Muennighoff/xP3x-sample huggingface dataset:
inputs: I found an interesting problem on https://atcoder.jp/contests/abc110/tasks/abc110_b:
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B.
The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y.
One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control.
If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out.
 - X < Z \leq Y
 - x_1, x_2, ..., x_N < Z
 - y_1, y_2, ..., y_M \geq Z
Determine if war will break out.

-----Constraints-----
 - All values in input are integers.
 - 1 \leq N, M \leq 100
 - -100 \leq X < Y \leq 100
 - -100 \leq x_i, y_i \leq 100
 - x_1, x_2, ..., x_N \neq X
 - x_i are all different.
 - y_1, y_2, ..., y_M \neq Y
 - y_i are all different.

-----Input-----
Input is given from Standard Input in the following format:
N M X Y
x_1 x_2 ... x_N
y_1 y_2 ... y_M

-----Output-----
If war will break out, print War; otherwise, print No War.

-----Sample Input-----
3 2 10 20
8 15 13
16 22

-----Sample Output-----
No War

The choice Z = 16 satisfies all of the three conditions as follows, thus they will come to an agreement.
 - X = 10 < 16 \leq 20 = Y
 - 8, 15, 13 < 16
 - 16, 22 \geq 16

I tried it in Python, but could not do it. Can you solve it?
targets: n,m,x,y = map(int, input().split())
xl = list(map(int, input().split()))
yl = list(map(int, input().split()))
z = 'War'
for i in range(x+1,y+1):
    if max(xl) < i and i <= min(yl):
        z = 'No War'
        break
print(z)
inputs: Solve in Python:
Further research on zombie thought processes yielded interesting results. As we know from the previous problem, the nervous system of a zombie consists of n brains and m brain connectors joining some pairs of brains together. It was observed that the intellectual abilities of a zombie depend mainly on the topology of its nervous system. More precisely, we define the distance between two brains u and v (1 ≤ u, v ≤ n) as the minimum number of brain connectors used when transmitting a thought between these two brains. The brain latency of a zombie is defined to be the maximum distance between any two of its brains. Researchers conjecture that the brain latency is the crucial parameter which determines how smart a given zombie is. Help them test this conjecture by writing a program to compute brain latencies of nervous systems.

In this problem you may assume that any nervous system given in the input is valid, i.e., it satisfies conditions (1) and (2) from the easy version.


-----Input-----

The first line of the input contains two space-separated integers n and m (1 ≤ n, m ≤ 100000) denoting the number of brains (which are conveniently numbered from 1 to n) and the number of brain connectors in the nervous system, respectively. In the next m lines, descriptions of brain connectors follow. Every connector is given as a pair of brains a b it connects (1 ≤ a, b ≤ n and a ≠ b).


-----Output-----

Print one number – the brain latency.


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

Output
2
Input
5 4
1 2
2 3
3 4
3 5

Output
3
targets: n,m=map(int,input().split())

gr=[[] for i in range(n)]

for i in range(m):
	u,v=map(int,input().split())
	gr[v-1].append(u-1)
	gr[u-1].append(v-1)

v=[False for i in range(n)]
s=[0]
tr={}
tr[0]=0
while s:
	x=s.pop()
	v[x]=True
	for j in gr[x]:
		if v[j]:continue
		s.append(j)
		tr[j]=tr[x]+1
va=0
ma=0
for j in tr.keys():
	if ma<tr[j]:
		ma=tr[j]
		va=j
v=[False for i in range(n)]
s=[va]
tr={}
tr[va]=0
while s:
	x=s.pop()
	v[x]=True
	for i in gr[x]:
		if v[i]:continue
		s.append(i)
		tr[i]=tr[x]+1
print(max(tr.values()))
inputs: I found an interesting problem on https://www.codewars.com/kata/56c0ca8c6d88fdb61b000f06:
You're fed up about changing the version of your software manually. Instead, you will create a little script that will make it for you.

# Exercice

Create a function `nextVersion`, that will take a string in parameter, and will return a string containing the next version number.

For example:

# Rules

All numbers, except the first one, must be lower than 10: if there are, you have to set them to 0 and increment the next number in sequence.

You can assume all tests inputs to be valid.

I tried it in Python, but could not do it. Can you solve it?
targets: def next_version(version):
    c, s = version.count(".") + 1, version.replace(".", "")
    n = f"{int(s)+1:0{len(s)}d}"[::-1]
    return (".".join(n[i] for i in range(c)) + n[c:])[::-1]
inputs: I found an interesting problem on https://www.codewars.com/kata/5a3141fe55519e04d90009d8:
Lеt's create function to play cards. Our rules:

We have the preloaded `deck`:

```
deck = ['joker','2♣','3♣','4♣','5♣','6♣','7♣','8♣','9♣','10♣','J♣','Q♣','K♣','A♣',
                '2♦','3♦','4♦','5♦','6♦','7♦','8♦','9♦','10♦','J♦','Q♦','K♦','A♦',
                '2♥','3♥','4♥','5♥','6♥','7♥','8♥','9♥','10♥','J♥','Q♥','K♥','A♥',
                '2♠','3♠','4♠','5♠','6♠','7♠','8♠','9♠','10♠','J♠','Q♠','K♠','A♠']
```

We have 3 arguments:

`card1` and `card2` - any card of our deck.

`trump` - the main suit of four ('♣', '♦', '♥', '♠').

If both cards have the same suit, the big one wins.

If the cards have different suits (and no one has trump) return 'Let's play again.'

If one card has `trump` unlike another, wins the first one.

If both cards have `trump`, the big one wins.

If `card1` wins, return 'The first card won.' and vice versa.

If the cards are equal, return 'Someone cheats.'

A few games:

```
('3♣', 'Q♣', '♦') -> 'The second card won.'

('5♥', 'A♣', '♦') -> 'Let us play again.'

('8♠', '8♠', '♣') -> 'Someone cheats.'

('2♦', 'A♠', '♦') -> 'The first card won.'

('joker', 'joker', '♦') -> 'Someone cheats.'

```
P.S. As a card you can also get the string 'joker' - it means this card always wins.

I tried it in Python, but could not do it. Can you solve it?
targets: deck = ['joker','2♣','3♣','4♣','5♣','6♣','7♣','8♣','9♣','10♣','J♣','Q♣','K♣','A♣',
                '2♦','3♦','4♦','5♦','6♦','7♦','8♦','9♦','10♦','J♦','Q♦','K♦','A♦',
                '2♥','3♥','4♥','5♥','6♥','7♥','8♥','9♥','10♥','J♥','Q♥','K♥','A♥',
                '2♠','3♠','4♠','5♠','6♠','7♠','8♠','9♠','10♠','J♠','Q♠','K♠','A♠']

def card_game(card_1, card_2, trump):
    if card_1 == card_2:
        return "Someone cheats."
    ordinal, trumps = ["first", "second"], f"{card_1}{card_2}".count(trump)
    if "joker" in (card_1, card_2):
        winner = ordinal[card_2 == "joker"]
    elif trumps == 1:
        winner = ordinal[trump in card_2]
    elif card_1[-1] == card_2[-1]:
        winner = ordinal[deck.index(card_2) > deck.index(card_1)]
    elif trumps == 0:
        return "Let us play again."
    return f"The {winner} card won."
inputs: Solve in Python:
Dasha logged into the system and began to solve problems. One of them is as follows:

Given two sequences a and b of length n each you need to write a sequence c of length n, the i-th element of which is calculated as follows: c_{i} = b_{i} - a_{i}.

About sequences a and b we know that their elements are in the range from l to r. More formally, elements satisfy the following conditions: l ≤ a_{i} ≤ r and l ≤ b_{i} ≤ r. About sequence c we know that all its elements are distinct.

 [Image] 

Dasha wrote a solution to that problem quickly, but checking her work on the standard test was not so easy. Due to an error in the test system only the sequence a and the compressed sequence of the sequence c were known from that test.

Let's give the definition to a compressed sequence. A compressed sequence of sequence c of length n is a sequence p of length n, so that p_{i} equals to the number of integers which are less than or equal to c_{i} in the sequence c. For example, for the sequence c = [250, 200, 300, 100, 50] the compressed sequence will be p = [4, 3, 5, 2, 1]. Pay attention that in c all integers are distinct. Consequently, the compressed sequence contains all integers from 1 to n inclusively.

Help Dasha to find any sequence b for which the calculated compressed sequence of sequence c is correct.


-----Input-----

The first line contains three integers n, l, r (1 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ 10^9) — the length of the sequence and boundaries of the segment where the elements of sequences a and b are.

The next line contains n integers a_1,  a_2,  ...,  a_{n} (l ≤ a_{i} ≤ r) — the elements of the sequence a.

The next line contains n distinct integers p_1,  p_2,  ...,  p_{n} (1 ≤ p_{i} ≤ n) — the compressed sequence of the sequence c.


-----Output-----

If there is no the suitable sequence b, then in the only line print "-1".

Otherwise, in the only line print n integers — the elements of any suitable sequence b.


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

Output
3 1 5 4 2 
Input
4 2 9
3 4 8 9
3 2 1...
targets: R= lambda: list(map(int,input().split()))
n,l,r= R()
a,d=list(R()),list(R())
b=[l for _ in range(n)]
c=[0 for _ in range(n)]
for i in range(n):
    c[d[i]-1]=i
dif=l-a[c[0]]
for x in c[1:]:
    if b[x]-a[x]<=dif:  
        b[x]=a[x]+dif+1
    dif=b[x]-a[x]
    if b[x]>r:
        print("-1")
        return
print(" ".join(map(str,b)))
inputs: I found an interesting problem on https://www.codewars.com/kata/549c7ae26d86c7c3ed000b87:
For all x in the range of integers [0, 2 ** n), let y[x] be the binary exclusive-or of x and x // 2. Find the sum of all numbers in y.

Write a function sum_them that, given n, will return the value of the above sum.

This can be implemented a simple loop as shown in the initial code. But once n starts getting to higher numbers, such as 2000 (which will be tested), the loop is too slow.

There is a simple solution that can quickly find the sum. Find it!

Assume that n is a nonnegative integer.

Hint: The complete solution can be written in two lines.

I tried it in Python, but could not do it. Can you solve it?
targets: def sum_them(n):
    return  2 ** (n*2-1) - 2 ** (n-1)
inputs: Solve in Python:
Given two integer arrays where the second array is a shuffled duplicate of the first array with one element missing, find the missing element.

Please note, there may be duplicates in the arrays, so checking if a numerical value exists in one and not the other is not a valid solution.

```
find_missing([1, 2, 2, 3], [1, 2, 3]) => 2
```
```
find_missing([6, 1, 3, 6, 8, 2], [3, 6, 6, 1, 2]) => 8
```

The first array will always have at least one element.
targets: def find_missing(a, b):
    for x in b:
        a.remove(x)

    return a[0]
inputs: I found an interesting problem on https://leetcode.com/problems/decode-string/:
Given an encoded string, return it's decoded string.


The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.


You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc.

Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there won't be input like 3a or 2[4].


Examples:

s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2[c]]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef".

I tried it in Python, but could not do it. Can you solve it?
targets: class Solution:
     def decodeString(self, s):
         """
         :type s: str
         :rtype: str
         """
         stack_num = []
         stack_str = []
         num = ''
         string = ''
         for c in s:
             if c.isdigit():
                 if num == '':
                     stack_str.append(string)
                     string = ''
                 num += c
             elif c == '[':
                 stack_num.append(int(num))
                 num = ''
             elif c == ']':
                 string = stack_str.pop() + string * stack_num.pop()
             else:
                 string += c
         return string
inputs: I found an interesting problem on https://www.codechef.com/problems/CHEFADV:
Mysterious Chefland… Recently, Chef realised that Discuss, the educational system of Chefland, is out of date. Therefore, he is trying to find ways to update the infrastructure in the country. One possible way is to move all materials from Discuss to Discourse.
Chef will have access to Discourse if his knowledge and power become exactly equal to $N$ and $M$ respectively. Initially, he has power $1$ and knowledge $1$.
Chef can perform actions of the following types to improve his skills:
- solve a problem — increase his knowledge by $X$
- do a push-up — increase his power by $Y$
- install ShareChat to keep in touch with friends — increase both knowledge and power by $1$
Chef can only install ShareChat at most once. The remaining actions may be performed any number of times and the actions may be performed in any order.
Help Chef find out whether it is possible to move from Discuss to Discourse.

-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first and only line of each test case contains four space-separated integers $N$, $M$, $X$ and $Y$.

-----Output-----
For each test case, print a single line containing the string "Chefirnemo" if it is possible to reach the required knowledge and power or "Pofik" if it is impossible.

-----Constraints-----
- $1 \le T \le 1,000$
- $1 \le N, M, X, Y \le 10^9$

-----Subtasks-----
Subtask #1 (30 points): $1 \le N, M, X, Y \le 100$
Subtask #2 (70 points): original constraints

-----Example Input-----
5
2 2 1 2
11 10 5 9
11 11 5 9
12 11 5 9
1 2 1 100

-----Example Output-----
Chefirnemo
Chefirnemo
Pofik
Chefirnemo
Pofik

-----Explanation-----
Example case 2: We add $Y=9$ once to the power to get power $10$. We add $X=5$ twice to the knowledge to get knowledge $11$.
Example case 3: We can see that it is impossible to reach power $M=11$ no matter which or how many operations we do. Note that the ShareChat operation will increase both knowledge and power by $1$, and...

I tried it in Python, but could not do it. Can you solve it?
targets: t = int(input())
for _ in range(t):
 n, m, x, y = (int(x) for x in input().split())
 a = (n-1)%x == 0 and (m-1)%y == 0
 b = (n-2)%x == 0 and (m-2)%y == 0 and min(n,m)>1
 if a or b :
  print("Chefirnemo")
 else:
  print("Pofik")
inputs: Solve in Python:
You are given two strings s and t of the same length. You want to change s to t. Changing the i-th character of s to i-th character of t costs |s[i] - t[i]| that is, the absolute difference between the ASCII values of the characters.
You are also given an integer maxCost.
Return the maximum length of a substring of s that can be changed to be the same as the corresponding substring of twith a cost less than or equal to maxCost.
If there is no substring from s that can be changed to its corresponding substring from t, return 0.
 
Example 1:
Input: s = "abcd", t = "bcdf", maxCost = 3
Output: 3
Explanation: "abc" of s can change to "bcd". That costs 3, so the maximum length is 3.
Example 2:
Input: s = "abcd", t = "cdef", maxCost = 3
Output: 1
Explanation: Each character in s costs 2 to change to charactor in t, so the maximum length is 1.

Example 3:
Input: s = "abcd", t = "acde", maxCost = 0
Output: 1
Explanation: You can't make any change, so the maximum length is 1.

 
Constraints:

1 <= s.length, t.length <= 10^5
0 <= maxCost <= 10^6
s and t only contain lower case English letters.
targets: class Solution:
    def equalSubstring(self, s: str, t: str, maxCost: int) -> int:
        i = 0
        for j in range(len(s)):
            maxCost -= abs(ord(s[j]) - ord(t[j]))
            if maxCost < 0:
                maxCost += abs(ord(s[i]) - ord(t[i]))
                i += 1
        return j - i + 1
inputs: I found an interesting problem on https://www.codewars.com/kata/5720a81309e1f9b232001c5b:
# RegExp Fun #1 - When I miss few days of gym

## Disclaimer

The background story of this Kata is 100% fiction.  Any resemblance to real people or real events is **nothing more than a coincidence** and should be regarded as such.

## Background Story

You are a person who loves to go to the gym everyday with the squad of people that you've known since early childhood.  However, you recently contracted a sickness that forced you to stay at home for over a week.  As you see your body getting weaker and weaker every day and as you see your biceps and triceps disappearing, you can't help but lay in bed and cry.  You're usually an optimistic person but this time negative thoughts come to your head ... 

![When I miss few days of gym](https://pics.onsizzle.com/Instagram-faf8c9.png)


## Task

As can be seen from the funny image above (or am I the only person to find the picture above hilarious?) there is lots of slang.  Your task is to define a function ```gymSlang``` which accepts a string argument and does the following:

1. Replace *all* instances of ```"probably"``` to ```"prolly"```
2. Replace *all* instances of ```"i am"``` to ```"i'm"```
3. Replace *all* instances of ```"instagram"``` to ```"insta"```
4. Replace *all* instances of ```"do not"``` to ```"don't"```
5. Replace *all* instances of ```"going to"``` to ```"gonna"```
6. Replace *all* instances of ```"combination"``` to ```"combo"```

Your replacement regexes **should be case-sensitive**, only replacing the words above with slang if the detected pattern is in **lowercase**.  However, please note that apart from 100% lowercase matches, you will **also have to replace matches that are correctly capitalized** (e.g. ```"Probably" => "Prolly"``` or ```"Instagram" => "Insta"```).

Finally, your code will be tested to make sure that you have used **RegExp** replace in your code.

Enjoy :D

I tried it in Python, but could not do it. Can you solve it?
targets: from functools import reduce
REPLS = (
    ('probably', 'prolly'), ('Probably', 'Prolly'), ('i am', "i'm"),
    ('I am', "I'm"), ('instagram', 'insta'), ('Instagram', 'Insta'),
    ('do not', "don't"), ('Do not', "Don't"), ('going to', 'gonna'),
    ('Going to', 'Gonna'), ('combination', 'combo'), ('Combination', 'Combo')
)


def gym_slang(s, _=None):
    return reduce(lambda a, kv: a.replace(*kv), REPLS, s)
inputs: Solve in Python:
In this Kata, you will create a function that converts a string with letters and numbers to the inverse of that string (with regards to Alpha and Numeric characters). So, e.g. the letter `a` will become `1` and number `1` will become `a`; `z` will become `26` and `26` will become `z`.

Example: `"a25bz"` would become `"1y226"`


Numbers representing letters (`n <= 26`) will always be separated by letters, for all test cases: 

* `"a26b"` may be tested, but not `"a262b"`
* `"cjw9k"` may be tested, but not `"cjw99k"`

A list named `alphabet` is preloaded for you: `['a', 'b', 'c', ...]`

A dictionary of letters and their number equivalent is also preloaded for you called `alphabetnums = {'a': '1', 'b': '2', 'c': '3', ...}`
targets: import re

def AlphaNum_NumAlpha(string):
    return ''.join(chr(int(e)+96) if e.isdigit() else str(ord(e)-96) for e in re.split('([a-z])', string) if e)
inputs: Solve in Python:
Although Inzane successfully found his beloved bone, Zane, his owner, has yet to return. To search for Zane, he would need a lot of money, of which he sadly has none. To deal with the problem, he has decided to hack the banks. [Image] 

There are n banks, numbered from 1 to n. There are also n - 1 wires connecting the banks. All banks are initially online. Each bank also has its initial strength: bank i has initial strength a_{i}.

Let us define some keywords before we proceed. Bank i and bank j are neighboring if and only if there exists a wire directly connecting them. Bank i and bank j are semi-neighboring if and only if there exists an online bank k such that bank i and bank k are neighboring and bank k and bank j are neighboring.

When a bank is hacked, it becomes offline (and no longer online), and other banks that are neighboring or semi-neighboring to it have their strengths increased by 1.

To start his plan, Inzane will choose a bank to hack first. Indeed, the strength of such bank must not exceed the strength of his computer. After this, he will repeatedly choose some bank to hack next until all the banks are hacked, but he can continue to hack bank x if and only if all these conditions are met:  Bank x is online. That is, bank x is not hacked yet.  Bank x is neighboring to some offline bank.  The strength of bank x is less than or equal to the strength of Inzane's computer. 

Determine the minimum strength of the computer Inzane needs to hack all the banks.


-----Input-----

The first line contains one integer n (1 ≤ n ≤ 3·10^5) — the total number of banks.

The second line contains n integers a_1, a_2, ..., a_{n} ( - 10^9 ≤ a_{i} ≤ 10^9) — the strengths of the banks.

Each of the next n - 1 lines contains two integers u_{i} and v_{i} (1 ≤ u_{i}, v_{i} ≤ n, u_{i} ≠ v_{i}) — meaning that there is a wire directly connecting banks u_{i} and v_{i}.

It is guaranteed that the wires connect the banks in such a way that Inzane can somehow hack all the banks using a computer with appropriate...
targets: import sys
def solve():
	n=int(sys.stdin.readline())
	d=list(map(int,sys.stdin.readline().split()))
	s=[[] for g in d]
	mx_tmp=max(d)
	mx_tmp2=max(g for g in d+[-2e9] if g<mx_tmp)
	mxpt=[mx_tmp,mx_tmp2]
	mxcnt=[d.count(mx_tmp),d.count(mx_tmp2)]
	for i in range(1,n):
		a,b=map(int,sys.stdin.readline().split())
		a-=1;
		b-=1;
		s[a]+=[b]
		s[b]+=[a]

	mx=int(2e9)
	for i in range(n):
		nmx=[]+mxcnt
		tmpmax=d[i]
		for k in s[i]:
			if d[k]==mxpt[0]:			
				nmx[0]-=1
			elif d[k]==mxpt[1]:			
				nmx[1]-=1

		if nmx[0]!=mxcnt[0]:
			tmpmax=mxpt[0]+1
		elif nmx[1]!=mxcnt[1]:
			tmpmax=max(tmpmax,mxpt[1]+1)

		if d[i]==mxpt[0]:
			nmx[0]-=1
		elif d[i]==mxpt[1]:
			nmx[1]-=1

		if nmx[0]:
			tmpmax=mxpt[0]+2
		elif nmx[1]:
			tmpmax=max(mxpt[1]+2,tmpmax)
		mx=min(mx,tmpmax)
	print(mx)

def __starting_point():
    solve()
__starting_point()
inputs: Solve in Python:
Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.
targets: class Solution:
     def maxArea(self, height):
         """
         :type height: List[int]
         :rtype: int
         """
         left_index = 0
         right_index = len(height) - 1
         water = 0
         while True:
             if left_index >= right_index:
                 break
             left_height = height[left_index]
             right_height = height[right_index]
             water = max(water, (right_index - left_index) * min(left_height, right_height))
             if left_height < right_height:
                 left_index += 1
             else:
                 right_index -= 1
         return water
inputs: I found an interesting problem on https://www.codewars.com/kata/56044de2aa75e28875000017:
You have to create a method "compoundArray" which should take as input two int arrays of different length and return one int array with numbers of both arrays shuffled one by one. 
```Example: 
Input - {1,2,3,4,5,6} and {9,8,7,6} 
Output - {1,9,2,8,3,7,4,6,5,6}
```

I tried it in Python, but could not do it. Can you solve it?
targets: def compound_array(a, b):
    answer = []
    while a or b:
        if a: answer.append(a.pop(0))
        if b: answer.append(b.pop(0))
    return answer
inputs: I found an interesting problem on https://www.codewars.com/kata/54df2067ecaa226eca000229:
Due to another of his misbehaved, 
the primary school's teacher of the young Gauß,  Herr J.G. Büttner, to keep the bored and unruly young schoolboy Karl Friedrich Gauss busy for a good long time, while he teaching arithmetic to his mates,
assigned him the problem of adding up all the whole numbers from 1 through a given number `n`.

Your task is to help the young Carl Friedrich to solve this problem as quickly as you can; so, he can astonish his teacher and rescue his recreation interval.

Here's, an example:

```
f(n=100) // returns 5050 
```

It's your duty to verify that n is a valid positive integer number. If not, please, return false (None for Python, null for C#).

> **Note:** the goal of this kata is to invite you to think about some 'basic' mathematic formula and how you can do performance optimization on your code. 

> Advanced - experienced users should try to solve it in one line, without loops, or optimizing the code as much as they can.

-----

**Credits:** this kata was inspired by the farzher's kata 'Sum of large ints' . In fact, it can be seen as a sort of prep kata for that one.

I tried it in Python, but could not do it. Can you solve it?
targets: def f(n):
    
    if isinstance(n, int) and n >0:

        return (1/2)*n*(n+1)
    
    else:
        
        return None
inputs: I found an interesting problem on https://www.codechef.com/LTIME40/problems/LTM40AB:
Chef likes inequalities. Please help him to solve next one.
Given four integers a, b, c, d. Find number of solutions x < y, where a ≤ x ≤ b and c ≤ y ≤ d and x, y integers.

-----Input-----
The first line contains an integer T denoting number of tests.
First line of each test case contains four positive integer numbers a, b, c and d.

-----Output-----
For each test case, output a single number each in separate line denoting number of integer solutions as asked in the problem.

-----Constraints-----
- 1 ≤ T ≤ 20 
- 1 ≤ a, b, c, d ≤ 106 

-----Subtasks-----
- Subtask #1: (30 points)  1 ≤ a, b, c, d ≤ 103.
- Subtask #2: (70 points)  Original constraints.

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

Output:3

Input:1
2 999999 1 1000000

Output:499998500001

I tried it in Python, but could not do it. Can you solve it?
targets: # Akhilesh Ravi
# Codechef - Chef and Inequality

T = int(input())
l = []
for i in range(T):
 l += [[int(j) for j in input().split()]]

l1 = []
for i in l:
 a,b,c,d = tuple(i)
 if a >= d:
  l1 += [0]
  
 elif b < c:
  l1 += [(b-a+1)*(d-c+1)]
  
 elif c <= a <= d <= b:
  n = d-a
  l1 += [n*(n+1)/2]
  
 elif c <= a <= b <= d:
  l1 += [(d-a) * (d-a+1)/2
    - (d-b-1) * (d-b)/2]
  
 elif a < c <= d <= b:
  l1 += [ (d-c+1) * (c-a)
    + (d-c) * (d-c+1)/2 ]
  
 elif a < c <= b <= d:
  l1 += [ (d-c+1) * (c-a)
    + (d-c) * (d-c+1)/2
    - (d-b-1) * (d-b)/2]
  
 else:
  l1 += [0]

for i in l1:
 print(i)
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1165/E:
You are given two arrays $a$ and $b$, both of length $n$.

Let's define a function $f(l, r) = \sum\limits_{l \le i \le r} a_i \cdot b_i$.

Your task is to reorder the elements (choose an arbitrary order of elements) of the array $b$ to minimize the value of $\sum\limits_{1 \le l \le r \le n} f(l, r)$. Since the answer can be very large, you have to print it modulo $998244353$. Note that you should minimize the answer but not its remainder.


-----Input-----

The first line of the input contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in $a$ and $b$.

The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^6$), where $a_i$ is the $i$-th element of $a$.

The third line of the input contains $n$ integers $b_1, b_2, \dots, b_n$ ($1 \le b_j \le 10^6$), where $b_j$ is the $j$-th element of $b$.


-----Output-----

Print one integer — the minimum possible value of $\sum\limits_{1 \le l \le r \le n} f(l, r)$ after rearranging elements of $b$, taken modulo $998244353$. Note that you should minimize the answer but not its remainder.


-----Examples-----
Input
5
1 8 7 2 4
9 7 2 9 3

Output
646

Input
1
1000000
1000000

Output
757402647

Input
2
1 3
4 2

Output
20

I tried it in Python, but could not do it. Can you solve it?
targets: n = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
for i in range(n):
    A[i] *= (i + 1) * (n - i)
A.sort()
B.sort(reverse=True)
C = []
cnt = 0
M = 998244353
for i in range(n):
    cnt += A[i] * B[i]
    cnt %= M
print(cnt)
inputs: Solve in Python:
An integer partition of n is a weakly decreasing list of positive integers which sum to n.

For example, there are 7 integer partitions of 5:

[5], [4,1], [3,2], [3,1,1], [2,2,1], [2,1,1,1], [1,1,1,1,1].

Write a function named partitions which returns the number of integer partitions of n.  The function should be able to find the number of integer partitions of n for n as least as large as 100.
targets: def partitions(n, k=1, cache={}):
    if k > n: return 0
    if n == k: return 1
    if (n,k) in cache: return cache[n,k]
    return cache.setdefault((n,k), partitions(n, k+1) + partitions(n-k, k))
inputs: I found an interesting problem on https://www.codechef.com/problems/WGHTNUM:
VK gave a problem to Chef, but Chef is too lazy, so he asked you to solve the problem for him. The statement of the problem follows.
Consider an integer with $N$ digits (in decimal notation, without leading zeroes) $D_1, D_2, D_3, \dots, D_N$. Here, $D_1$ is the most significant digit and $D_N$ the least significant. The weight of this integer is defined as ∑i=2N(Di−Di−1).∑i=2N(Di−Di−1).\sum_{i=2}^N (D_i - D_{i-1})\,.
You are given integers $N$ and $W$. Find the number of positive integers with $N$ digits (without leading zeroes) and weight equal to $W$. Compute this number modulo $10^9+7$.

-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first and only line of each test case contains two space-separated integers $N$ and $W$ denoting the number of digits and the required weight.

-----Output-----
For each test case, print a single line containing one integer — the number of $N$-digit positive integers with weight $W$, modulo $10^9+7$.

-----Constraints-----
- $1 \le T \le 10^5$
- $2 \le N \le 10^{18}$
- $|W| \le 300$

-----Subtasks-----
Subtask #1 (20 points):
- $1 \le T \le 10^3$
- $2 \le N \le 10^3$
Subtask #2 (80 points): original constraints

-----Example Input-----

1
2 3

-----Example Output-----

6

-----Explanation-----
Example case 1: Remember that the digits are arranged from most significant to least significant as $D_1, D_2$. The two-digit integers with weight $3$ are $14, 25, 36, 47, 58, 69$. For example, the weight of $14$ is $D_2-D_1 = 4-1 = 3$. We can see that there are no other possible numbers.

I tried it in Python, but could not do it. Can you solve it?
targets: m = 10**9 + 7
def findNumbers(n, w): 
 x,s = 0,0 
 if (w >= 0 and w <= 8): 
  x = 9 - w; 
 elif (w >= -9 and w <= -1): 
  x = 10 + w;     
 s = (pow(10, n - 2,m)*(x%m));   
 return s%m; 
for _ in range(int(input())):
 n,w = map(int,input().split())
 print(findNumbers(n,w))
inputs: Solve in Python:
In this kata, the number 0 is infected. You are given a list. Every turn, any item in the list that is adjacent to a 0 becomes infected and transforms into a 0. How many turns will it take for the whole list to become infected?

```
[0,1,1,0] ==> [0,0,0,0] 
All infected in 1 turn.

[1,1,0,1,1] --> [1,0,0,0,1] --> [0,0,0,0,0]
All infected in 2 turns

[0,1,1,1] --> [0,0,1,1] --> [0,0,0,1] --> [0,0,0,0]
All infected in 3 turns.
```

All lists will contain at least one item, and at least one zero, and the only items will be 0s and 1s. Lists may be very very long, so pure brute force approach will not work.
targets: def infected_zeroes(lst):
    one = "".join(str(d) for d in lst).split("0")
    return max(len(one[0]), len(one[-1]), *((len(o) + 1)//2 for o in one[1:-1]))
inputs: Solve in Python:
Fox Ciel is playing a game with numbers now. 

Ciel has n positive integers: x_1, x_2, ..., x_{n}. She can do the following operation as many times as needed: select two different indexes i and j such that x_{i} > x_{j} hold, and then apply assignment x_{i} = x_{i} - x_{j}. The goal is to make the sum of all numbers as small as possible.

Please help Ciel to find this minimal sum.


-----Input-----

The first line contains an integer n (2 ≤ n ≤ 100). Then the second line contains n integers: x_1, x_2, ..., x_{n} (1 ≤ x_{i} ≤ 100).


-----Output-----

Output a single integer — the required minimal sum.


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

Output
2

Input
3
2 4 6

Output
6

Input
2
12 18

Output
12

Input
5
45 12 27 30 18

Output
15



-----Note-----

In the first example the optimal way is to do the assignment: x_2 = x_2 - x_1.

In the second example the optimal sequence of operations is: x_3 = x_3 - x_2, x_2 = x_2 - x_1.
targets: n = int(input())
a = list(map(int, input().split()))
while(1):
    T = True
    for i in range(n):
        for j in range(n):
            if a[i] > a[j]:
                a[i] = a[i] - a[j]
                T = False
            elif a[i] < a[j]:
                a[j] = a[j] - a[i]
                T = False
    if T:
        break
print(sum(a))
inputs: Solve in Python:
You are given an integer N. Find the number of strings of length N that satisfy the following conditions, modulo 10^9+7:
 - The string does not contain characters other than A, C, G and T.
 - The string does not contain AGC as a substring.
 - The condition above cannot be violated by swapping two adjacent characters once.

-----Notes-----
A substring of a string T is a string obtained by removing zero or more characters from the beginning and the end of T.
For example, the substrings of ATCODER include TCO, AT, CODER, ATCODER and  (the empty string), but not AC.

-----Constraints-----
 - 3 \leq N \leq 100

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

-----Output-----
Print the number of strings of length N that satisfy the following conditions, modulo 10^9+7.

-----Sample Input-----
3

-----Sample Output-----
61

There are 4^3 = 64 strings of length 3 that do not contain characters other than A, C, G and T. Among them, only AGC, ACG and GAC violate the condition, so the answer is 64 - 3 = 61.
targets: import numpy as np

N = int(input())
mod = 10**9+7

dp = np.zeros([N+1,4,4,4],dtype=int)
dp[0,-1,-1,-1] = 1

for i in range(N):
  for c1 in range(4):
    for c2 in range(4):
      for c3 in range(4):
        if not dp[i,c1,c2,c3]:continue
        for a in range(4):
          if c1 == 0 and c3 == 1 and a == 2:continue
          if c1 == 0 and c2 == 1 and a == 2:continue
          if c2 == 0 and c3 == 1 and a == 2:continue
          if c2 == 0 and c3 == 2 and a == 1:continue
          if c2 == 1 and c3 == 0 and a == 2:continue
          dp[i+1,c2,c3,a] += dp[i,c1,c2,c3]
          dp[i+1,c2,c3,a] %= mod

print(dp[-1].sum()%mod)
inputs: I found an interesting problem on https://leetcode.com/problems/search-a-2d-matrix/:
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:


       Integers in each row are sorted from left to right.
       The first integer of each row is greater than the last integer of the previous row.


Example 1:


Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true


Example 2:


Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false

I tried it in Python, but could not do it. Can you solve it?
targets: class Solution:
     def searchMatrix(self, matrix, target):
         if not matrix or target is None:
             return False
         n = len(matrix[0])
         lo, hi = 0, len(matrix) * n
         while lo < hi:
             mid = (lo + hi) / 2
             x = matrix[int(mid/n)][int(mid%n)]
             if x < target:
                 lo = mid + 1
             elif x > target:
                 hi = mid
             else:
                 return True
         return False
         """
         :type matrix: List[List[int]]
         :type target: int
         :rtype: bool
         """
inputs: I found an interesting problem on https://www.codewars.com/kata/5711fc7c159cde6ac70003e2:
You are given an array of positive and negative integers and a number ```n``` and ```n > 1```. The array may have elements that occurs more than once.
Find all the combinations of n elements of the array that their sum are 0.
```python
arr = [1, -1, 2, 3, -2]
n = 3
find_zero_sum_groups(arr, n) == [-2, -1, 3] # -2 - 1 + 3 = 0
```
The function should ouput every combination or group in increasing order.

We may have more than one group:
```python
arr = [1, -1, 2, 3, -2, 4, 5, -3 ]
n = 3
find_zero_sum_groups(arr, n) == [[-3, -2, 5], [-3, -1, 4], [-3, 1, 2], [-2, -1, 3]]
```
In the case above the function should output a sorted 2D array.

The function will not give a group twice, or more, only once.

```python
arr = [1, -1, 2, 3, -2, 4, 5, -3, -3, -1, 2, 1, 4, 5, -3 ]
n = 3
find_zero_sum_groups(arr, n) == [[-3, -2, 5], [-3, -1, 4], [-3, 1, 2], [-2, -1, 3]]
```
If there are no combinations with sum equals to 0, the function will output an alerting message.

```python
arr = [1, 1, 2, 3]
n = 2
find_zero_sum_groups(arr, n) == "No combinations"
```
If the function receives an empty array will output an specific alert:
```python
arr = []
n = 2
find_zero_sum_groups(arr, n) == "No elements to combine"
```
As you have seen the solutions may have a value occurring only once.
Enjoy it!

I tried it in Python, but could not do it. Can you solve it?
targets: from itertools import combinations
def find_zero_sum_groups(a, n):
    li = sorted(map(sorted,filter(lambda x:sum(x)==0,combinations(set(a),n))))
    return li if len(li)>1 else li[0] if li else "No combinations" if a else "No elements to combine"
inputs: I found an interesting problem on https://www.codechef.com/problems/SEBIHWY:
Sebi goes to school daily with his father. They cross a big highway in the car to reach to the school. Sebi sits in front seat beside his father at driving seat. To kill boredom, they play a game of guessing speed of other cars on the highway. Sebi makes a guess of other car's speed being SG kph, his father FG kph. 

The highway is usually empty, so the drivers use cruise control, i.e. vehicles run at a constant speed. There are markers on the highway at a gap of 50 meters. Both father-son duo wants to check the accuracy of their guesses. For that, they start a timer at the instant at which their car and the other car (which speed they are guessing) are parallel to each other (they need not to be against some marker, they can be in between the markers too). After some T seconds, they observe that both the cars are next to some markers and the number of markers in between the markers of their car and the other car is D - 1 (excluding the markers next to both the cars). Also, they can observe these markers easily because the other car is faster than their. Speed of Sebi's father's car is S. Using this information, one can find the speed of the other car accurately.

An example situation when Sebi's father starts the timer. Notice that both the car's are parallel to each other.

Example situation after T seconds. The cars are next to the markers. Here the value of D is 1. The green car is Sebi's and the other car is of blue color.

Sebi's a child, he does not know how to find the check whose guess is close to the real speed of the car. He does not trust his father as he thinks that he might cheat. Can you help to resolve this issue between them by telling whose guess is closer. If Sebi's guess is better, output "SEBI". If his father's guess is better, output "FATHER". If both the guess are equally close, then output "DRAW".

-----Input-----
The first line of the input contains an integer T denoting the number of test cases. 
Each of the next T lines contain five space separated integers S, SG, FG, D, T...

I tried it in Python, but could not do it. Can you solve it?
targets: # cook your dish here
n=int(input())
for i in range(n):
    S, SG, FG, D, T = map(int, input().split())
    speed = (D*180)/T + S
    if abs(SG-speed) == abs(FG-speed):
        print('DRAW')
    elif abs(SG-speed) > abs(FG-speed):
        print('FATHER')
    else:
        print('SEBI')
inputs: Solve in Python:
A *[Hamming number][1]* is a positive integer of the form 2*i*3*j*5*k*, for some non-negative integers *i*, *j*, and *k*.

Write a function that computes the *n*th smallest Hamming number. 

Specifically:

  - The first smallest Hamming number is 1 = 2^(0)3^(0)5^(0)
  - The second smallest Hamming number is 2 = 2^(1)3^(0)5^(0)
  - The third smallest Hamming number is 3 = 2^(0)3^(1)5^(0)
  - The fourth smallest Hamming number is 4 = 2^(2)3^(0)5^(0)
  - The fifth smallest Hamming number is 5 = 2^(0)3^(0)5^(1)

The 20 smallest Hamming numbers are given in example test fixture.

Your code should be able to compute all of the smallest 5,000 (Clojure: 2000, NASM: 13282) Hamming numbers without timing out.

[1]:https://en.wikipedia.org/wiki/Regular_number
targets: def hamming(n):
    bases = [2, 3, 5]
    expos = [0, 0, 0]
    hamms = [1]
    for _ in range(1, n):
        next_hamms = [bases[i] * hamms[expos[i]] for i in range(3)]
        next_hamm = min(next_hamms)
        hamms.append(next_hamm)
        for i in range(3):
            expos[i] += int(next_hamms[i] == next_hamm)
    return hamms[-1]
inputs: Solve in Python:
Takahashi is standing on a multiplication table with infinitely many rows and columns.
The square (i,j) contains the integer i \times j. Initially, Takahashi is standing at (1,1).
In one move, he can move from (i,j) to either (i+1,j) or (i,j+1).
Given an integer N, find the minimum number of moves needed to reach a square that contains N.

-----Constraints-----
 - 2 \leq N \leq 10^{12}
 - N is an integer.

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

-----Output-----
Print the minimum number of moves needed to reach a square that contains the integer N.

-----Sample Input-----
10

-----Sample Output-----
5

(2,5) can be reached in five moves. We cannot reach a square that contains 10 in less than five moves.
targets: N = int(input())
N_ri = round(pow(N, 1/2))

for i in range(N_ri, 0, -1):
  if N % i == 0:
    j = N // i
    break
print(i + j - 2)
inputs: Solve in Python:
[Image] 


-----Input-----

The input contains a single integer $a$ ($0 \le a \le 15$).


-----Output-----

Output a single integer.


-----Example-----
Input
3

Output
13
targets: n = int(input())

a = [0] * 4

i = 0
for i in range(4):
    a[i] = n % 2
    n = n // 2


a[3] = a[3] ^ 1
a[2] = a[3] ^ a[2]
a[1] = (a[3] and a[2]) ^ a[1]
a[0] = (a[3] and a[2] and a[1]) ^ a[0]

sum_ = 0
for i in range(4):
    sum_ += a[i] * 2 ** i

print(sum_)
inputs: Solve in Python:
# Task
 Consider the following algorithm for constructing 26 strings S(1) .. S(26):
```
S(1) = "a";
For i in [2, 3, ..., 26]:
S(i) = S(i - 1) + character(i) + S(i - 1).```

 For example:
```
S(1) = "a"
S(2) = S(1) + "b" + S(1) = "a" + "b" + "a" = "aba"
S(3) = S(2) + "c" + S(2) = "aba" + "c" +"aba" = "abacaba"
...
S(26) = S(25) + "z" + S(25)```
Finally, we got a long string S(26). Your task is to find the `k`th symbol (indexing from 1) in the string S(26). All strings consist of lowercase letters only.

# Input / Output


 - `[input]` integer `k`

  1 ≤ k < 2^(26)


 - `[output]` a string(char in C#)

  the `k`th symbol of S(26)
targets: def abacaba(k):
    alph = 'abcdefghijklmnopqrstuvwxyz'
    s = ''
    for i in range(26):
        s = s + alph[i] + s
    return s[k-1]
inputs: I found an interesting problem on https://atcoder.jp/contests/abc177/tasks/abc177_b:
Given are two strings S and T.
Let us change some of the characters in S so that T will be a substring of S.
At least how many characters do we need to change?
Here, a substring is a consecutive subsequence. For example, xxx is a substring of yxxxy, but not a substring of xxyxx.

-----Constraints-----
 - The lengths of S and T are each at least 1 and at most 1000.
 - The length of T is at most that of S.
 - S and T consist of lowercase English letters.

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

-----Output-----
Print the minimum number of characters in S that need to be changed.

-----Sample Input-----
cabacc
abc

-----Sample Output-----
1

For example, changing the fourth character a in S to c will match the second through fourth characters in S to T.
Since S itself does not have T as its substring, this number of changes - one - is the minimum needed.

I tried it in Python, but could not do it. Can you solve it?
targets: s=input()
t=input()
ls=len(s)
lt=len(t)
ans=ls
for i in range(ls-lt+1):
    cnt=0
    for j in range(lt):
        if s[i+j]!=t[j]:
            cnt+=1
    ans=min(ans,cnt)
            
print(ans)
inputs: Solve in Python:
Vasya and Petya wrote down all integers from 1 to n to play the "powers" game (n can be quite large; however, Vasya and Petya are not confused by this fact).

Players choose numbers in turn (Vasya chooses first). If some number x is chosen at the current turn, it is forbidden to choose x or all of its other positive integer powers (that is, x^2, x^3, ...) at the next turns. For instance, if the number 9 is chosen at the first turn, one cannot choose 9 or 81 later, while it is still allowed to choose 3 or 27. The one who cannot make a move loses.

Who wins if both Vasya and Petya play optimally?


-----Input-----

Input contains single integer n (1 ≤ n ≤ 10^9).


-----Output-----

Print the name of the winner — "Vasya" or "Petya" (without quotes).


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

Output
Vasya

Input
2

Output
Petya

Input
8

Output
Petya



-----Note-----

In the first sample Vasya will choose 1 and win immediately.

In the second sample no matter which number Vasya chooses during his first turn, Petya can choose the remaining number and win.
targets: from sys import stdin, stdout
import math, collections
mod = 10**9+7

def isPower(n):
    if (n <= 1):
        return True
    for x in range(2, (int)(math.sqrt(n)) + 1):
        p = x
        while (p <= n):
            p = p * x
            if (p == n):
                return True

    return False
n = int(input())
arr = [0,1,2,1,4,3,2,1,5,6,2,1,8,7,5,9,8,7,3,4,7,4,2,1,10,9,3,6,11,12]
ans = arr[int(math.log(n, 2))]
s = int(math.log(n, 2))
for i in range(3, int(n**0.5)+1):
    if not isPower(i):
        ans^=arr[int(math.log(n, i))]
        s+=int(math.log(n, i))
ans^=((n-s)%2)
print("Vasya" if ans else "Petya")