or equal to the woman's "needs".

The husband's "usefulness" is the SUM of his cooking, cleaning and childcare abilities and takes the form of an array .

usefulness example --> [15, 26, 19]    (15 + 26 + 19) = 60

Every woman that signs up, begins with a "needs" rating of 100. However, it's realised that the longer women wait for their husbands, the more dissatisfied they become with our service. They also become less picky, therefore their needs are subject to exponential decay of 15% per month. https://en.wikipedia.org/wiki/Exponential_decay

Given the number of months since sign up, write a function that returns "Match!" if the husband is useful enough, or "No match!" if he's not.

I tried it in Python, but could not do it. Can you solve it?
targets: def match(a, n):
    return "Match!" if sum(a) >= 100 * 0.85**n else "No match!"
inputs: Solve in Python:
Complete the function which returns the weekday according to the input number:

* `1` returns `"Sunday"`
* `2` returns `"Monday"`
* `3` returns `"Tuesday"`
* `4` returns `"Wednesday"`
* `5` returns `"Thursday"`
* `6` returns `"Friday"`
* `7` returns `"Saturday"`
* Otherwise returns `"Wrong, please enter a number between 1 and 7"`
targets: def whatday(num):
    dict={
        2:"Monday",
        3:"Tuesday",
        4:"Wednesday",
        5:"Thursday",
        6:"Friday",
        7:"Saturday",
        1:"Sunday"
    }
    return dict.get(num,'Wrong, please enter a number between 1 and 7')
inputs: Solve in Python:
Bizon the Champion isn't just attentive, he also is very hardworking.

Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represented as n vertical planks, put in a row. Adjacent planks have no gap between them. The planks are numbered from the left to the right starting from one, the i-th plank has the width of 1 meter and the height of a_{i} meters.

Bizon the Champion bought a brush in the shop, the brush's width is 1 meter. He can make vertical and horizontal strokes with the brush. During a stroke the brush's full surface must touch the fence at all the time (see the samples for the better understanding). What minimum number of strokes should Bizon the Champion do to fully paint the fence? Note that you are allowed to paint the same area of the fence multiple times.


-----Input-----

The first line contains integer n (1 ≤ n ≤ 5000) — the number of fence planks. The second line contains n space-separated integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ 10^9).


-----Output-----

Print a single integer — the minimum number of strokes needed to paint the whole fence.


-----Examples-----
Input
5
2 2 1 2 1

Output
3

Input
2
2 2

Output
2

Input
1
5

Output
1



-----Note-----

In the first sample you need to paint the fence in three strokes with the brush: the first stroke goes on height 1 horizontally along all the planks. The second stroke goes on height 2 horizontally and paints the first and second planks and the third stroke (it can be horizontal and vertical) finishes painting the fourth plank.

In the second sample you can paint the fence with two strokes, either two horizontal or two vertical strokes.

In the third sample there is only one plank that can be painted using a single vertical stroke.
targets: import sys
oo=1000000000000
ar=[]
def solve(l, r, val):
	if(r<l): return 0
	indx=l+ar[l:r+1].index(min(ar[l:r+1]))
	tot=r-l+1
	cur=ar[indx]-val+solve(l, indx-1, ar[indx])+solve(indx+1, r, ar[indx])
	return min(tot, cur)
sys.setrecursionlimit(10000)
n=int(input())
ar=list(map(int, input().split()))
print(solve(0, n-1, 0))
inputs: I found an interesting problem on https://www.codechef.com/APRIL14/problems/BINTREE:
Consider an infinite full binary tree (each node has two children except the leaf nodes) defined as follows. For a node labelled v its left child will be labelled 2*v and its right child will be labelled 2*v+1. The root is labelled as 1.
You are given N queries of the form i j. For each query, you have to print the length of the shortest path between node labelled i and  node labelled j.

-----Input-----
First line contains N, the number of queries. Each query consists of two space separated integers i and j in one line.

-----Output-----
For each query, print the required answer in one line.

-----Constraints-----
- 1 ≤ N ≤ 105
- 1 ≤ i,j ≤ 109

-----Example-----
Input:
3
1 2
2 3
4 3

Output:
1
2
3

-----Explanation-----
For first query, 1 is directly connected to 2 by an edge. Hence distance 1.

I tried it in Python, but could not do it. Can you solve it?
targets: import math
import sys
import time


def getLevel(number):
 _level = math.floor(math.log(number,2))
 return int(_level)

def isPair(number):
 return (number%2 == 0)
 
def getInterval(a,b, max):
 _max = max
 _loop = -1
 _apart = 0
 
 while (_apart == 0) and (_max != 1):
  _loop += 1
  _max = _max >> 1
  if (a < _max) and (b >= _max):
   _apart = 1
  elif (a >= _max):
   a = a - _max
   b = b - _max
 
 return _loop

def resolve2(nodes):
 
 if nodes[0] == nodes[1]:
  return 0

 l = 0
 k = 0
 
 while (nodes[0] != nodes[1]) :
  if (nodes[1] < nodes[0]) :
   if (nodes[0] > 1):
    nodes[0] = nodes[0]/2 if isPair(nodes[0]) else (nodes[0] -1)/2
    l += 1
  else:
   if (nodes[1] > 1):
    nodes[1] = nodes[1]/2 if isPair(nodes[1]) else (nodes[1] -1)/2
    k += 1
  
 return (l + k)
 

def __starting_point():

 step = 0
 
 for line in sys.stdin:
  
  if (step==0):
   step = 1
   continue
  nodes =[int(nums) for nums in line.strip().split()]
  
  
  if nodes[0] == nodes[1]:
   print(0)
   continue
 
  _temp = nodes[0]
  if nodes[0] < nodes[1] :
   nodes[0] = nodes[1]
   nodes[1] = _temp
  
  _max_level = getLevel(nodes[0])
  if (nodes[1] == 1):
   print(_max_level)
   continue
  
  _min_level = getLevel(nodes[1])
  substract = _max_level - _min_level
  
  
  if (substract != 0):
   nodes[0] = nodes[0] >> substract
  
  _temp = nodes[0]
  if nodes[0] < nodes[1] :
   nodes[0] = nodes[1]
   nodes[1] = _temp
  
  
  if (nodes[1] == nodes[0]):
   print( substract)
   continue

  _max = 2**_min_level
  
  apart = getInterval(nodes[1] - _max, nodes[0] - _max,_max)
  
  
  print( 2*(_min_level - apart) + substract)
  
  
  #print( resolve2(nodes) )

__starting_point()
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1105/A:
Salem gave you $n$ sticks with integer positive lengths $a_1, a_2, \ldots, a_n$.

For every stick, you can change its length to any other positive integer length (that is, either shrink or stretch it). The cost of changing the stick's length from $a$ to $b$ is $|a - b|$, where $|x|$ means the absolute value of $x$.

A stick length $a_i$ is called almost good for some integer $t$ if $|a_i - t| \le 1$.

Salem asks you to change the lengths of some sticks (possibly all or none), such that all sticks' lengths are almost good for some positive integer $t$ and the total cost of changing is minimum possible. The value of $t$ is not fixed in advance and you can choose it as any positive integer. 

As an answer, print the value of $t$ and the minimum cost. If there are multiple optimal choices for $t$, print any of them.


-----Input-----

The first line contains a single integer $n$ ($1 \le n \le 1000$) — the number of sticks.

The second line contains $n$ integers $a_i$ ($1 \le a_i \le 100$) — the lengths of the sticks.


-----Output-----

Print the value of $t$ and the minimum possible cost. If there are multiple optimal choices for $t$, print any of them.


-----Examples-----
Input
3
10 1 4

Output
3 7

Input
5
1 1 2 2 3

Output
2 0



-----Note-----

In the first example, we can change $1$ into $2$ and $10$ into $4$ with cost $|1 - 2| + |10 - 4| = 1 + 6 = 7$ and the resulting lengths $[2, 4, 4]$ are almost good for $t = 3$.

In the second example, the sticks lengths are already almost good for $t = 2$, so we don't have to do anything.

I tried it in Python, but could not do it. Can you solve it?
targets: n = int(input())

arr = list(map(int, input().split()))
arr.sort()

a = []
for t in range(1, 101):
    tot = 0
    for item in arr:
        if (abs(item - t) >= 1):
            tot += abs(item - t) - 1
    
    a.append((tot, t))

a.sort()

print(a[0][1], a[0][0])
inputs: Solve in Python:
You are given a set $S$ and $Q$ queries. Initially, $S$ is empty. In each query:
- You are given a positive integer $X$.
- You should insert $X$ into $S$.
- For each $y \in S$ before this query such that $y \neq X$, you should also insert $y \oplus X$ into $S$ ($\oplus$ denotes the XOR operation).
- Then, you should find two values $E$ and $O$: the number of elements of $S$ with an even number of $1$-s and with an odd number of $1$-s in the binary representation, respectively.
Note that a set cannot have duplicate elements, so if you try to insert into $S$ an element that is already present in $S$, then nothing happens.

-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first line of each test case contains a single integer $Q$.
- Each of the next $Q$ lines contains a single integer $X$ describing a query.

-----Output-----
For each query, print a single line containing two space-separated integers $E$ and $O$.

-----Constraints-----
- $1 \le T \le 5$
- $1 \le Q, X \le 10^5$

-----Subtasks-----
Subtask #1 (30 points):
- $1 \le Q \le 1,000$
- $1 \le X \le 128$
Subtask #2 (70 points): original constraints

-----Example Input-----
1
3
4
2
7

-----Example Output-----
0 1
1 2
3 4

-----Explanation-----
Example case 1:
- Initially, the set is empty: $S = \{\}$.
- After the first query, $S = \{4\}$, so there is only one element with an odd number of $1$-s in the binary representation ("100").
- After the second query, $S = \{4,2,6\}$, there is one element with an even number of $1$-s in the binary representation ($6$ is "110") and the other two elements have an odd number of $1$-s.
- After the third query, $S = \{4,2,6,7,3,5,1\}$.
targets: import collections
def bits(x):
 return bin(x).count('1')
t=int(input())
for _ in range(t):
 q=int(input())
 s=[]
 d=collections.defaultdict(lambda:-1)
 e=0
 o=0
 p=-1
 for i in range(q):
  x=int(input())
  # print(d)
  if d[x]!=-1:
   print(e,o)
   continue
  s.append(x)
  d[x]=1
  #z=len(s)
  for j in s:
   if j!=x:
    l=j^x
    if d[l]!=-1:
     continue
    s.append(l)
    d[l]=1
   
  for j in range(p+1,len(s)):
   p=j
   if bits(s[j])%2==0:
    e=e+1
   elif bits(s[j])%2!=0:
    o=o+1
  print(e,o)
inputs: I found an interesting problem on https://www.codechef.com/IGNS2012/problems/IG04:
In a fictitious city of CODASLAM there were many skyscrapers. The mayor of the city decided to make the city beautiful and for this he decided to arrange the skyscrapers in descending order of their height, and the order must be strictly decreasing but he also didn’t want to waste much money so he decided to get the minimum cuts possible. Your job is to output the minimum value of cut that is possible to arrange the skyscrapers in descending order.

-----Input-----

*First line of input is the number of sky-scrappers in the city
*Second line of input is the height of the respective sky-scrappers


-----Output-----

* Your output should be the minimum value of cut required to arrange these sky-scrappers in descending order.

-----Example-----
Input:
5
1 2 3 4 5

Output:
8

By:
Chintan,Asad,Ashayam,Akanksha

I tried it in Python, but could not do it. Can you solve it?
targets: import sys
 
num=int(sys.stdin.readline())
s=sys.stdin.readline().split()
sky=list(map(int,s))
sky.reverse()
cuts=0
change=0
t=False
i=1
 
while i<len(sky):
 if sky[i]<=sky[i-1]:
  for j in range(i-1,-1,-1):
   
   if sky[j]<=sky[i]-(i-j):
    break
   else:
    change+=sky[j]-(sky[i]-(i-j))
    
   if change>=sky[i]:
    change=sky[i]
    t=True
    break
    
  cuts+=change
  
  if t:
   del sky[i]
   t=False
   i-=1
   
  else:
   for j in range(i-1,-1,-1):
    if sky[j]<sky[i]-(i-j):
     break
    else:
     sky[j]=sky[i]-(i-j)
 i+=1
    
 change=0
     
print(cuts)
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1205/A:
You are given integer $n$. You have to arrange numbers from $1$ to $2n$, using each of them exactly once, on the circle, so that the following condition would be satisfied:

For every $n$ consecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard $2n$ numbers differ not more than by $1$.

For example, choose $n = 3$. On the left you can see an example of a valid arrangement: $1 + 4 + 5 = 10$, $4 + 5 + 2 = 11$, $5 + 2 + 3 = 10$, $2 + 3 + 6 = 11$, $3 + 6 + 1 = 10$, $6 + 1 + 4 = 11$, any two numbers differ by at most $1$. On the right you can see an invalid arrangement: for example, $5 + 1 + 6 = 12$, and $3 + 2 + 4 = 9$, $9$ and $12$ differ more than by $1$.

 [Image] 


-----Input-----

The first and the only line contain one integer $n$ ($1 \le n \le 10^5$).


-----Output-----

If there is no solution, output "NO" in the first line. 

If there is a solution, output "YES" in the first line. In the second line output $2n$ numbers — numbers from $1$ to $2n$ in the order they will stay in the circle. Each number should appear only once. If there are several solutions, you can output any of them.


-----Examples-----
Input
3

Output
YES
1 4 5 2 3 6 
Input
4

Output
NO


-----Note-----

Example from the statement is shown for the first example. 

It can be proved that there is no solution in the second example.

I tried it in Python, but could not do it. Can you solve it?
targets: # 1205A Almost Equal
n = int(input())

if n % 2 == 0:
    print("NO")
else:
    print("YES")
    d = 1
    for i in range(n):
        print(d, end=' ')
        d += 3 if i % 2 == 0 else 1
    d = 2
    for i in range(n):
        print(d, end=' ')
        d += 1 if i % 2 == 0 else 3
inputs: I found an interesting problem on https://atcoder.jp/contests/abc077/tasks/abc077_a:
You are given a grid with 2 rows and 3 columns of squares.
The color of the square at the i-th row and j-th column is represented by the character C_{ij}.
Write a program that prints YES if this grid remains the same when rotated 180 degrees, and prints NO otherwise.

-----Constraints-----
 - C_{i,j}(1 \leq i \leq 2, 1 \leq j \leq 3) is a lowercase English letter.

-----Input-----
Input is given from Standard Input in the following format:
C_{11}C_{12}C_{13}
C_{21}C_{22}C_{23}

-----Output-----
Print YES if this grid remains the same when rotated 180 degrees; print NO otherwise.

-----Sample Input-----
pot
top

-----Sample Output-----
YES

This grid remains the same when rotated 180 degrees.

I tried it in Python, but could not do it. Can you solve it?
targets: c1 = input()
c2 = input()
print("YES" if c1 == c2[::-1] else "NO")
inputs: Solve in Python:
Chef bought a huge (effectively infinite) planar island and built $N$ restaurants (numbered $1$ through $N$) on it. For each valid $i$, the Cartesian coordinates of restaurant $i$ are $(X_i, Y_i)$.
Now, Chef wants to build $N-1$ straight narrow roads (line segments) on the island. The roads may have arbitrary lengths; restaurants do not have to lie on the roads. The slope of each road must be $1$ or $-1$, i.e. for any two points $(x_1, y_1)$ and $(x_2, y_2)$ on the same road, $|x_1-x_2| = |y_1-y_2|$ must hold.
Let's denote the minimum distance Chef has to walk from restaurant $i$ to reach a road by $D_i$. Then, let's denote $a = \mathrm{max}\,(D_1, D_2, \ldots, D_N)$; Chef wants this distance to be minimum possible.
Chef is a busy person, so he decided to give you the job of building the roads. You should find a way to build them that minimises $a$ and compute $a \cdot \sqrt{2}$.

-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first line of each test case contains a single integer $N$.
- $N$ lines follow. For each valid $i$, the $i$-th of these lines contains two space-separated integers $X_i$ and $Y_i$.

-----Output-----
For each test case, print a single line containing one real number — the minimum distance $a$ multiplied by $\sqrt{2}$. Your answer will be considered correct if its absolute or relative error does not exceed $10^{-6}$.

-----Constraints-----
- $1 \le T \le 100$
- $2 \le N \le 10^4$
- $|X_i|, |Y_i| \le 10^9$ for each valid $i$

-----Subtasks-----
Subtask #1 (10 points):
- $1 \le T \le 10$
- $2 \le N \le 5$
- $|X_i|, |Y_i| \le 10$ for each valid $i$
- $a \cdot \sqrt{2}$ is an integer
Subtask #2 (90 points): original constraints

-----Example Input-----
2
3
0 0
0 1
0 -1
3
0 1
1 0
-1 0

-----Example Output-----
0.5
0

-----Explanation-----
Example case 1: We should build roads described by equations $y-x+0.5 = 0$ and $y-x-0.5 = 0$.
Example case 2: We should build roads described by...
targets: t=int(input())
for i in range(t):
 n=int(input())
 x=[]
 y=[]
 dx=[]
 dy=[]
 for j in range(n):
  c=list(map(int,input().split()))
  x.append(c[0]-c[1])
  y.append(c[0]+c[1])
 x.sort()
 y.sort()
 for j in range(1,n):
  dx.append(x[j]-x[j-1])
  dy.append(y[j]-y[j-1])
 a=min(dx)
 b=min(dy)
 print('{0:.6f}'.format(min(a,b)/2))
inputs: Solve in Python:
*** No Loops Allowed ***

You will be given an array (a) and a limit value (limit). You must check that all values in the array are below or equal to the limit value. If they are, return true. Else, return false.

You can assume all values in the array are numbers.

Do not use loops. Do not modify input array.

Looking for more, loop-restrained fun? Check out the other kata in the series:

 https://www.codewars.com/kata/no-loops-2-you-only-need-one
 https://www.codewars.com/kata/no-loops-3-copy-within
targets: small_enough = lambda a, x: False not in map(lambda i: i <= x, a)
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/758/D:
Alexander is learning how to convert numbers from the decimal system to any other, however, he doesn't know English letters, so he writes any number only as a decimal number, it means that instead of the letter A he will write the number 10. Thus, by converting the number 475 from decimal to hexadecimal system, he gets 11311 (475 = 1·16^2 + 13·16^1 + 11·16^0). Alexander lived calmly until he tried to convert the number back to the decimal number system.

Alexander remembers that he worked with little numbers so he asks to find the minimum decimal number so that by converting it to the system with the base n he will get the number k.


-----Input-----

The first line contains the integer n (2 ≤ n ≤ 10^9). The second line contains the integer k (0 ≤ k < 10^60), it is guaranteed that the number k contains no more than 60 symbols. All digits in the second line are strictly less than n.

Alexander guarantees that the answer exists and does not exceed 10^18.

The number k doesn't contain leading zeros.


-----Output-----

Print the number x (0 ≤ x ≤ 10^18) — the answer to the problem.


-----Examples-----
Input
13
12

Output
12
Input
16
11311

Output
475
Input
20
999

Output
3789
Input
17
2016

Output
594


-----Note-----

In the first example 12 could be obtained by converting two numbers to the system with base 13: 12 = 12·13^0 or 15 = 1·13^1 + 2·13^0.

I tried it in Python, but could not do it. Can you solve it?
targets: n=int(input())
#print(n)
d=input()
l=len(d)
dp=[[0,0] for i in range(0,l+1)]
dp[l-1]=[ord(d[l-1])-ord('0'),1]
for i in range(1,l):
	w=l-1-i;
	m=(ord(d[w])-ord('0'))*(n**i)+dp[w+1][0]
	dp[w]=[m,i+1];
	if (d[w]=='0'):
		dp[w][0]=dp[w+1][0]
		dp[w][1]=dp[w+1][1]+1
		continue
	for j in range(w,l):
		subs=int(d[w:j+1])
		u=dp[j+1]
		if subs<n:
			re=subs*(n**u[1])+u[0]
			if re<dp[w][0] or (re==dp[w][0] and 1+u[1]<dp[w][1]):
				dp[w][0]=re
				dp[w][1]=1+u[1]
		else:
			break
print(dp[0][0])
inputs: Solve in Python:
Ashish has an array $a$ of size $n$.

A subsequence of $a$ is defined as a sequence that can be obtained from $a$ by deleting some elements (possibly none), without changing the order of the remaining elements.

Consider a subsequence $s$ of $a$. He defines the cost of $s$ as the minimum between:   The maximum among all elements at odd indices of $s$.  The maximum among all elements at even indices of $s$. 

Note that the index of an element is its index in $s$, rather than its index in $a$. The positions are numbered from $1$. So, the cost of $s$ is equal to $min(max(s_1, s_3, s_5, \ldots), max(s_2, s_4, s_6, \ldots))$.

For example, the cost of $\{7, 5, 6\}$ is $min( max(7, 6), max(5) ) = min(7, 5) = 5$.

Help him find the minimum cost of a subsequence of size $k$.


-----Input-----

The first line contains two integers $n$ and $k$ ($2 \leq k \leq n \leq 2 \cdot 10^5$)  — the size of the array $a$ and the size of the subsequence.

The next line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$)  — the elements of the array $a$.


-----Output-----

Output a single integer  — the minimum cost of a subsequence of size $k$.


-----Examples-----
Input
4 2
1 2 3 4

Output
1
Input
4 3
1 2 3 4

Output
2
Input
5 3
5 3 4 2 6

Output
2
Input
6 4
5 3 50 2 4 5

Output
3


-----Note-----

In the first test, consider the subsequence $s$ = $\{1, 3\}$. Here the cost is equal to $min(max(1), max(3)) = 1$.

In the second test, consider the subsequence $s$ = $\{1, 2, 4\}$. Here the cost is equal to $min(max(1, 4), max(2)) = 2$.

In the fourth test, consider the subsequence $s$ = $\{3, 50, 2, 4\}$. Here the cost is equal to $min(max(3, 2), max(50, 4)) = 3$.
targets: def check(num):
	count = 0
	flag = -1
	s = 0
	for i in range(n):
		if a[i]<=num:
			count += 1
			s += 1
		else:
			if flag == -1:
				flag = s%2
				count += 1
				s += 1
			else:
				if (s+1)%2!=flag:
					count += 1
					s += 1
		if count==k:
			return True
	return False

n,k = map(int,input().split())
a = list(map(int,input().split()))
if n==k:
	m1 = 0
	m2 = 0
	for i in range(0,n,2):
		m1 = max(m1,a[i])
	for i in range(1,n,2):
		m2 = max(m2,a[i])
	print (min(m1,m2))
	return
s = set(a)
minn,maxx = min(a),max(a)
low = minn
high = maxx
while low<high:
	mid = (low+high)//2
	if check(mid):
		high = mid-1
	else:
		low = mid+1
if check(low):
	print (low)
else:
	print (low+1)
inputs: I found an interesting problem on https://www.hackerrank.com/challenges/word-order/problem:
=====Problem Statement=====
You are given n words. Some words may repeat. For each word, output its number of occurrences. The output order should correspond with the input order of appearance of the word. See the sample input/output for clarification.

Note: Each input line ends with a "\n" character.

=====Constraints=====
1≤n≤10^5
The sum of the lengths of all the words do not exceed 10^6
All the words are composed of lowercase English letters only.

=====Input Format=====
The first line contains the integer, n.
The next n lines each contain a word.

=====Output Format=====
Output 2 lines.
On the first line, output the number of distinct words from the input.
On the second line, output the number of occurrences for each distinct word according to their appearance in the input.

I tried it in Python, but could not do it. Can you solve it?
targets: #!/usr/bin/env python3

from collections import OrderedDict

def __starting_point():
    num = int(input().strip())
    history = OrderedDict()
    
    for _ in range(num):
        word = str(input().strip().split())
        if word not in list(history.keys()):
            history[word] = 1
        else:
            history[word] += 1
            
    print((len(list(history.keys()))))
    print((" ".join(map(str, list(history.values())))))

__starting_point()
inputs: Solve in Python:
Limak is a little polar bear. He plays by building towers from blocks. Every block is a cube with positive integer length of side. Limak has infinitely many blocks of each side length.

A block with side a has volume a^3. A tower consisting of blocks with sides a_1, a_2, ..., a_{k} has the total volume a_1^3 + a_2^3 + ... + a_{k}^3.

Limak is going to build a tower. First, he asks you to tell him a positive integer X — the required total volume of the tower. Then, Limak adds new blocks greedily, one by one. Each time he adds the biggest block such that the total volume doesn't exceed X.

Limak asks you to choose X not greater than m. Also, he wants to maximize the number of blocks in the tower at the end (however, he still behaves greedily). Secondarily, he wants to maximize X.

Can you help Limak? Find the maximum number of blocks his tower can have and the maximum X ≤ m that results this number of blocks.


-----Input-----

The only line of the input contains one integer m (1 ≤ m ≤ 10^15), meaning that Limak wants you to choose X between 1 and m, inclusive.


-----Output-----

Print two integers — the maximum number of blocks in the tower and the maximum required total volume X, resulting in the maximum number of blocks.


-----Examples-----
Input
48

Output
9 42

Input
6

Output
6 6



-----Note-----

In the first sample test, there will be 9 blocks if you choose X = 23 or X = 42. Limak wants to maximize X secondarily so you should choose 42.

In more detail, after choosing X = 42 the process of building a tower is:  Limak takes a block with side 3 because it's the biggest block with volume not greater than 42. The remaining volume is 42 - 27 = 15.  The second added block has side 2, so the remaining volume is 15 - 8 = 7.  Finally, Limak adds 7 blocks with side 1, one by one. 

So, there are 9 blocks in the tower. The total volume is is 3^3 + 2^3 + 7·1^3 = 27 + 8 + 7 = 42.
targets: def g(m, s, n):
    k = int(m ** (1 / 3))
    if m == 0 or k == 1: return s + m, n + m
    x, y = k ** 3, (k - 1) ** 3
    a = g(m - x, s + x, n + 1)
    b = g(x - y - 1, s + y, n + 1)
    return b if a[1] < b[1] else a
s, n = g(int(input()), 0, 0)
print(n, s)
inputs: Solve in Python:
Task
======

Make a custom esolang interpreter for the language [InfiniTick](https://esolangs.org/wiki/InfiniTick). InfiniTick is a descendant of [Tick](https://esolangs.org/wiki/tick) but also very different. Unlike Tick, InfiniTick has 8 commands instead of 4. It also runs infinitely, stopping the program only when an error occurs. You may choose to do the [Tick](https://www.codewars.com/kata/esolang-tick) kata first.

Syntax/Info
======

InfiniTick runs in an infinite loop. This makes it both harder and easier to program in. It has an infinite memory of bytes and an infinite output amount. The program only stops when an error is reached. The program is also supposed to ignore non-commands.

Commands
======

`>`: Move data selector right.

`<`: Move data selector left.

`+`: Increment amount of memory cell. Truncate overflow: 255+1=0.

`-`: Decrement amount of memory cell. Truncate overflow: 0-1=255.

`*`: Add ascii value of memory cell to the output tape.

`&`: Raise an error, ie stop the program.

`/`: Skip next command if cell value is zero.

`\`: Skip next command if cell value is nonzero.

Examples
======

**Hello world!**

The following is a valid hello world program in...
targets: from collections import defaultdict
from itertools import cycle

def interpreter(tape):
    i, skip, D, res = 0, False, defaultdict(int), []
    for c in cycle(tape):
        if skip: skip = False
        elif c == '>': i += 1
        elif c == '<': i -= 1
        elif c == '+': D[i] = (D[i] + 1)%256
        elif c == '-': D[i] = (D[i] - 1)%256
        elif c == '*': res.append(chr(D[i]))
        elif c == '&': break
        elif c == '/': skip = not D[i]
        elif c == '\\': skip = D[i]
    return ''.join(res)
inputs: I found an interesting problem on https://www.codewars.com/kata/5a32526ae1ce0ec0f10000b2:
Given an integer, take the (mean) average of each pair of consecutive digits. Repeat this process until you have a single integer, then return that integer. e.g.

Note: if the average of two digits is not an integer, round the result **up** (e.g. the average of 8 and 9 will be 9)

## Examples

```
digitsAverage(246)  ==>  4

original: 2   4   6
           \ / \ /
1st iter:   3   5
             \ /
2nd iter:     4


digitsAverage(89)  ==>  9

original: 8   9
           \ /
1st iter:   9
```

p.s. for a bigger challenge, check out the [one line version](https://www.codewars.com/kata/one-line-task-digits-average) of this kata by myjinxin2015!

I tried it in Python, but could not do it. Can you solve it?
targets: import math
def digits_average(input):
    m = str(input)
    for i in range(len(str(input))-1):
        m = ''.join(str(math.ceil(int(m[i])/2 + int(m[i+1])/2)) for  i in range(len(m) - 1))
    return int(m)
inputs: I found an interesting problem on https://www.codewars.com/kata/5ecc1d68c6029000017d8aaf:
In this kata, your task is to find the maximum sum of any straight "beam" on a hexagonal grid, where its cell values are determined by a finite integer sequence seq.
In this context, a beam is a linear sequence of cells in any of the 3 pairs of opposing sides of a hexagon. We'll refer to the sum of a beam's integer values as the "beam value".Refer to the example below for further clarification.
Input
Your function will receive two arguments:

n : the length of each side of the hexagonal grid, where 2 <= n < 100
seq : a finite sequence of (positive and/or nonpositive) integers with a length >= 1The sequence is used to populate the cells of the grid and should be repeated as necessary.The sequence type will be dependent on the language (e.g. array for JavaScript, tuple for Python, etc.).

Output
Your function should return the largest beam value as an integer.
Test Example

In our test example, we have the following arguments:
n = 4
seq = [2, 4, 6, 8]

Below is the hexagonal grid determined by our example arguments;
the sequence repeats itself from left to right, then top to bottom.

   2 4 6 8
  2 4 6 8 2
 4 6 8 2 4 6
8 2 4 6 8 2 4
 6 8 2 4 6 8
  2 4 6 8 2
   4 6 8 2

The three diagrams below illustrate the "beams" in the hexagonal grid above.
In the grid on the left, the horizontal beams are highlighted by their likewise colors,
and the value of each beam is given to its right.
In the center grid, the beams highlighted go from upper-right to bottom-left (and vice-versa).
In the grid on the right, the beams highlighted go from upper-left to bottom-right (and vice-versa).

   2 4 6 8 -> 20           2 4 6 8           2 4 6 8
  2 4 6 8 2 -> 22         2 4 6 8 2         2 4 6 8 2
 4 6 8 2 4 6 -> 30       4 6 8 2 4 6       4 6 8 2 4 6
8 2 4 6 8 2 4 -> 34     8 2 4 6 8 2 4     8 2 4 6 8 2 4
 6 8 2 4 6 8 -> 34       6 8 2 4 6 8       6 8 2 4 6 8
  2 4 6 8 2 -> 22         2 4 6 8 2         2 4 6 8 2
   4 6 8 2 -> 20           4 6 8 2           4 6 8 2

The maximum beam value in our example is 34.

Test...

I tried it in Python, but could not do it. Can you solve it?
targets: from itertools import cycle

def max_hexagon_beam(n: int,seq: tuple):
    l=n*2-1                                  #the number of rows of the hexagon
    ll =[l-abs(n-i-1) for i in range(l)]     #the lengths of each row
    c=cycle(seq)
    hex = [[next(c) for i in range(j)] for j in ll]    # the hexagon
    sums = [sum(i)for i in hex] # the straight lines
    for index, i in enumerate(ll):
        start_row = [0, index%n + 1][index>=n]
        hex_row=[]
        hex_row2=[]
        for j in range(i):
            y=j+start_row #the y-axis or the row used
            x=index-[0, y%n + 1][y>=n] # the x-axis or the position in the row.
            hex_row.append(hex[y][x]) # the line going right-up
            hex_row2.append(hex[y][-1-x]) # the line going right-down
        sums +=[sum(hex_row), sum(hex_row2)]
    sums.sort()
    return sums[-1]
    
# I also considered extending the hexagon with edges of zeros so the x and y would have less possibility to go out of the list.
inputs: Solve in Python:
Well, those numbers were right and we're going to feed their ego.

Write a function, isNarcissistic, that takes in any amount of numbers and returns true if all the numbers are narcissistic. Return false for invalid arguments (numbers passed in as strings are ok).

For more information about narcissistic numbers (and believe me, they love it when you read about them) follow this link: https://en.wikipedia.org/wiki/Narcissistic_number
targets: is_narcissistic=lambda*a:all(n.isdigit()and int(n)==sum(int(d)**len(n)for d in n)for n in map(str,a))
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1144/E:
You are given two strings $s$ and $t$, both consisting of exactly $k$ lowercase Latin letters, $s$ is lexicographically less than $t$.

Let's consider list of all strings consisting of exactly $k$ lowercase Latin letters, lexicographically not less than $s$ and not greater than $t$ (including $s$ and $t$) in lexicographical order. For example, for $k=2$, $s=$"az" and $t=$"bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].

Your task is to print the median (the middle element) of this list. For the example above this will be "bc".

It is guaranteed that there is an odd number of strings lexicographically not less than $s$ and not greater than $t$.


-----Input-----

The first line of the input contains one integer $k$ ($1 \le k \le 2 \cdot 10^5$) — the length of strings.

The second line of the input contains one string $s$ consisting of exactly $k$ lowercase Latin letters.

The third line of the input contains one string $t$ consisting of exactly $k$ lowercase Latin letters.

It is guaranteed that $s$ is lexicographically less than $t$.

It is guaranteed that there is an odd number of strings lexicographically not less than $s$ and not greater than $t$.


-----Output-----

Print one string consisting exactly of $k$ lowercase Latin letters — the median (the middle element) of list of strings of length $k$ lexicographically not less than $s$ and not greater than $t$.


-----Examples-----
Input
2
az
bf

Output
bc

Input
5
afogk
asdji

Output
alvuw

Input
6
nijfvj
tvqhwp

Output
qoztvz

I tried it in Python, but could not do it. Can you solve it?
targets: import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools

sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)


def main():
    n = I()
    s = [ord(c)-ord('a') for c in S()]
    t = [ord(c)-ord('a') for c in S()]
    r = [s[i]+t[i] for i in range(n)]
    for i in range(n-1,0,-1):
        if r[i] >= 26:
            r[i] -= 26
            r[i-1] += 1
    k = 0
    rs = []
    for i in range(n):
        b = r[i] + k
        rs.append(chr(b // 2 + ord('a')))
        if b % 2 == 0:
            k = 0
        else:
            k = 26

    return ''.join(rs)


print(main())
inputs: Solve in Python:
In some other world, today is Christmas Eve.
There are N trees planted in Mr. Takaha's garden. The height of the i-th tree (1 \leq i \leq N) is h_i meters.
He decides to choose K trees from these trees and decorate them with electric lights. To make the scenery more beautiful, the heights of the decorated trees should be as close to each other as possible.
More specifically, let the height of the tallest decorated tree be h_{max} meters, and the height of the shortest decorated tree be h_{min} meters. The smaller the value h_{max} - h_{min} is, the better. What is the minimum possible value of h_{max} - h_{min}?

-----Constraints-----
 - 2 \leq K < N \leq 10^5
 - 1 \leq h_i \leq 10^9
 - h_i is an integer.

-----Input-----
Input is given from Standard Input in the following format:
N K
h_1
h_2
:
h_N

-----Output-----
Print the minimum possible value of h_{max} - h_{min}.

-----Sample Input-----
5 3
10
15
11
14
12

-----Sample Output-----
2

If we decorate the first, third and fifth trees, h_{max} = 12, h_{min} = 10 so h_{max} - h_{min} = 2. This is optimal.
targets: n,k = map(int,input().split())
s = []
for i in range(n):
    s.append(int(input()))
s.sort()
s.reverse()
ans = s[0]-s[-1]
for i in range(n-k+1):
    ans = min(ans,s[i]-s[i+k-1])
print(ans)