Muennighoff/xP3x-sample huggingface dataset:
inputs: I found an interesting problem on https://leetcode.com/problems/valid-anagram/:
Given two strings s and t , write a function to determine if t is an anagram of s.

Example 1:


Input: s = "anagram", t = "nagaram"
Output: true


Example 2:


Input: s = "rat", t = "car"
Output: false


Note:
You may assume the string contains only lowercase alphabets.

Follow up:
What if the inputs contain unicode characters? How would you adapt your solution to such case?

I tried it in Python, but could not do it. Can you solve it?
targets: class Solution:
     def isAnagram(self, s, t):
         """
         :type s: str
         :type t: str
         :rtype: bool
         """
         if len(s)!=len(t):
             return False
         ssort=sorted(list(s))
         tsort=sorted(list(t))
         return ssort==tsort
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/702/A:
You are given array consisting of n integers. Your task is to find the maximum length of an increasing subarray of the given array.

A subarray is the sequence of consecutive elements of the array. Subarray is called increasing if each element of this subarray strictly greater than previous.


-----Input-----

The first line contains single positive integer n (1 ≤ n ≤ 10^5) — the number of integers.

The second line contains n positive integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ 10^9).


-----Output-----

Print the maximum length of an increasing subarray of the given array.


-----Examples-----
Input
5
1 7 2 11 15

Output
3

Input
6
100 100 100 100 100 100

Output
1

Input
3
1 2 3

Output
3

I tried it in Python, but could not do it. Can you solve it?
targets: n=int(input())
l=input().split()
best=1
curr=1
for i in range (n-1):
    if int(l[i])<int(l[i+1]):
        curr+=1
    else:
        curr=1
    best=max(best,curr)
print (best)
inputs: Solve in Python:
PolandBall has such a convex polygon with n veritces that no three of its diagonals intersect at the same point. PolandBall decided to improve it and draw some red segments. 

He chose a number k such that gcd(n, k) = 1. Vertices of the polygon are numbered from 1 to n in a clockwise way. PolandBall repeats the following process n times, starting from the vertex 1: 

Assume you've ended last operation in vertex x (consider x = 1 if it is the first operation). Draw a new segment from vertex x to k-th next vertex in clockwise direction. This is a vertex x + k or x + k - n depending on which of these is a valid index of polygon's vertex.

Your task is to calculate number of polygon's sections after each drawing. A section is a clear area inside the polygon bounded with drawn diagonals or the polygon's sides.


-----Input-----

There are only two numbers in the input: n and k (5 ≤ n ≤ 10^6, 2 ≤ k ≤ n - 2, gcd(n, k) = 1).


-----Output-----

You should print n values separated by spaces. The i-th value should represent number of polygon's sections after drawing first i lines.


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

Output
2 3 5 8 11 
Input
10 3

Output
2 3 4 6 9 12 16 21 26 31 


-----Note-----

The greatest common divisor (gcd) of two integers a and b is the largest positive integer that divides both a and b without a remainder.

For the first sample testcase, you should output "2 3 5 8 11". Pictures below correspond to situations after drawing lines. [Image] [Image] [Image] [Image] [Image] [Image]
targets: def __starting_point():
    n,k = list(map(int, input().strip().split()))

    if k > n//2:
        k = n - k

    intersection = n * [0]

    count = 1

    done = False
    i=0
    result = []
    for i in range(1,n+1):
        nn = (i*k) // n
        j = (i*k)%n
        if j < k:
            count += (2*nn )
        else:
            count += (2*nn +1)
        result.append(count)

    result[-1] -= 1
    print(" ".join([str(r) for r in result]))

__starting_point()
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/875/A:
Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number n. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that n is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer x was given. The task was to add x to the sum of the digits of the number x written in decimal numeral system.

Since the number n on the board was small, Vova quickly guessed which x could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number n for all suitable values of x or determine that such x does not exist. Write such a program for Vova.


-----Input-----

The first line contains integer n (1 ≤ n ≤ 10^9).


-----Output-----

In the first line print one integer k — number of different values of x satisfying the condition. 

In next k lines print these values in ascending order.


-----Examples-----
Input
21

Output
1
15

Input
20

Output
0



-----Note-----

In the first test case x = 15 there is only one variant: 15 + 1 + 5 = 21.

In the second test case there are no such x.

I tried it in Python, but could not do it. Can you solve it?
targets: n=int(input())
m=[]
if n<=18:
    a=0
else:
    a=n-len(str(n))*9
for i in range(a,n):
    x=i
    for j in str(i):
        x+=int(j)
    if n==x:
        m.append(i)
print(len(m))
[print(i) for i in m]
inputs: Solve in Python:
Let's call an array $t$ dominated by value $v$ in the next situation.

At first, array $t$ should have at least $2$ elements. Now, let's calculate number of occurrences of each number $num$ in $t$ and define it as $occ(num)$. Then $t$ is dominated (by $v$) if (and only if) $occ(v) > occ(v')$ for any other number $v'$. For example, arrays $[1, 2, 3, 4, 5, 2]$, $[11, 11]$ and $[3, 2, 3, 2, 3]$ are dominated (by $2$, $11$ and $3$ respectevitely) but arrays $[3]$, $[1, 2]$ and $[3, 3, 2, 2, 1]$ are not.

Small remark: since any array can be dominated only by one number, we can not specify this number and just say that array is either dominated or not.

You are given array $a_1, a_2, \dots, a_n$. Calculate its shortest dominated subarray or say that there are no such subarrays.

The subarray of $a$ is a contiguous part of the array $a$, i. e. the array $a_i, a_{i + 1}, \dots, a_j$ for some $1 \le i \le j \le n$.


-----Input-----

The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. Each test case consists of two lines.

The first line contains single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array $a$.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le n$) — the corresponding values of the array $a$.

It's guaranteed that the total length of all arrays in one test doesn't exceed $2 \cdot 10^5$.


-----Output-----

Print $T$ integers — one per test case. For each test case print the only integer — the length of the shortest dominated subarray, or $-1$ if there are no such subarrays.


-----Example-----
Input
4
1
1
6
1 2 3 4 5 1
9
4 1 2 4 5 4 3 2 1
4
3 3 3 3

Output
-1
6
3
2



-----Note-----

In the first test case, there are no subarrays of length at least $2$, so the answer is $-1$.

In the second test case, the whole array is dominated (by $1$) and it's the only dominated subarray.

In the third test case, the subarray $a_4, a_5, a_6$ is the shortest dominated subarray.

In the fourth test case, all subarrays of length more...
targets: import sys
T = int(sys.stdin.readline())
for i in range(T):
	n = int(sys.stdin.readline())
	array = sys.stdin.readline().split(" ")
	lastSeen = [-10**6 for i in range(n + 1)]
	best = 10**6
	for j in range(len(array)):
		if j - lastSeen[int(array[j])] < best:
			best = j - lastSeen[int(array[j])]
		lastSeen[int(array[j])] = j
	if best == 10**6:
		print(-1)
	else:
		print(best + 1)
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1081/A:
Chouti was doing a competitive programming competition. However, after having all the problems accepted, he got bored and decided to invent some small games.

He came up with the following game. The player has a positive integer $n$. Initially the value of $n$ equals to $v$ and the player is able to do the following operation as many times as the player want (possibly zero): choose a positive integer $x$ that $x<n$ and $x$ is not a divisor of $n$, then subtract $x$ from $n$. The goal of the player is to minimize the value of $n$ in the end.

Soon, Chouti found the game trivial. Can you also beat the game?


-----Input-----

The input contains only one integer in the first line: $v$ ($1 \le v \le 10^9$), the initial value of $n$.


-----Output-----

Output a single integer, the minimum value of $n$ the player can get.


-----Examples-----
Input
8

Output
1

Input
1

Output
1



-----Note-----

In the first example, the player can choose $x=3$ in the first turn, then $n$ becomes $5$. He can then choose $x=4$ in the second turn to get $n=1$ as the result. There are other ways to get this minimum. However, for example, he cannot choose $x=2$ in the first turn because $2$ is a divisor of $8$.

In the second example, since $n=1$ initially, the player can do nothing.

I tried it in Python, but could not do it. Can you solve it?
targets: n = int(input())
if n == 2:
    print(2)
else:
    print(1)
inputs: I found an interesting problem on https://www.codewars.com/kata/550cc572b9e7b563be00054f:
Scheduling is how the processor decides which jobs(processes) get to use the processor and for how long. This can cause a lot of problems. Like a really long process taking the entire CPU and freezing all the other processes. One solution is Shortest Job First(SJF), which today you will be implementing.

SJF works by, well, letting the shortest jobs take the CPU first. If the jobs are the same size then it is First In First Out (FIFO). The idea is that the shorter jobs will finish quicker, so theoretically jobs won't get frozen because of large jobs. (In practice they're frozen because of small jobs).

You will be implementing:
```python
  def SJF(jobs, index)
```

It takes in:

1. "jobs" a non-empty array of positive integers. They represent the clock-cycles(cc) needed to finish the job.
2. "index" a positive integer. That represents the job we're interested in.

SJF returns:

1. A positive integer representing the cc it takes to complete the job at index.

Here's an example:
```
SJF([3, 10, 20, 1, 2], 0)
at 0cc [3, 10, 20, 1, 2] jobs[3] starts
at 1cc [3, 10, 20, 0, 2] jobs[3] finishes, jobs[4] starts
at 3cc [3, 10, 20, 0, 0] jobs[4] finishes, jobs[0] starts
at 6cc [0, 10, 20, 0, 0] jobs[0] finishes
```

so:
```
SJF([3,10,20,1,2], 0) == 6
```

I tried it in Python, but could not do it. Can you solve it?
targets: def SJF(jobs, index):
    return sum(n for i, n in enumerate(jobs) if n < jobs[index] + (i <= index))
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1152/D:
Neko is playing with his toys on the backyard of Aki's house. Aki decided to play a prank on him, by secretly putting catnip into Neko's toys. Unfortunately, he went overboard and put an entire bag of catnip into the toys...

It took Neko an entire day to turn back to normal. Neko reported to Aki that he saw a lot of weird things, including a trie of all correct bracket sequences of length $2n$.

The definition of correct bracket sequence is as follows:  The empty sequence is a correct bracket sequence,  If $s$ is a correct bracket sequence, then $(\,s\,)$ is a correct bracket sequence,  If $s$ and $t$ are a correct bracket sequence, then $st$ is also a correct bracket sequence. 

For example, the strings "(())", "()()" form a correct bracket sequence, while ")(" and "((" not.

Aki then came up with an interesting problem: What is the size of the maximum matching (the largest set of edges such that there are no two edges with a common vertex) in this trie? Since the answer can be quite large, print it modulo $10^9 + 7$.


-----Input-----

The only line contains a single integer $n$ ($1 \le n \le 1000$).


-----Output-----

Print exactly one integer — the size of the maximum matching in the trie. Since the answer can be quite large, print it modulo $10^9 + 7$.


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

Output
1

Input
2

Output
3

Input
3

Output
9



-----Note-----

The pictures below illustrate tries in the first two examples (for clarity, the round brackets are replaced with angle brackets). The maximum matching is highlighted with blue. [Image] [Image]

I tried it in Python, but could not do it. Can you solve it?
targets: ans = [...
inputs: I found an interesting problem on https://www.codewars.com/kata/5a53a17bfd56cb9c14000003:
# Definition

**_Disarium number_** is the number that *The sum of its digits powered with their respective positions is equal to the number itself*.

____

# Task

**_Given_** a number, **_Find if it is Disarium or not_** . 
____

# Warm-up (Highly recommended)

# [Playing With Numbers Series](https://www.codewars.com/collections/playing-with-numbers)
___

# Notes 

* **_Number_** *passed is always*  **_Positive_** .
* **_Return_** *the result as* **_String_**
___

# Input >> Output Examples

```
disariumNumber(89) ==> return "Disarium !!"
```
## **_Explanation_**:

* Since , **_8^(1) + 9^(2) = 89_** , thus *output* is `"Disarium !!"`
___

```
disariumNumber(564) ==> return "Not !!"
```
## **_Explanation_**:

Since , **_5^(1) + 6^(2) + 4^(3) = 105 != 564_** ,  thus *output* is `"Not !!"`

___
___
___

# [Playing with Numbers Series](https://www.codewars.com/collections/playing-with-numbers)

# [Playing With Lists/Arrays Series](https://www.codewars.com/collections/playing-with-lists-slash-arrays)

# [For More Enjoyable Katas](http://www.codewars.com/users/MrZizoScream/authored)
___

## ALL translations are welcomed

## Enjoy Learning !!
# Zizou

I tried it in Python, but could not do it. Can you solve it?
targets: def disarium_number(number):
    newNumber = 0
    for index, value in enumerate(str(number)):
        newNumber += int(value)**(int(index)+1)
    if newNumber == number:
        return "Disarium !!"
    return "Not !!"
inputs: I found an interesting problem on https://www.codewars.com/kata/5c46ea433dd41b19af1ca3b3:
### Description
As hex values can include letters `A` through to `F`, certain English words can be spelled out, such as `CAFE`, `BEEF`, or `FACADE`.
This vocabulary can be extended by using numbers to represent other letters, such as `5EAF00D`, or `DEC0DE5`.

Given a string, your task is to return the decimal sum of all words in the string that can be interpreted as such hex values.


### Example

Working with the string `BAG OF BEES`:  
* `BAG` ==> `0` as it is not a valid hex value  
* `OF` ==> `0F` ==> `15`  
* `BEES` ==> `BEE5` ==> `48869`

So `hex_word_sum('BAG OF BEES')` returns the sum of these, `48884`.


### Notes
* Inputs are all uppercase and contain no punctuation
* `0` can be substituted for `O`
* `5` can be substituted for `S`

I tried it in Python, but could not do it. Can you solve it?
targets: def hex_word_sum(string):
    string = string.upper().replace('S', '5')
    string = string.replace('O', '0')
    sum = 0
    for word in string.split(' '):
        try:
            sum += int(word.strip(), 16)
        except:
            pass
    return sum
inputs: Solve in Python:
You are given three positive integers $n$, $a$ and $b$. You have to construct a string $s$ of length $n$ consisting of lowercase Latin letters such that each substring of length $a$ has exactly $b$ distinct letters. It is guaranteed that the answer exists.

You have to answer $t$ independent test cases.

Recall that the substring $s[l \dots r]$ is the string $s_l, s_{l+1}, \dots, s_{r}$ and its length is $r - l + 1$. In this problem you are only interested in substrings of length $a$.


-----Input-----

The first line of the input contains one integer $t$ ($1 \le t \le 2000$) — the number of test cases. Then $t$ test cases follow.

The only line of a test case contains three space-separated integers $n$, $a$ and $b$ ($1 \le a \le n \le 2000, 1 \le b \le \min(26, a)$), where $n$ is the length of the required string, $a$ is the length of a substring and $b$ is the required number of distinct letters in each substring of length $a$.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2000$ ($\sum n \le 2000$).


-----Output-----

For each test case, print the answer — such a string $s$ of length $n$ consisting of lowercase Latin letters that each substring of length $a$ has exactly $b$ distinct letters. If there are multiple valid answers, print any of them. It is guaranteed that the answer exists.


-----Example-----
Input
4
7 5 3
6 1 1
6 6 1
5 2 2

Output
tleelte
qwerty
vvvvvv
abcde



-----Note-----

In the first test case of the example, consider all the substrings of length $5$:  "tleel": it contains $3$ distinct (unique) letters,  "leelt": it contains $3$ distinct (unique) letters,  "eelte": it contains $3$ distinct (unique) letters.
targets: t = int(input(''))
c = []
for i in range(97,123,1):
    c.append(chr(i))
for _ in range(t):
    n,a,b = list(map(int,input('').split(' ')))
    s = ''
    for i in range(n):
        s = s+c[i%b]
    print(s)
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1313/B:
Nikolay has only recently started in competitive programming, but already qualified to the finals of one prestigious olympiad. There going to be $n$ participants, one of whom is Nikolay. Like any good olympiad, it consists of two rounds. Tired of the traditional rules, in which the participant who solved the largest number of problems wins, the organizers came up with different rules.

Suppose in the first round participant A took $x$-th place and in the second round — $y$-th place. Then the total score of the participant A is sum $x + y$. The overall place of the participant A is the number of participants (including A) having their total score less than or equal to the total score of A. Note, that some participants may end up having a common overall place. It is also important to note, that in both the first and the second round there were no two participants tying at a common place. In other words, for every $i$ from $1$ to $n$ exactly one participant took $i$-th place in first round and exactly one participant took $i$-th place in second round.

Right after the end of the Olympiad, Nikolay was informed that he got $x$-th place in first round and $y$-th place in the second round. Nikolay doesn't know the results of other participants, yet he wonders what is the minimum and maximum place he can take, if we consider the most favorable and unfavorable outcome for him. Please help Nikolay to find the answer to this question.


-----Input-----

The first line contains an integer $t$ ($1 \le t \le 100$) — the number of test cases to solve.

Each of the following $t$ lines contains integers $n$, $x$, $y$ ($1 \leq n \leq 10^9$, $1 \le x, y \le n$) — the number of participants in the olympiad, the place that Nikolay took in the first round and the place that Nikolay took in the second round.


-----Output-----

Print two integers — the minimum and maximum possible overall place Nikolay could take.


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

Output
1 3

Input
1
6 3 4

Output
2 6



-----Note-----

Explanation for the first...

I tried it in Python, but could not do it. Can you solve it?
targets: t = int(input())
for _ in range(t):
    n, x, y = [int(item) for item in input().split()]
    total = x + y
    maximum = min(x + y - 1, n)
    minimum = min(n - min((2 * n - total), n) + 1, n)
    print(minimum, maximum)
inputs: Solve in Python:
Mike and Joe are fratboys that love beer and games that involve drinking. They play the following game: Mike chugs one beer, then Joe chugs 2 beers, then Mike chugs 3 beers, then Joe chugs 4 beers, and so on. Once someone can't drink what he is supposed to drink, he loses.

Mike can chug at most A beers in total (otherwise he would pass out), while Joe can chug at most B beers in total. Who will win the game? 

Write the function ```game(A,B)``` that returns the winner, ```"Mike"``` or ```"Joe"``` accordingly, for any given integer values of A and B.

Note: If either Mike or Joe cannot drink at least 1 beer, return the string  ```"Non-drinkers can't play"```.
targets: def game(a, b):
    if a * b:
        c = int(a ** 0.5)
        return ('Mike', 'Joe')[c * (c + 1) <= b]
    return "Non-drinkers can't play"
inputs: Solve in Python:
Consider the sequence `a(1) = 7, a(n) = a(n-1) + gcd(n, a(n-1)) for n >= 2`: 

`7, 8, 9, 10, 15, 18, 19, 20, 21, 22, 33, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 69, 72, 73...`.

Let us take the differences between successive elements of the sequence and
get a second sequence `g: 1, 1, 1, 5, 3, 1, 1, 1, 1, 11, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 23, 3, 1...`.

For the sake of uniformity of the lengths of sequences **we add** a `1` at the head of g:

`g: 1, 1, 1, 1, 5, 3, 1, 1, 1, 1, 11, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 23, 3, 1...`

Removing the 1s gives a third sequence:
`p: 5, 3, 11, 3, 23, 3...`
where you can see prime numbers.

#Task:
Write functions:
```
1: an(n) with parameter n: returns the first n terms of the series a(n) (not tested)

2: gn(n) with parameter n: returns the first n terms of the series g(n) (not tested)

3: countOnes(n) with parameter n: returns the number of 1 in g(n) 
    (don't forget to add a `1` at the head) # (tested)
    
4: p(n) with parameter n: returns an array of n unique prime numbers (not tested)

5: maxp(n) with parameter n: returns the biggest prime number of the sequence pn(n) # (tested)

6: anOver(n) with parameter n: returns an array (n terms) of the a(i)/i for every i such g(i) != 1 (not tested but interesting result)

7: anOverAverage(n) with parameter n: returns as an *integer* the average of anOver(n)  (tested)
```

#Note:
You can write directly functions `3:`, `5:` and `7:`. There is no need to write functions `1:`, `2:`, `4:` `6:`
except out of pure curiosity.
targets: def count_ones(n):
    a, ones = 7, 1
    for i in range(2, n+1):
        b = a + gcd(i, a)
        if b == a + 1: ones += 1
        a = b

    return ones

def max_pn(n):
    a, p, i = 7, {1}, 1
    
    while len(p) < n + 1:
        i += 1
        b = a + gcd(i, a)
        p.add(b - a)
        a = b
    return max(p)

def an_over_average(n):
    return 3

def gcd(a, b):
    while b:
        a, b = b, a % b
    return a
inputs: Solve in Python:
A palindrome is a word, phrase, number, or other sequence of characters which reads the same backward as forward. Examples of numerical palindromes are: 

* 232
* 110011
* 54322345

Complete the function to test if the given number (`num`) **can be rearranged** to form a numerical palindrome or not. Return a boolean (`true` if it can be rearranged to a palindrome, and `false` if it cannot). Return `"Not valid"` if the input is not an integer or is less than 0.

For this kata, single digit numbers are **NOT** considered numerical palindromes.  


## Examples

```
5        =>  false
2121     =>  true
1331     =>  true 
3357665  =>  true 
1294     =>  false 
"109982" =>  "Not valid"
-42      =>  "Not valid"
```
targets: def palindrome(num):
    s = str(num)
    return "Not valid" if not isinstance(num, int) or num < 0 \
        else num > 10 and sum(s.count(d) % 2 > 0 for d in set(s)) < 2
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1183/E:
The only difference between the easy and the hard versions is constraints.

A subsequence is a string that can be derived from another string by deleting some or no symbols without changing the order of the remaining symbols. Characters to be deleted are not required to go successively, there can be any gaps between them. For example, for the string "abaca" the following strings are subsequences: "abaca", "aba", "aaa", "a" and "" (empty string). But the following strings are not subsequences: "aabaca", "cb" and "bcaa".

You are given a string $s$ consisting of $n$ lowercase Latin letters.

In one move you can take any subsequence $t$ of the given string and add it to the set $S$. The set $S$ can't contain duplicates. This move costs $n - |t|$, where $|t|$ is the length of the added subsequence (i.e. the price equals to the number of the deleted characters).

Your task is to find out the minimum possible total cost to obtain a set $S$ of size $k$ or report that it is impossible to do so.


-----Input-----

The first line of the input contains two integers $n$ and $k$ ($1 \le n, k \le 100$) — the length of the string and the size of the set, correspondingly.

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


-----Output-----

Print one integer — if it is impossible to obtain the set $S$ of size $k$, print -1. Otherwise, print the minimum possible total cost to do it.


-----Examples-----
Input
4 5
asdf

Output
4

Input
5 6
aaaaa

Output
15

Input
5 7
aaaaa

Output
-1

Input
10 100
ajihiushda

Output
233



-----Note-----

In the first example we can generate $S$ = { "asdf", "asd", "adf", "asf", "sdf" }. The cost of the first element in $S$ is $0$ and the cost of the others is $1$. So the total cost of $S$ is $4$.

I tried it in Python, but could not do it. Can you solve it?
targets: import sys
input = sys.stdin.readline

n,W=list(map(int,input().split()))
s=input().strip()

NEXTLIST=[[n]*26 for i in range(n+1)]

for i in range(n-1,-1,-1):
    for j in range(26):
        NEXTLIST[i][j]=NEXTLIST[i+1][j]
    NEXTLIST[i][ord(s[i])-97]=i

DP=[[0]*(n+1) for i in range(n+1)]

DP[0][0]=1

for i in range(n):
    for j in range(26):
        if NEXTLIST[i][j]!=n:
            for k in range(n):
                DP[NEXTLIST[i][j]+1][k+1]+=DP[i][k]

#print(DP)

HLIST=[0]*(n+1)

for i in range(n+1):
    for j in range(n+1):
        HLIST[j]+=DP[i][j]

#print(HLIST)

ANS=0
for i in range(n,-1,-1):
    #print(i,W)
    if W>HLIST[i]:
        ANS+=(n-i)*HLIST[i]
        W-=HLIST[i]
    else:
        ANS+=W*(n-i)
        print(ANS)
        return

else:
    print(-1)
inputs: Solve in Python:
Helen works in Metropolis airport. She is responsible for creating a departure schedule. There are n flights that must depart today, the i-th of them is planned to depart at the i-th minute of the day.

Metropolis airport is the main transport hub of Metropolia, so it is difficult to keep the schedule intact. This is exactly the case today: because of technical issues, no flights were able to depart during the first k minutes of the day, so now the new departure schedule must be created.

All n scheduled flights must now depart at different minutes between (k + 1)-th and (k + n)-th, inclusive. However, it's not mandatory for the flights to depart in the same order they were initially scheduled to do so — their order in the new schedule can be different. There is only one restriction: no flight is allowed to depart earlier than it was supposed to depart in the initial schedule.

Helen knows that each minute of delay of the i-th flight costs airport c_{i} burles. Help her find the order for flights to depart in the new schedule that minimizes the total cost for the airport.


-----Input-----

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 300 000), here n is the number of flights, and k is the number of minutes in the beginning of the day that the flights did not depart.

The second line contains n integers c_1, c_2, ..., c_{n} (1 ≤ c_{i} ≤ 10^7), here c_{i} is the cost of delaying the i-th flight for one minute.


-----Output-----

The first line must contain the minimum possible total cost of delaying the flights.

The second line must contain n different integers t_1, t_2, ..., t_{n} (k + 1 ≤ t_{i} ≤ k + n), here t_{i} is the minute when the i-th flight must depart. If there are several optimal schedules, print any of them.


-----Example-----
Input
5 2
4 2 1 10 2

Output
20
3 6 7 4 5 



-----Note-----

Let us consider sample test. If Helen just moves all flights 2 minutes later preserving the order, the total cost of delaying the flights would be (3 - 1)·4 + (4 - 2)·2 + (5 - 3)·1 + (6 - 4)·10 +...
targets: from heapq import heappush,heappop,heapify
n,k=list(map(int,input().split()))
*l,=list(map(int,input().split()))
q=[(-l[i],i)for i in range(k)]
heapify(q)
a=[0]*n
s=0
for i in range(k,n) :
    heappush(q,(-l[i],i))
    x,j=heappop(q)
    s-=x*(i-j)
    a[j]=i+1
for i in range(n,n+k) :
    x,j=heappop(q)
    s-=x*(i-j)
    a[j]=i+1
print(s)
print(' '.join(map(str,a)))
inputs: Solve in Python:
You are given an array A of strings.
A move onto S consists of swapping any two even indexed characters of S, or any two odd indexed characters of S.
Two strings S and T are special-equivalent if after any number of moves onto S, S == T.
For example, S = "zzxy" and T = "xyzz" are special-equivalent because we may make the moves "zzxy" -> "xzzy" -> "xyzz" that swap S[0] and S[2], then S[1] and S[3].
Now, a group of special-equivalent strings from A is a non-empty subset of A such that:

Every pair of strings in the group are special equivalent, and;
The group is the largest size possible (ie., there isn't a string S not in the group such that S is special equivalent to every string in the group)

Return the number of groups of special-equivalent strings from A.
 

Example 1:
Input: ["abcd","cdab","cbad","xyzz","zzxy","zzyx"]
Output: 3
Explanation: 
One group is ["abcd", "cdab", "cbad"], since they are all pairwise special equivalent, and none of the other strings are all pairwise special equivalent to these.

The other two groups are ["xyzz", "zzxy"] and ["zzyx"].  Note that in particular, "zzxy" is not special equivalent to "zzyx".


Example 2:
Input: ["abc","acb","bac","bca","cab","cba"]
Output: 3
 






Note:

1 <= A.length <= 1000
1 <= A[i].length <= 20
All A[i] have the same length.
All A[i] consist of only lowercase letters.
targets: class Solution:
    def numSpecialEquivGroups(self, A: List[str]) -> int:
        
        # set1 = set()
        even_set = []
        odd_set = []
        for i in A:
            even = []
            odd = []
            for j in range(len(i)):
                if j %2 == 0:
                    even.append(i[j])
                else:
                    odd.append(i[j])
            even.sort()
            odd.sort()
            if even in even_set:
                k = []
                for p,values in enumerate(even_set):
                    if even == values:
                        k += [p]
                flag = 0
                for p in k:
                    if odd_set[p] == odd:
                        flag = 1 
                        break
                if flag == 0:
                    even_set.append(even) 
                    odd_set.append(odd)
            else:
                even_set.append(even) 
                odd_set.append(odd)
                
        return len(even_set)
inputs: Solve in Python:
Mobile Display Keystrokes

Do you remember the old mobile display keyboards? Do you also remember how inconvenient it was to write on it?
Well, here you have to calculate how much keystrokes you have to do for a specific word.

This is the layout:



Return the amount of keystrokes of input str (! only letters, digits and special characters in lowercase included in layout without whitespaces !)

e.g:


mobileKeyboard("123") => 3 (1+1+1)
mobileKeyboard("abc") => 9 (2+3+4)
mobileKeyboard("codewars") => 26 (4+4+2+3+2+2+4+5)
targets: def mobile_keyboard(s):
    a = "12abc3def4ghi5jkl6mno7pqrs8tuv9wxyz*0#"
    b = "11234123412341234123412345123412345111"
    d = {x: int(y) for x, y in zip(a, b)}
    return sum(d[x] for x in s)
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1417/B:
RedDreamer has an array $a$ consisting of $n$ non-negative integers, and an unlucky integer $T$.

Let's denote the misfortune of array $b$ having length $m$ as $f(b)$ — the number of pairs of integers $(i, j)$ such that $1 \le i < j \le m$ and $b_i + b_j = T$. RedDreamer has to paint each element of $a$ into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays $c$ and $d$ so that all white elements belong to $c$, and all black elements belong to $d$ (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that $f(c) + f(d)$ is minimum possible.

For example:

  if $n = 6$, $T = 7$ and $a = [1, 2, 3, 4, 5, 6]$, it is possible to paint the $1$-st, the $4$-th and the $5$-th elements white, and all other elements black. So $c = [1, 4, 5]$, $d = [2, 3, 6]$, and $f(c) + f(d) = 0 + 0 = 0$;  if $n = 3$, $T = 6$ and $a = [3, 3, 3]$, it is possible to paint the $1$-st element white, and all other elements black. So $c = [3]$, $d = [3, 3]$, and $f(c) + f(d) = 0 + 1 = 1$. 

Help RedDreamer to paint the array optimally!


-----Input-----

The first line contains one integer $t$ ($1 \le t \le 1000$) — the number of test cases. Then $t$ test cases follow.

The first line of each test case contains two integers $n$ and $T$ ($1 \le n \le 10^5$, $0 \le T \le 10^9$) — the number of elements in the array and the unlucky integer, respectively. 

The second line contains $n$ integers $a_1$, $a_2$, ..., $a_n$ ($0 \le a_i \le 10^9$) — the elements of the array. 

The sum of $n$ over all test cases does not exceed $10^5$.


-----Output-----

For each test case print $n$ integers: $p_1$, $p_2$, ..., $p_n$ (each $p_i$ is either $0$ or $1$) denoting the colors. If $p_i$ is $0$, then $a_i$ is white and belongs to the array $c$, otherwise it is black and belongs to the array $d$.

If there are multiple answers that minimize the value of $f(c) + f(d)$, print any of them.


-----Example-----
Input
2
6 7
1 2 3 4 5 6
3 6
3...

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

t = int(input())
for _ in range(t):
    n,k = list(map(int, input().split()))
    arr = list(map(int, input().split()))
    
    alt = 0
    ans = []
    for i in range(len(arr)):
        if k%2==1:
            if arr[i] < k/2:
                ans.append(0)
            else:
                ans.append(1)
        else:
            if arr[i] == k//2:
                ans.append(alt%2)
                alt += 1
            elif arr[i] < k//2:
                ans.append(0)
            else:
                ans.append(1)
                
    print(*ans)
inputs: I found an interesting problem on https://www.codewars.com/kata/57f4ccf0ab9a91c3d5000054:
Lot of junior developer can be stuck when they need to change the access permission to a file or a directory in an Unix-like operating systems.

To do that they can use the `chmod` command and with some magic trick they can change the permissionof a file or a directory. For more information about the `chmod` command you can take a look at the [wikipedia page](https://en.wikipedia.org/wiki/Chmod).

`chmod` provides two types of syntax that can be used for changing permissions. An absolute form using octal to denote which permissions bits are set e.g: 766.
Your goal in this kata is to define the octal you need to use in order to set yout permission correctly.

Here is the list of the permission you can set with the octal representation of this one.

- User
  - read (4)
  - write (2)
  - execute (1)
- Group
  - read (4)
  - write (2)
  - execute (1)
- Other
  - read (4)
  - write (2)
  - execute (1)
  
The method take a hash in argument this one can have a maximum of 3 keys (`owner`,`group`,`other`). Each key will have a 3 chars string to represent the permission, for example the string `rw-` say that the user want the permission `read`, `write` without the `execute`.
If a key is missing set the permission to `---`

**Note**: `chmod` allow you to set some special flags too (`setuid`, `setgid`, `sticky bit`) but to keep some simplicity for this kata we will ignore this one.

I tried it in Python, but could not do it. Can you solve it?
targets: def chmod_calculator(perm):    
    return str(sum([wbin(perm[k]) * {"user":100, "group":10, "other":1}[k] for k in perm])).zfill(3)

def wbin(w):
    return int(w.translate(str.maketrans('rwx-', '1110')), 2)
inputs: Solve in Python:
One of Arkady's friends works at a huge radio telescope. A few decades ago the telescope has sent a signal $s$ towards a faraway galaxy. Recently they've received a response $t$ which they believe to be a response from aliens! The scientists now want to check if the signal $t$ is similar to $s$.

The original signal $s$ was a sequence of zeros and ones (everyone knows that binary code is the universe-wide language). The returned signal $t$, however, does not look as easy as $s$, but the scientists don't give up! They represented $t$ as a sequence of English letters and say that $t$ is similar to $s$ if you can replace all zeros in $s$ with some string $r_0$ and all ones in $s$ with some other string $r_1$ and obtain $t$. The strings $r_0$ and $r_1$ must be different and non-empty.

Please help Arkady's friend and find the number of possible replacements for zeros and ones (the number of pairs of strings $r_0$ and $r_1$) that transform $s$ to $t$.


-----Input-----

The first line contains a string $s$ ($2 \le |s| \le 10^5$) consisting of zeros and ones — the original signal.

The second line contains a string $t$ ($1 \le |t| \le 10^6$) consisting of lowercase English letters only — the received signal.

It is guaranteed, that the string $s$ contains at least one '0' and at least one '1'.


-----Output-----

Print a single integer — the number of pairs of strings $r_0$ and $r_1$ that transform $s$ to $t$.

In case there are no such pairs, print $0$.


-----Examples-----
Input
01
aaaaaa

Output
4

Input
001
kokokokotlin

Output
2



-----Note-----

In the first example, the possible pairs $(r_0, r_1)$ are as follows:  "a", "aaaaa"  "aa", "aaaa"  "aaaa", "aa"  "aaaaa", "a"  

The pair "aaa", "aaa" is not allowed, since $r_0$ and $r_1$ must be different.

In the second example, the following pairs are possible:   "ko", "kokotlin"  "koko", "tlin"
targets: import sys
from math import *

def minp():
	return sys.stdin.readline().strip()

def mint():
	return int(minp())

def mints():
	return map(int, minp().split())

def add(a,b):
	return (a+b)%1000000007

def sub(a,b):
	return (a+1000000007-b)%1000000007

def mul(a,b):
	return (a*b)%1000000007

p = 102367
s = list(map(int,minp()))
t = list(map(ord,minp()))
h = [0]*(len(t)+1)
pp = [1]*(len(t)+1)
for i in range(len(t)):
	h[i+1] = add(mul(h[i], p), t[i])
	pp[i+1] = mul(pp[i], p)

def cmp(a, b, l):
	if a > b:
		a, b = b, a
	h1 = sub(h[a+l], mul(h[a], pp[l]))
	h2 = sub(h[b+l], mul(h[b], pp[l]))
	return h2 == h1

c = [0,0]
idx = [-1,-1]
for i in range(len(s)):
	c[s[i]] += 1
	if idx[s[i]] < 0:
		idx[s[i]] = i
Mv = max(c)
mv = min(c)
Mi = c.index(Mv)
mi = (Mi^1)
lt = len(t)
sp = [0,0]
res = 0
for k in range(1,lt//Mv+1):
	l = [0,0]
	x = (lt-k*Mv)//mv
	if x > 0 and x*mv + k*Mv == lt:
		l[Mi] = k
		l[mi] = x
		if idx[0] < idx[1]:
			sp[0] = 0
			sp[1] = idx[1]*l[0]
		else:
			sp[1] = 0
			sp[0] = idx[0]*l[1]
		ok = True
		j = 0
		for i in range(len(s)):
			if not cmp(sp[s[i]], j, l[s[i]]):
				ok = False
				break
			j += l[s[i]]
		if l[0] == l[1] and cmp(sp[0], sp[1], l[0]):
			ok = False
		if ok:
			res += 1
print(res)
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/979/C:
Kuro is living in a country called Uberland, consisting of $n$ towns, numbered from $1$ to $n$, and $n - 1$ bidirectional roads connecting these towns. It is possible to reach each town from any other. Each road connects two towns $a$ and $b$. Kuro loves walking and he is planning to take a walking marathon, in which he will choose a pair of towns $(u, v)$ ($u \neq v$) and walk from $u$ using the shortest path to $v$ (note that $(u, v)$ is considered to be different from $(v, u)$).

Oddly, there are 2 special towns in Uberland named Flowrisa (denoted with the index $x$) and Beetopia (denoted with the index $y$). Flowrisa is a town where there are many strong-scent flowers, and Beetopia is another town where many bees live. In particular, Kuro will avoid any pair of towns $(u, v)$ if on the path from $u$ to $v$, he reaches Beetopia after he reached Flowrisa, since the bees will be attracted with the flower smell on Kuro’s body and sting him.

Kuro wants to know how many pair of city $(u, v)$ he can take as his route. Since he’s not really bright, he asked you to help him with this problem.


-----Input-----

The first line contains three integers $n$, $x$ and $y$ ($1 \leq n \leq 3 \cdot 10^5$, $1 \leq x, y \leq n$, $x \ne y$) - the number of towns, index of the town Flowrisa and index of the town Beetopia, respectively.

$n - 1$ lines follow, each line contains two integers $a$ and $b$ ($1 \leq a, b \leq n$, $a \ne b$), describes a road connecting two towns $a$ and $b$.

It is guaranteed that from each town, we can reach every other town in the city using the given roads. That is, the given map of towns and roads is a tree.


-----Output-----

A single integer resembles the number of pair of towns $(u, v)$ that Kuro can use as his walking route.


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

Output
5
Input
3 1 3
1 2
1 3

Output
4


-----Note-----

On the first example, Kuro can choose these pairs:   $(1, 2)$: his route would be $1 \rightarrow 2$,  $(2, 3)$: his route would be $2 \rightarrow 3$,  $(3, 2)$: his route...

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

n,x,y = list(map(int,input().split()))
graph = defaultdict(list)
vis = [False for i in range(n+1)]
mat = [False for i in range(n+1)]
subtree = [0 for i in range(n+1)]

for i in range(n-1):
	u,v = list(map(int,input().split()))
	graph[u].append(v)
	graph[v].append(u)
q = []
cur = 0
for v in graph[x]:
	if v!=y:
		q.append([v,v])
	else:
		cur = v
vis[x] = 1
while q!=[]:
	temp = q.pop()
	u,v = temp
	vis[u] = True
	subtree[v]+=1
	for node in graph[u]:
		if vis[node]==False:
			if node!=y:
				q.append([node,v])
			else:
				cur = v
val = sum(subtree)
val1 = (val+1-subtree[cur])
val2 = n-(sum(subtree)+1)
val = val1*val2
print(n*(n-1)-val)
inputs: Solve in Python:
Vasya has a sequence $a$ consisting of $n$ integers $a_1, a_2, \dots, a_n$. Vasya may pefrom the following operation: choose some number from the sequence and swap any pair of bits in its binary representation. For example, Vasya can transform number $6$ $(\dots 00000000110_2)$ into $3$ $(\dots 00000000011_2)$, $12$ $(\dots 000000001100_2)$, $1026$ $(\dots 10000000010_2)$ and many others. Vasya can use this operation any (possibly zero) number of times on any number from the sequence.

Vasya names a sequence as good one, if, using operation mentioned above, he can obtain the sequence with bitwise exclusive or of all elements equal to $0$.

For the given sequence $a_1, a_2, \ldots, a_n$ Vasya'd like to calculate number of integer pairs $(l, r)$ such that $1 \le l \le r \le n$ and sequence $a_l, a_{l + 1}, \dots, a_r$ is good.


-----Input-----

The first line contains a single integer $n$ ($1 \le n \le 3 \cdot 10^5$) — length of the sequence.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^{18}$) — the sequence $a$.


-----Output-----

Print one integer — the number of pairs $(l, r)$ such that $1 \le l \le r \le n$ and the sequence $a_l, a_{l + 1}, \dots, a_r$ is good.


-----Examples-----
Input
3
6 7 14

Output
2

Input
4
1 2 1 16

Output
4



-----Note-----

In the first example pairs $(2, 3)$ and $(1, 3)$ are valid. Pair $(2, 3)$ is valid since $a_2 = 7 \rightarrow 11$, $a_3 = 14 \rightarrow 11$ and $11 \oplus 11 = 0$, where $\oplus$ — bitwise exclusive or. Pair $(1, 3)$ is valid since $a_1 = 6 \rightarrow 3$, $a_2 = 7 \rightarrow 13$, $a_3 = 14 \rightarrow 14$ and $3 \oplus 13 \oplus 14 = 0$.

In the second example pairs $(1, 2)$, $(2, 3)$, $(3, 4)$ and $(1, 4)$ are valid.
targets: #!/usr/bin/env python3
import sys

def rint():
    return list(map(int, sys.stdin.readline().split()))
#lines = stdin.readlines()

def get_num1(i):
    cnt = 0
    while i:
        if i%2:
            cnt +=1
        i //=2
    return cnt

n = int(input())

a = list(rint())

b = [get_num1(aa) for aa in a]

ans = 0
#S0[i] : 1 if sum of 1s in ragne (0, i) is odd, else 0
S0 = [0]*n
S0[0] = b[0]%2
for i in range(1, n):
    S0[i] = (S0[i-1] + b[i])%2

#total even pairs in (0, n)
even_cnt = S0.count(0)

ans = even_cnt

# check total even pairs in (i, n)
for i in range(1, n):
    if b[i-1] %2:
        even_cnt = n - i - even_cnt
    else:
        even_cnt -= 1
    ans += even_cnt

for i in range(n):
    max_value = 0
    sum_value = 0
    for j in range(1, 62):
        if i + j > n:
            break
        sum_value += b[i+j-1]
        max_value = max(max_value, b[i+j-1])
        if 2 * max_value > sum_value and sum_value%2 == 0:
            ans -= 1

print(ans)
inputs: Solve in Python:
You are given an n × m rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table

abcd

edfg

hijk



 

we obtain the table:

acd

efg

hjk



 

A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.


-----Input-----

The first line contains two integers  — n and m (1 ≤ n, m ≤ 100).

Next n lines contain m small English letters each — the characters of the table.


-----Output-----

Print a single number — the minimum number of columns that you need to remove in order to make the table good.


-----Examples-----
Input
1 10
codeforces

Output
0

Input
4 4
case
care
test
code

Output
2

Input
5 4
code
forc
esco
defo
rces

Output
4



-----Note-----

In the first sample the table is already good.

In the second sample you may remove the first and third column.

In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).

Let strings s and t have equal length. Then, s is lexicographically larger than t if they are not equal and the character following the largest common prefix of s and t (the prefix may be empty) in s is alphabetically larger than the corresponding character of t.
targets: import sys
n, m = [int(i) for i in input().split()]
a = [[0] * n for i in range(m)]
for i in range(n):
    temp = input()
    for j in range(m):
        a[j][i] = temp[j]

c = 0

def blatant(a):
    for i in range(1,len(a)):
        if a[i] < a[i-1]:
            return True
    return False

match = []
while True:
    if len(a) == 0:
        print(m)
        return
    elif blatant(a[0]):
        del a[0]
        c += 1
    else:
        break

for i in range(1,n):
    if a[0][i] == a[0][i-1]:
        match.append(i)
        
def rm(a,match):
    c = 0
    for m in match:
        i = 1
        while i < len(a):
            if a[i][m] < a[i][m-1]:
                del a[i]
                c += 1
            elif a[i][m] == a[i][m-1]:
                i += 1
            else:
                break
    return c

while True:
    temp = rm(a,match)
    if temp == 0:
        break
    c += temp
    
print(c)