Muennighoff/xP3x-sample huggingface dataset:
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1342/B:
Let's say string $s$ has period $k$ if $s_i = s_{i + k}$ for all $i$ from $1$ to $|s| - k$ ($|s|$ means length of string $s$) and $k$ is the minimum positive integer with this property.

Some examples of a period: for $s$="0101" the period is $k=2$, for $s$="0000" the period is $k=1$, for $s$="010" the period is $k=2$, for $s$="0011" the period is $k=4$.

You are given string $t$ consisting only of 0's and 1's and you need to find such string $s$ that:  String $s$ consists only of 0's and 1's;  The length of $s$ doesn't exceed $2 \cdot |t|$;  String $t$ is a subsequence of string $s$;  String $s$ has smallest possible period among all strings that meet conditions 1—3. 

Let us recall that $t$ is a subsequence of $s$ if $t$ can be derived from $s$ by deleting zero or more elements (any) without changing the order of the remaining elements. For example, $t$="011" is a subsequence of $s$="10101".


-----Input-----

The first line contains single integer $T$ ($1 \le T \le 100$) — the number of test cases.

Next $T$ lines contain test cases — one per line. Each line contains string $t$ ($1 \le |t| \le 100$) consisting only of 0's and 1's.


-----Output-----

Print one string for each test case — string $s$ you needed to find. If there are multiple solutions print any one of them.


-----Example-----
Input
4
00
01
111
110

Output
00
01
11111
1010


-----Note-----

In the first and second test cases, $s = t$ since it's already one of the optimal solutions. Answers have periods equal to $1$ and $2$, respectively.

In the third test case, there are shorter optimal solutions, but it's okay since we don't need to minimize the string $s$. String $s$ has period equal to $1$.

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


def read_ints():
    return list(map(int, input().split(' ')))


t = read_int()
for case_num in range(t):
    a = input()
    n = len(a)
    one = 0
    for c in a:
        if c == '1':
            one += 1
    if one == 0 or one == n:
        print(a)
    else:
        print(''.join(['01' for i in range(n)]))
inputs: Solve in Python:
Zonal Computing Olympiad 2013, 10 Nov 2012

Little Red Riding Hood is carrying a basket with berries through the forest to her grandmother's house.  The forest is arranged in the form of a square N × N grid of cells.  The top left corner cell, where Little Red Riding Hood starts her journey, is numbered (1,1) and the bottom right corner cell, where her grandmother lives, is numbered (N,N).  In each step, she can move either one position right or one position down.

The forest is full of dangerous wolves and she is looking for a safe path to reach her destination.  Little Red Riding Hood's fairy godmother has placed some special anti-wolf magical charms in some of the cells in the grid.  Each charm has a strength.  If the charm in cell (i,j) has strength k then its zone of influence is all the cells within k steps of (i,j); that is, all cells (i',j') such that |i - i'| + |j - j'| ≤ k.  A cell within the zone of influence of a charm is safe from wolves.  A safe path from (1,1) to (N,N) is one in which every cell along the path is safe.

Little Red Riding Hood is carrying a basket with berries. In each cell, she drops some berries while pushing her way through the thick forest.  However, sometimes she is also able to pick up fresh berries.  Each cell is labelled with an integer that indicates the net change in the number of berries in her basket on passing through the cell; that is, the number of berries she picks up in that cell minus the number of berries she drops.  You can assume that there are enough berries in her basket to start with so that the basket never becomes empty.

Little Red Riding Hood knows the positions and strengths of all the magic charms and is looking for a safe path along which the number of berries she has in the basket when she reaches her grandmother's house is maximized.

As an example consider the following grid:

3  3  2  4  3 
2  1 -1 -2  2  
-1  2  4  3 -3  
-2  2  3  2  1  
3 -1  2 -1  2  

Suppose there are 3 magic charms, at position (1,2) with strength 2, at position (4,5)...
targets: def main():
	from sys import stdin, stdout
	rl = stdin.readline

	n, m = (int(x) for x in rl().split())
	grid = [[int(x) for x in rl().split()] for _ in range(n)]
	charms = [[int(x) for x in rl().split()] for _ in range(m)]

	n_inf = -9999
	dp = [[n_inf] * n for _ in range(n)]

	for c in charms:
		c[0] -= 1
		c[1] -= 1

		x = max(0, c[0] - c[2])
		while x < n and x <= c[0] + c[2]:
			off = abs(x - c[0])
			y = max(0, c[1] - c[2] + off)
			while y < n and y <= c[1] + c[2] - off:
				dp[x][y] = grid[x][y]
				y += 1
			x += 1

	for x in range(1, n):
		if dp[x - 1][0] == n_inf:
			for x1 in range(x, n):
				dp[x1][0] = n_inf
			break
		dp[x][0] += dp[x - 1][0]

	for y in range(1, n):
		if dp[0][y - 1] == n_inf:
			for y1 in range(y, n):
				dp[0][y1] = n_inf
			break
		dp[0][y] += dp[0][y - 1]

	for x in range(1, n):
		for y in range(1, n):
			m = max(dp[x - 1][y], dp[x][y - 1])
			dp[x][y] = m + dp[x][y] if m != n_inf else n_inf

	if dp[-1][-1] != n_inf:
		stdout.write('YES\n')
		stdout.write(str(dp[-1][-1]))
	else:
		stdout.write('NO')


main()
inputs: I found an interesting problem on https://www.codewars.com/kata/58a6a56942fd72afdd000161:
# Task
 In a black and white image we can use `1` instead of black pixels and `0` instead of white pixels. 
 
 For compression image file we can reserve pixels by consecutive pixels who have the same color. 
 
 Your task is to determine how much of black and white pixels is in each row sorted by place of those.
 
# Example:

 For `height=2,width=100 and compressed=[40,120,40,0]`

 The result should be `[[40,60],[0,60,40,0]]`
 
 Sum of compressed array always divisible by width and height of image.
 
 Pixels available is in Compressed array in this order:
 
 `[Black,White,Black,White,Black,White,...]`
 
 ![](http://s3.picofile.com/file/8194350650/Compressed_Image_1_.png)

 For `height=2,width=100 and compressed=[0, 180, 20, 0]`

 The result should be `[[0,100],[0,80,20,0]]`

 ![](http://s6.picofile.com/file/8194348868/Compressed_Image.png)


# Input/Output


- `[input]` integer `height`

  Height of image


- `[input]` integer `width`

  Width of image


- `[input]` integer array `compressed`

  Consecutive pixels


- `[output]` 2D integer array

 The black and white pixels in each row.

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

def black_and_white(height, width, compressed):
    
    compr = ''.join(('0' if i % 2 else '1') * v for i, v in enumerate(compressed))
    
    res = []
    
    for i in range(0, len(compr), width): 
        row, temp, black = compr[i: i + width], [], True
        
        m = [(c, len(g)) for g, c in findall(r'((.)\2*)', row)]
        for dig, n in m:
            if black and dig == '0':
                temp.extend([0, n])
                black == False
            elif not black and dig == '1':
                temp.extend([n, 0])
                black = True
            else: 
                temp.append(n)
                black = not black
        
        if len(temp) % 2: temp.append(0)
        res.append(temp)
    
    return res
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/646/A:
Три брата договорились о встрече. Пронумеруем братьев следующим образом: пусть старший брат имеет номер 1, средний брат имеет номер 2, а младший брат — номер 3. 

Когда пришло время встречи, один из братьев опоздал. По заданным номерам двух братьев, которые пришли вовремя, вам предстоит определить номер опоздавшего брата.


-----Входные данные-----

В первой строке входных данных следуют два различных целых числа a и b (1 ≤ a, b ≤ 3, a ≠ b) — номера братьев, которые пришли на встречу вовремя. Номера даны в произвольном порядке.


-----Выходные данные-----

Выведите единственное целое число — номер брата, который опоздал на встречу.


-----Пример-----
Входные данные
3 1

Выходные данные
2

I tried it in Python, but could not do it. Can you solve it?
targets: a, b = [int(x) for x in input().split()]
for i in range(1, 4):
    if i != a and i != b:
        print(i)
inputs: I found an interesting problem on https://www.codechef.com/LTIME39/problems/ALPHABET:
Not everyone probably knows that Chef has younder brother Jeff. Currently Jeff learns to read.
He knows some subset of the letter of Latin alphabet. In order to help Jeff to study, Chef gave him a book with the text consisting of N words. Jeff can read a word iff it consists only of the letters he knows.
Now Chef is curious about which words his brother will be able to read, and which are not. Please help him!

-----Input-----
The first line of the input contains a lowercase Latin letter string S, consisting of the letters Jeff can read. Every letter will appear in S no more than once.
The second line of the input contains an integer N denoting the number of words in the book.
Each of the following N lines contains a single lowecase Latin letter string Wi, denoting the ith word in the book.

-----Output-----
For each of the words, output "Yes" (without quotes) in case Jeff can read it, and "No" (without quotes) otherwise.

-----Constraints-----
- 1 ≤ |S| ≤ 26
- 1 ≤ N ≤ 1000
- 1 ≤ |Wi| ≤ 12
- Each letter will appear in S no more than once.
- S, Wi consist only of lowercase Latin letters.

-----Subtasks-----
- Subtask #1 (31 point): |S| = 1, i.e. Jeff knows only one letter.
- Subtask #2 (69 point)	: no additional constraints

-----Example-----
Input:act
2
cat
dog

Output:Yes
No

-----Explanation-----
The first word can be read.
The second word contains the letters d, o and g that aren't known by Jeff.

I tried it in Python, but could not do it. Can you solve it?
targets: S = input()
N = int(input())
while N != 0:
 W = input()
 flag = 0
 for i in range(0, len(W)):
  if W[i] in S:
   continue
  else:
   flag = 1
   break
 if flag == 0:
  print("Yes")
 else:
  print("No")
 N -= 1
inputs: I found an interesting problem on https://www.codewars.com/kata/559379505c859be5a9000034:
< PREVIOUS KATA
NEXT KATA >

## Task:

You have to write a function `pattern` which returns the following Pattern(See Examples) upto desired number of rows. 

* Note:`Returning` the pattern is not the same as `Printing` the pattern.

## Parameters:
    
    pattern(        n        ,        y        );
                    ^                 ^         
                    |                 |         
             Term upto which   Number of times  
              Basic Pattern     Basic Pattern   
                should be         should be     
                 created          repeated      
                                 vertically    
    
* Note: `Basic Pattern` means what we created in Complete The Pattern #12 i.e. a `simple X`.

## Rules/Note:

* The pattern should be created using only unit digits.
* If `n < 1` then it should return "" i.e. empty string.
* If `y <= 1` then the basic pattern should not be repeated vertically.
* `The length of each line is same`, and is equal to the length of longest line in the pattern.
* Range of Parameters (for the sake of CW Compiler) :
  + `n ∈ (-∞,50]`
  + `y ∈ (-∞,25]`
* If only one argument is passed then the function `pattern` should run as if `y <= 1`.
* The function `pattern` should work when extra arguments are passed, by ignoring the extra arguments.
   
   
## Examples:

#### Having Two Arguments:

##### pattern(4,3):

        1     1
         2   2 
          3 3  
           4   
          3 3  
         2   2 
        1     1
         2   2 
          3 3  
           4   
          3 3  
         2   2 
        1     1
         2   2 
          3 3  
           4   
          3 3  
         2   2 
        1     1
       
##### pattern(10,2):

        1                 1
         2               2 
          3             3  
           4           4   
            5         5    
             6       6     
              7     7      
               8   8       
                9 9        
                 0         
                9 9        
        ...

I tried it in Python, but could not do it. Can you solve it?
targets: def pattern(n, y=1, *_):
    if n < 1:
        return ''
    ans, p, y = [], [], max(y, 1)
    for i in range(1, n+1):
        t = [' '] * (n * 2 - 1)
        t[i - 1] = t[n * 2 - 1 - 1 - i + 1] = str(i % 10)
        p.append(''.join(t))
    p += p[:-1][::-1]
    for _ in range(y):
        ans += p[1:] if ans else p
    return '\n'.join(ans)
inputs: I found an interesting problem on https://leetcode.com/problems/minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix/:
Given a m x n binary matrix mat. In one step, you can choose one cell and flip it and all the four neighbours of it if they exist (Flip is changing 1 to 0 and 0 to 1). A pair of cells are called neighboors if they share one edge.
Return the minimum number of steps required to convert mat to a zero matrix or -1 if you cannot.
Binary matrix is a matrix with all cells equal to 0 or 1 only.
Zero matrix is a matrix with all cells equal to 0.
 
Example 1:

Input: mat = [[0,0],[0,1]]
Output: 3
Explanation: One possible solution is to flip (1, 0) then (0, 1) and finally (1, 1) as shown.

Example 2:
Input: mat = [[0]]
Output: 0
Explanation: Given matrix is a zero matrix. We don't need to change it.

Example 3:
Input: mat = [[1,1,1],[1,0,1],[0,0,0]]
Output: 6

Example 4:
Input: mat = [[1,0,0],[1,0,0]]
Output: -1
Explanation: Given matrix can't be a zero matrix

 
Constraints:

m == mat.length
n == mat[0].length
1 <= m <= 3
1 <= n <= 3
mat[i][j] is 0 or 1.

I tried it in Python, but could not do it. Can you solve it?
targets: class Solution:
    def minFlips(self, mat: List[List[int]]) -> int:
        m = len(mat)
        n = len(mat[0])
        
        start = sum(val << (i*n + j) for i, row in enumerate(mat) for j, val in enumerate(row))
        
        queue = collections.deque([(start, 0)])
        seen = { start }
        
        dirs = [[0, 0], [0,1], [1, 0], [0, -1], [-1, 0]]
        while queue:
            # print(queue)
            current, d = queue.popleft()
            if current == 0:
                return d
            
            # for each index in matrix find neighbour
            for i in range(len(mat)):
                for j in range(len(mat[0])):
                    next_state = current
                    
                    # importants dirs has [0, 0] we need flip the current element and neigbour
                    for dir_ in dirs:
                        new_i = i + dir_[0]
                        new_j = j + dir_[1]
                        
                        if new_i >= 0 and new_i < len(mat) and new_j >= 0 and new_j < len(mat[0]):
                            next_state ^= (1 << (new_i * n + new_j )) # 0 xor 1 = 1, 1 xor 1 = 0
                    
                    if next_state not in seen:
                        seen.add(next_state)
                        queue.append((next_state, d + 1))
        
        return -1
inputs: Solve in Python:
Mahmoud and Ehab play a game called the even-odd game. Ehab chooses his favorite integer n and then they take turns, starting from Mahmoud. In each player's turn, he has to choose an integer a and subtract it from n such that:  1 ≤ a ≤ n.  If it's Mahmoud's turn, a has to be even, but if it's Ehab's turn, a has to be odd. 

If the current player can't choose any number satisfying the conditions, he loses. Can you determine the winner if they both play optimally?


-----Input-----

The only line contains an integer n (1 ≤ n ≤ 10^9), the number at the beginning of the game.


-----Output-----

Output "Mahmoud" (without quotes) if Mahmoud wins and "Ehab" (without quotes) otherwise.


-----Examples-----
Input
1

Output
Ehab
Input
2

Output
Mahmoud


-----Note-----

In the first sample, Mahmoud can't choose any integer a initially because there is no positive even integer less than or equal to 1 so Ehab wins.

In the second sample, Mahmoud has to choose a = 2 and subtract it from n. It's Ehab's turn and n = 0. There is no positive odd integer less than or equal to 0 so Mahmoud wins.
targets: def li():
    return list(map(int, input().split()))
n = int(input())
if n % 2 == 0:
    print('Mahmoud')
else:
    print('Ehab')
inputs: Solve in Python:
Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.

 [Image] 

The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.


-----Input-----

The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.

The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.

The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.


-----Output-----

If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".


-----Examples-----
Input
3 3
><>
v^v

Output
NO

Input
4 6
<><>
v^v^v^

Output
YES



-----Note-----

The figure above shows street directions in the second sample test case.
targets: def main():
    n, m = map(int, input().split())
    nm = n * m
    neigh = [[] for _ in range(nm)]
    for y, c in enumerate(input()):
        for x in range(y * m + 1, y * m + m):
            if c == '<':
                neigh[x].append(x - 1)
            else:
                neigh[x - 1].append(x)
    for x, c in enumerate(input()):
        for y in range(m + x, nm, m):
            if c == '^':
                neigh[y].append(y - m)
            else:
                neigh[y - m].append(y)

    def dfs(yx):
        l[yx] = False
        for yx1 in neigh[yx]:
            if l[yx1]:
                dfs(yx1)

    for i in range(nm):
        l = [True] * nm
        dfs(i)
        if any(l):
            print('NO')
            return
    print('YES')


def __starting_point():
    main()
__starting_point()
inputs: Solve in Python:
Hi guys, welcome to introduction to DocTesting.

The kata is composed of two parts; in part (1) we write three small functions, and in part (2) we write a few doc tests for those functions.

Lets talk about the functions first...

    The reverse_list function takes a list and returns the reverse of it.
    If given an empty list, simply return an empty list.

The second function...

    The sum_list function takes a list as input and adds up all the values, 
    returning an integer. If the list is empty, return 0.

The third function...

    The head_of_list function simply returns the first item in the list.
    If the list is empty return None.

Each of these functions can be easily written with a single line of code; there are some tests for correctness but no tests for effciency.

Once you have implemented all three of these functions you can move onto phase two, which is writing doc tests. If you haven't written doc tests before then I suggest you check out the following documentation: 
https://docs.python.org/3/library/doctest.html

To complete this kata all you have to do is write **EXACTLY TWO** doc tests for each of the three functions (any more/less than that and you will fail the tests).

Here is an example:

    def double(y):
      """Function returns y * 2
      >>> double(2)
      4
      """
      return y * 2

In the example above we have a function called 'double' and a single doctest. When we run the doctest module Python will check if double(2) equals 4. If it does not, then the doctest module will flag up an error.

Please note that this is intended as a beginners introduction to docstrings, if you try to do something clever (such as writing doc tests to catch exceptions, or tinkering with the 'option flags'), you will probably fail a test or two. This is due to how the tests are written.  

Oh and one last thing, don't try and get too 'cheeky' and try something like:
    
    """
    >>> True
    True
    """

such a solution is (a) not in the spirit of things and (b) I got tests for...
targets: def reverse_list(a):
    """Takes an list and returns the reverse of it. 
    If x is empty, return [].
    
    >>> reverse_list([1])
    [1]
    >>> reverse_list([])
    []
    """
    return a[::-1]

def sum_list(a):
    """Takes a list, and returns the sum of that list.
    If x is empty list, return 0.

    >>> sum_list([1])
    1
    >>> sum_list([])
    0
    """
    return sum(a)

def head_of_list(a):
    """Takes a list, returns the first item in that list.
    If x is empty, return None

    >>> head_of_list([1])
    1
    >>> head_of_list([]) is None
    True
    """
    if a:
        return a[0]
inputs: Solve in Python:
Given a string, return the minimal number of parenthesis reversals needed to make balanced parenthesis. 

For example:
```Javascript
solve(")(") = 2 Because we need to reverse ")" to "(" and "(" to ")". These are 2 reversals. 
solve("(((())") = 1 We need to reverse just one "(" parenthesis to make it balanced.
solve("(((") = -1 Not possible to form balanced parenthesis. Return -1.
```

Parenthesis will be either `"("` or `")"`. 

More examples in the test cases. 

Good luck.
targets: class Stack:
    def __init__(self):
        self.items = []

    def is_empty(self):
        return self.items == []

    def push(self, item):
        self.items.append(item)

    def pop(self):
        return self.items.pop()

    def peek(self):
        return self.items[-1]

    def size(self):
        return len(self.items)


def solve(s):
    l = len(s)
    if l % 2 != 0: return -1
    stack = Stack(); count = 0; i = 0
    while i < len(s):
        if s[i] == "(":
            stack.push(s[i])
        else:
            if stack.is_empty():
                count += 1
            else: stack.pop()
        i += 1
    q = (count + stack.size())//2
    return q if count % 2 == 0 and stack.size() % 2 == 0 else q + 1
inputs: Solve in Python:
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
 - A level-0 burger is a patty.
 - A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like BPPPB and BBPPPBPBPPPBB (rotated 90 degrees), where B and P stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?

-----Constraints-----
 - 1 \leq N \leq 50
 - 1 \leq X \leq ( the total number of layers in a level-N burger )
 - N and X are integers.

-----Input-----
Input is given from Standard Input in the following format:
N X

-----Output-----
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.

-----Sample Input-----
2 7

-----Sample Output-----
4

There are 4 patties in the bottom-most 7 layers of a level-2 burger (BBPPPBPBPPPBB).
targets: n,x=map(int,input().split())

def func(m,y):
  if y==0:return 0
  ret=0
  t=4*pow(2,m)-3
  if y>=t-1:
    ret=pow(2,m+1)-1
  elif y==(t-1)//2:
    ret=pow(2,m)-1
  elif y==(t-1)//2+1:
    ret=pow(2,m)
  elif y>(t-1)//2+1:
    ret=pow(2,m)
    ret+=func(m-1,y-(t-1)//2-1)
  else:
    ret=func(m-1,y-1)
  return ret

print(func(n,x))
inputs: Solve in Python:
An Ironman Triathlon is one of a series of long-distance triathlon races organized by the World Triathlon Corporaion (WTC).
It consists of a 2.4-mile swim, a 112-mile bicycle ride and a marathon (26.2-mile) (run, raced in that order and without a break. It hurts... trust me.

Your task is to take a distance that an athlete is through the race, and return one of the following:

If the distance is zero, return `'Starting Line... Good Luck!'`.

If the athlete will be swimming, return an object with `'Swim'` as the key, and the remaining race distance as the value.

If the athlete will be riding their bike, return an object with `'Bike'` as the key, and the remaining race distance as the value.

If the athlete will be running, and has more than 10 miles to go, return an object with `'Run'` as the key, and the remaining race distance as the value.

If the athlete has 10 miles or less to go, return return an object with `'Run'` as the key, and `'Nearly there!'` as the value.

Finally, if the athlete has completed te distance, return `"You're done! Stop running!"`.

All distance should be calculated to two decimal places.
targets: def i_tri(s):
    out = 'Starting Line... Good Luck!'
    if s>0.0    and s<=2.4   : out = {'Swim': f'{140.6 - s:.2f} to go!'}
    if s>2.4    and s<=114.4 : out = {'Bike': f'{140.6 - s:.2f} to go!'}
    if s>114.4  and s<130.6  : out = {'Run' : f'{140.6 - s:.2f} to go!'}
    if s>=130.6 and s<140.6  : out = {'Run' : 'Nearly there!'}
    if s>140.6               : out = "You're done! Stop running!"
    return out
inputs: Solve in Python:
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it?

The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper.

There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.


-----Input-----

The first line contains a non-empty name that needs to be checked. The name contains at most 10^5 large English letters. The name will be written with the next sans serif font: $\text{ABCDEFGHI JKLMNOPQRSTUVWXYZ}$


-----Output-----

Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).


-----Examples-----
Input
AHA

Output
YES

Input
Z

Output
NO

Input
XO

Output
NO
targets: good = set('AHIMOTUVWXY')

in_str = input("")

str_len = len(in_str)

result = ""

i = 0

while(i < str_len/2):
    if((in_str[i] in good) and (in_str[i] == in_str[-1-i])):
        i += 1
    else:
        result = "NO"
        break

if(result != "NO"):
    result = "YES"

print(result)
inputs: I found an interesting problem on https://www.codewars.com/kata/5b7bae02402fb702ce000159:
Assume we take a number `x` and perform any one of the following operations:
```Pearl
a) Divide x by 3 (if it is divisible by 3), or
b) Multiply x by 2
```
After each operation, we write down the result. If we start with `9`, we can get a sequence such as:
```
[9,3,6,12,4,8] -- 9/3=3 -> 3*2=6 -> 6*2=12 -> 12/3=4 -> 4*2=8
```
You will be given a shuffled sequence of integers and your task is to reorder them so that they conform to the above sequence. There will always be an answer. 
```
For the above example:
solve([12,3,9,4,6,8]) = [9,3,6,12,4,8].
```

More examples in the test cases. Good luck!

I tried it in Python, but could not do it. Can you solve it?
targets: def solve(lst):
    return sorted(lst, key=factors_count)
    
def factors_count(n):
    result = []
    for k in (2, 3):
        while n % k == 0:
            n //= k
            result.append(k)
    return -result.count(3), result.count(2)
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1430/E:
You are given a string $s$. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string. 

Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.


-----Input-----

The first line contains one integer $n$ ($2 \le n \le 200\,000$) — the length of $s$.

The second line contains $s$ — a string consisting of $n$ lowercase Latin letters.


-----Output-----

Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.


-----Examples-----
Input
5
aaaza

Output
2

Input
6
cbaabc

Output
0

Input
9
icpcsguru

Output
30



-----Note-----

In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.

Since the string in the second example is a palindrome, you don't have to do anything to reverse it.

I tried it in Python, but could not do it. Can you solve it?
targets: N = int(input())
S = input()
T = S[::-1]

from collections import defaultdict
dic = defaultdict(lambda: [])
for i,t in enumerate(T):
    dic[t].append(i)
for k in list(dic.keys()):
    dic[k].reverse()

arr = []
for c in S:
    arr.append(dic[c].pop())

def inversion(inds):
    bit = [0] * (N+1)
    def bit_add(x,w):
        while x <= N:
            bit[x] += w
            x += (x & -x)
    def bit_sum(x):
        ret = 0
        while x > 0:
            ret += bit[x]
            x -= (x & -x)
        return ret
    inv = 0
    for ind in reversed(inds):
        inv += bit_sum(ind + 1)
        bit_add(ind + 1, 1)
    return inv

print(inversion(arr))
inputs: I found an interesting problem on https://www.codewars.com/kata/588e27b7d1140d31cb000060:
Write a function `generatePairs` that accepts an integer argument `n` and generates an array containing the pairs of integers `[a, b]` that satisfy the following conditions:
```
0 <= a <= b <= n
```

The pairs should be sorted by increasing values of `a` then increasing values of `b`.

For example, `generatePairs(2)` should return
```
[ [0, 0], [0, 1], [0, 2], [1, 1], [1, 2], [2, 2] ]
```

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


def generate_pairs(n):
    return [list(a) for a in combinations_with_replacement(range(n + 1), 2)]
inputs: I found an interesting problem on https://www.codewars.com/kata/55b6a3a3c776ce185c000021:
*You are a composer who just wrote an awesome piece of music. Now it's time to present it to a band that will perform your piece, but there's a problem! The singers vocal range doesn't stretch as your piece requires, and you have to transpose the whole piece.*

# Your task
Given a list of notes (represented as strings) and an interval, output a list of transposed notes in *sharp* notation.

**Input notes may be represented both in flat and sharp notations (more on that below).**

**For this kata, assume that input is always valid and the song is *at least* 1 note long.**

**Assume that interval is an integer between -12 and 12.**

# Short intro to musical notation
Transposing a single note means shifting its value by a certain interval.

The notes are as following:

    A, A#, B, C, C#, D, D#, E, F, F#, G, G#.

This is using *sharp* notation, where '#' after a note means that it is one step higher than the note. So A# is one step higher than A.

An alternative to sharp notation is the *flat* notation:

    A, Bb, B, C, Db, D, Eb, E, F, Gb, G, Ab.

The 'b' after a note means that it is one step lower than the note.

# Examples

    ['G'] -> 5 steps -> ['C']
    ['Db'] -> -4 steps -> ['A#']
    ['E', 'F'] -> 1 step -> ['F', 'F#']

I tried it in Python, but could not do it. Can you solve it?
targets: sharp = ['A', 'A#', 'B', 'C', 'C#', 'D', 'D#', 'E', 'F', 'F#', 'G', 'G#']
flat = ['A', 'Bb', 'B', 'C', 'Db', 'D', 'Eb', 'E', 'F', 'Gb', 'G', 'Ab']
def transpose(song, interval):
    return [sharp[((sharp.index(note) if note in sharp else flat.index(note)) + interval) % 12] for note in song]
inputs: I found an interesting problem on https://www.codechef.com/COOK23/problems/NOKIA:
To protect people from evil, 
a long and tall wall was constructed a few years ago. 
But just a wall is not safe, there should also be soldiers on it, 
always keeping vigil. 
The wall is very long and connects the left and the right towers. 
There are exactly N spots (numbered 1 to N) on the wall for soldiers. 
The Kth spot is K miles far from the left tower and (N+1-K) miles from the right tower.

Given a permutation of spots P of {1, 2, ..., N}, soldiers occupy the N spots in that order. 
The P[i]th spot is occupied before the P[i+1]th spot. 
When a soldier occupies a spot, he is connected to his nearest soldier already placed to his left. 
If there is no soldier to his left, he is connected to the left tower. The same is the case with right side. 
A connection between two spots requires a wire of length equal to the distance between the two.

The realm has already purchased a wire of M miles long from Nokia, 
possibly the wire will be cut into smaller length wires. 
As we can observe, the total length of the used wire depends on the permutation of the spots P. Help the realm in minimizing the length of the unused wire. If there is not enough wire, output -1.

-----Input-----
First line contains an integer T (number of test cases, 1 ≤ T ≤ 10 ). Each of the next T lines contains two integers N M, as explained in the problem statement (1 ≤ N ≤ 30 , 1 ≤ M ≤ 1000).

-----Output-----
For each test case, output the minimum length of the unused wire, or -1 if the the wire is not sufficient.

-----Example-----
Input:
4
3 8
3 9
2 4
5 25

Output:
0
0
-1
5

Explanation:

In the 1st case, for example, the permutation P = {2, 1, 3} will use the exact 8 miles wires in total.

In the 2nd case, for example, the permutation P = {1, 3, 2} will use the exact 9 miles wires in total.

To understand the first two cases, you can see the following figures:



In the 3rd case, the minimum length of wire required is 5, for any of the permutations {1,2} or {2,1}, so length 4 is not sufficient.

In the 4th case, for the permutation...

I tried it in Python, but could not do it. Can you solve it?
targets: import sys
first = True
memmin = {}
memmax = {}

def fmax(n):
 if n == 1: return 2
 if n == 0: return 0
 if n in memmax: return memmax[n]

 res = 0
 for i in range(n):
  cur = i + 1 + n - i + fmax(i) + fmax(n-i-1)
  if cur > res: res = cur

 memmax[n] = res
 return res

def fmin(n):
 if n == 1: return 2
 if n == 0: return 0
 if n in memmin: return memmin[n]

 res = 10 ** 9
 for i in range(n):
  cur = i + 1 + n - i + fmin(i) + fmin(n-i-1)
  if cur < res: res = cur

 memmin[n] = res
 return res 

for line in sys.stdin:
 if first:
  first = False
  tc = int(line)
  continue

 tc -= 1
 if tc < 0: break

 n, m = list(map(int, line.split()))

 val1 = fmin(n)
 val2 = fmax(n)

 if m < val1: print(-1)
 else:
  if m > val2: print(m - val2)
  else: print(0)
inputs: I found an interesting problem on https://www.codewars.com/kata/5912701a89fc3d0a6a000169:
# Task
Given array of integers `sequence` and some integer `fixedElement`, output the number of `even` values in sequence before the first occurrence of `fixedElement` or `-1` if and only if `fixedElement` is not contained in sequence.



# Input/Output


`[input]` integer array `sequence`

A non-empty array of positive integers.

`4 ≤ sequence.length ≤ 100`

`1 ≤ sequence[i] ≤ 9`

`[input]` integer `fixedElement`

An positive integer

`1 ≤ fixedElement ≤ 9`

`[output]` an integer


# Example

For `sequence = [1, 4, 2, 6, 3, 1] and fixedElement = 6`, the output should be `2`.

There are `2` even numbers before `6`: `4 and 2`

For `sequence = [2, 2, 2, 1] and fixedElement = 3`, the output should be `-1`.

There is no `3` appears in `sequence`. So returns `-1`.

For `sequence = [1, 3, 4, 3] and fixedElement = 3`, the output should be `0`.

`3` appears in `sequence`, but there is no even number before `3`.

I tried it in Python, but could not do it. Can you solve it?
targets: def even_numbers_before_fixed(sequence, fixed_element):
    try:
        return sum(elem % 2 == 0 for elem in sequence[:sequence.index(fixed_element)])
    except ValueError:
        return -1
inputs: Solve in Python:
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one. $0$ 

You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.

Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.

Note that the final necklace should remain as one circular part of the same length as the initial necklace.


-----Input-----

The only line of input contains a string $s$ ($3 \leq |s| \leq 100$), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.


-----Output-----

Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".

You can print each letter in any case (upper or lower).


-----Examples-----
Input
-o-o--
Output
YES
Input
-o---

Output
YES
Input
-o---o-

Output
NO
Input
ooo

Output
YES
targets: s=input()
b=s.count('o')
n=s.count('-')
if n==0 or b==0:
    print('YES')
else:
    if n%b==0:
        print('YES')
    else:
        print('NO')
inputs: Solve in Python:
Let S be the concatenation of 10^{10} copies of the string 110. (For reference, the concatenation of 3 copies of 110 is 110110110.)
We have a string T of length N.
Find the number of times T occurs in S as a contiguous substring.

-----Constraints-----
 - 1 \leq N \leq 2 \times 10^5
 - T is a string of length N consisting of 0 and 1.

-----Input-----
Input is given from Standard Input in the following format:
N
T

-----Output-----
Print the number of times T occurs in S as a contiguous substring.

-----Sample Input-----
4
1011

-----Sample Output-----
9999999999

S is so long, so let us instead count the number of times 1011 occurs in the concatenation of 3 copies of 110, that is, 110110110. We can see it occurs twice:
 - 1 1011 0110
 - 1101 1011 0
targets: n=int(input())
s='110'
t=input()
sn=s*((n+2)//3+1)
ret=0
for i in range(3):
    if sn[i:i+n]==t:
        ret += pow(10,10)-(i+n-1)//3
print(ret)
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/805/B:
In the beginning of the new year Keivan decided to reverse his name. He doesn't like palindromes, so he changed Naviek to Navick.

He is too selfish, so for a given n he wants to obtain a string of n characters, each of which is either 'a', 'b' or 'c', with no palindromes of length 3 appearing in the string as a substring. For example, the strings "abc" and "abca" suit him, while the string "aba" doesn't. He also want the number of letters 'c' in his string to be as little as possible.


-----Input-----

The first line contains single integer n (1 ≤ n ≤ 2·10^5) — the length of the string.


-----Output-----

Print the string that satisfies all the constraints.

If there are multiple answers, print any of them.


-----Examples-----
Input
2

Output
aa

Input
3

Output
bba



-----Note-----

A palindrome is a sequence of characters which reads the same backward and forward.

I tried it in Python, but could not do it. Can you solve it?
targets: n=int(input())
c=n // 4
if n%4 == 1:
    print('aabb' * c,'a',sep='')
if n%4 == 2:
    print('aabb' * c,'aa',sep='')
if n%4 == 3:
    print('aabb' * c,'aab',sep='')
if n%4 == 0:
    print('aabb' * c)
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1374/D:
You are given an array $a$ consisting of $n$ positive integers.

Initially, you have an integer $x = 0$. During one move, you can do one of the following two operations:  Choose exactly one $i$ from $1$ to $n$ and increase $a_i$ by $x$ ($a_i := a_i + x$), then increase $x$ by $1$ ($x := x + 1$).  Just increase $x$ by $1$ ($x := x + 1$). 

The first operation can be applied no more than once to each $i$ from $1$ to $n$.

Your task is to find the minimum number of moves required to obtain such an array that each its element is divisible by $k$ (the value $k$ is given).

You have to answer $t$ independent test cases. 


-----Input-----

The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.

The first line of the test case contains two integers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5; 1 \le k \le 10^9$) — the length of $a$ and the required divisior. The second line of the test case 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 $a$.

It is guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$ ($\sum n \le 2 \cdot 10^5$).


-----Output-----

For each test case, print the answer — the minimum number of moves required to obtain such an array that each its element is divisible by $k$.


-----Example-----
Input
5
4 3
1 2 1 3
10 6
8 7 1 8 3 7 5 10 8 9
5 10
20 100 50 20 100500
10 25
24 24 24 24 24 24 24 24 24 24
8 8
1 2 3 4 5 6 7 8

Output
6
18
0
227
8



-----Note-----

Consider the first test case of the example:  $x=0$, $a = [1, 2, 1, 3]$. Just increase $x$;  $x=1$, $a = [1, 2, 1, 3]$. Add $x$ to the second element and increase $x$;  $x=2$, $a = [1, 3, 1, 3]$. Add $x$ to the third element and increase $x$;  $x=3$, $a = [1, 3, 3, 3]$. Add $x$ to the fourth element and increase $x$;  $x=4$, $a = [1, 3, 3, 6]$. Just increase $x$;  $x=5$, $a = [1, 3, 3, 6]$. Add $x$ to the first element and increase $x$;  $x=6$, $a = [6, 3, 3, 6]$. We obtained the required array. 

Note...

I tried it in Python, but could not do it. Can you solve it?
targets: t = int(input())
from collections import Counter

for case in range(t):
    n, k = list(map(int, input().split()))
    a = [int(x) for x in input().split()]
    w = Counter(x % k for x in a)
    v = 0
    for x, freq in list(w.items()):
        if x == 0: continue
        if freq == 0: continue
        
        r = (-x) % k
        v = max(v, r + (freq-1)*k+1)

    print(v)
inputs: I found an interesting problem on https://www.codechef.com/BRBG2020/problems/FTNUM:
A prime number is number x which has only divisors as 1 and x itself.
Harsh is playing a game with his friends, where his friends give him a few numbers claiming that they are divisors of some number x but divisor 1 and the number x itself are not being given as divisors.
You need to help harsh find which number's divisors are given here.
His friends can also give him wrong set of divisors as a trick question for which no number exists.
Simply, We are given the divisors of a number x ( divisors except 1 and x itself ) , you have to print the number if only it is possible.
You have to answer t queries.
(USE LONG LONG TO PREVENT OVERFLOW)

-----Input:-----
- First line is T queires.
- Next are T queries.
- First line is N ( No of divisors except 1 and the number itself )
- Next line has N integers or basically the divisors.

-----Output:-----
Print the minimum possible x which has such divisors and print -1 if not possible.

-----Constraints-----
- 1<= T <= 30
- 1<= N <= 350
- 2<= Di <=10^6

-----Sample Input:-----
3
2
2 3
2
4 2
3
12 3 2

-----Sample Output:-----
6
8
-1

-----EXPLANATION:-----
Query 1 : Divisors of 6 are ( 1,2,3,6) Therefore, Divisors except 1 and the number 6 itself are ( 2 , 3). Thus, ans = 6.
Query 2 : Divisors of 8 are ( 1,2,4,8) Therefore, Divisors except 1 and the number 8 itself are ( 2 , 4). Thus, ans = 8.
Query 3 : There is no such number x with only ( 1,2,3,12,x ) as the divisors.

I tried it in Python, but could not do it. Can you solve it?
targets: # cook your dish here
import math
 
 
def findX(list, int):
    # Sort the given array
    list.sort()
 
    # Get the possible X
    x = list[0]*list[int-1]
 
    # Container to store divisors
    vec = []
 
    # Find the divisors of x
    i = 2
    while(i * i <= x):
        # Check if divisor
        if(x % i == 0):
            vec.append(i)
            if ((x//i) != i):
                vec.append(x//i)
        i += 1
 
    # sort the vec because a is sorted
        # and we have to compare all the elements
    vec.sort()
    # if size of both vectors is not same
    # then we are sure that both vectors
    # can't be equal
    if(len(vec) != int):
        return -1
    else:
        # Check if a and vec have
        # same elements in them
        j = 0
        for it in range(int):
            if(list[j] != vec[it]):
                return -1
            else:
                j += 1
    return x
 
 
# Driver code
try:
    
    m = int(input())
    for i in range (m):
        n = int(input())
        x = list(map(int, input().split()))
 

        print(findX(x, n))
except EOFError as e: 
    print(e)
inputs: I found an interesting problem on https://www.codewars.com/kata/58b3c2bd917a5caec0000017:
# Task

Given an array of integers, sum consecutive even numbers and consecutive odd numbers. Repeat the process while it can be done and return the length of the final array.

# Example

For `arr = [2, 1, 2, 2, 6, 5, 0, 2, 0, 5, 5, 7, 7, 4, 3, 3, 9]`
 
The result should be `6`.

```
[2, 1, 2, 2, 6, 5, 0, 2, 0, 5, 5, 7, 7, 4, 3, 3, 9]  -->
         2+2+6       0+2+0     5+5+7+7       3+3+9
[2, 1,   10,    5,    2,        24,     4,   15   ] -->
                               2+24+4
[2, 1,   10,    5,             30,           15   ]
The length of final array is 6
```

# Input/Output

- `[input]` integer array `arr`

  A non-empty array, 

  `1 ≤ arr.length ≤ 1000`

  `0 ≤ arr[i] ≤ 1000`


- `[output]` an integer

  The length of the final array

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

def sum_groups(arr):
    while True:
        n = len(arr)
        arr = [sum(grp) for key, grp in groupby(arr, key=lambda x: x % 2)]
        if len(arr) == n:
            return n
inputs: I found an interesting problem on https://www.codewars.com/kata/587731fda577b3d1b0001196:
```if-not:swift
Write simple .camelCase method (`camel_case` function in PHP, `CamelCase` in C# or `camelCase` in Java) for strings. All words must have their first letter capitalized without spaces.
```
```if:swift
Write a simple `camelCase` function for strings. All words must have their first letter capitalized and all spaces removed.
```
For instance:

```python
camelcase("hello case") => HelloCase
camelcase("camel case word") => CamelCaseWord
```
```c#
using Kata;

"hello case".CamelCase(); // => "HelloCase"
"camel case word".CamelCase(); // => "CamelCaseWord"
```

Don't forget to rate this kata! Thanks :)

I tried it in Python, but could not do it. Can you solve it?
targets: def camel_case(string):
    #a = [string.split()]
    a = list(string)
    for i in range(0, len(a)):
        if i==0 or a[i-1]==' ':
            a[i] = a[i].upper()
    return ''.join(a).replace(' ','')