My answers for tests on http://codility.com that I passed for company http://toptal.com I use Python language to solve problems.

tests_for_toptal_on_codility.py

#!/usr/bin/env python
# -*- coding: utf-8 -*-

#
# Test that I passed on codility.com for TopTal company
#


# Task #1
def binary_gap(N):
    '''
    A binary gap within a positive integer N is any maximal
    sequence of consecutive zeros that is surrounded by ones
    at both ends in the binary representation of N.

    Args:
      - N: integer within the range [1..2,147,483,647]
    '''
    bin_representation = bin(N)[2:]
    max_gap = 0
    gap_counter = 0
    gap_started = False
    for symbol in bin_representation:
        if symbol == '1':
            if gap_counter > max_gap:
                max_gap = gap_counter
            gap_counter = 0
            gap_started = True
        elif gap_started:
            gap_counter += 1
    return max_gap

print binary_gap(1041)



# Task #2
def count_div(A, B, K):
    '''
    Returns the number of integers within the range [A..B] that are divisible by K.

    Used generators to save memory on large amounts of data.

    Args:
      - A: is an integer within the range [0..2,000,000,000]
      - B: is an integer within the range [0..2,000,000,000] and A <= B
      - K: is an integer within the range [1..2,000,000,000]
    '''
    divs_count = 0
    for x in xrange(A, B + 1):
        if (x % K) == 0:
            divs_count += 1
    return divs_count

print count_div(1, 200000000, 1000)



# Task #3
def triangle(A):
    '''
    Calculate triangel of integers, where sentense of numbers P, Q, R
    correspond to next rules:
     - P + Q > R
     - Q + R > P
     - R + P > Q

    Args:
      - A: list of integers, where we will search triangle

    Return: 1 - if triangle exists, and 0 - otherwise
    '''
    A = tuple(enumerate(A))
    for p, P in A:
        for q, Q in A[p + 1:]:
            for r, R in A[q + 1:]:
                if (P + Q > R) and (Q + R > P) and (R + P > Q):
                    return 1 
    return 0

print triangle([10, 2, 5, 1, 8, 20])

Alternative solution for Task 3:
def isTriangle (arr):

# If the number of elements  
# is less than 3, then  
# a triangle isn’t possible 

N = len(arr)  
if N< 3: 
    return False
  
# first sort the array 
arr.sort() 
  
# then loop for all  three 
# consecutive triplets 
for i in range(N - 2): 
      
    # Check if the triplet satisfies the triangle 
    # condition 
    if arr[i] + arr[i + 1] > arr[i + 2]: 
        return True

Driver Code

arr = [5, 4, 3, 1, 2]
print("This satisfies the triangle inequality theorem" if isTriangle(arr) else " This does not satisfy the triangle inequality theorem ")

Alternate solution for Task #2

def count_div(A, B, K):
    # find smallest_divisible between (A or K) to B
    try:
        smallest_divisible = next(
            a for a in xrange(A if A == 0 or A >= K else K, B+1)
            if a%K==0
        )
    except:
        # no number within range is divisible
        return 0
     
    return (B-smallest_divisible)/K + 1

Alternate solution to Task #1

import re

def gap(number):
    binary_number = bin(number)[2:]
    zeros = re.findall(r'0+', binary_number)
    if binary_number[0] != '1':
        zeros = zeros[1:]
    if binary_number[-1] != '1':
        zeros = zeros[:-1:]
    if not zeros:
        return 0
    return len(max(zeros))

Solution for task 2: O(1)

// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');

function solution(A, B, K) {
// write your code in JavaScript (Node.js 8.9.4)

return Math.floor(B / K) - Math.floor((A - 1) / K)

}

For number one Ruby two liner that took me far longer than it should have because I'm weak with regex:

def solution(n)
  split = n.to_s(2).scan(/(?=(10+1))/)
  split.any? ? split.flatten.sort.first.length - 2 : 0
end

Here is one-liner for the problem 1:

import re
def gap(number):
    return len(max(re.findall(r'0+',bin(number)[2:]),default=[]))

Doesn't this incorrectly return 5 for '11100000'?

len(max(re.findall(r'0+', '11100000'),default=[]))
5

Doesn't this incorrectly return 5 for '11100000'?

@haveaguess, you are correct, the oneliner with re for problem 1 does not work for edge cases. You need the checking for the one's as in the post written by Odame.

binary gap

import re

def solution(n):
    bin_n = bin(n)[2:]
    gap = re.findall(r'(?=(10+1))', bin_n)
    return 0 if len(gap) == 0 else len(max(gap,key=len))-2

Are you sure this was the toptal test and not you just practising the lessons. Mine was much more harder than this with optimal solutions involving dynamic programming.

for task 3 another java imp

class Solution {
  public int solution(int[] A) {

    Arrays.sort(A);
    for (int i = 0; i < A.length-2; i++) {
      long i1 = A[i];
      long i2 = A[i+1];
      long i3 = A[i+2];
      if((i1+i2>i3 && i2+i3>i1 && i3+i1>i2))
        return 1;

    }

    return 0;
  }
}

These are the training question company asked the same questions which are provided in the traininng course of the codility website
my code for big binary gap

def DecimalToBinary(num):
S = bin(num).replace("0b", "")
res = [int(x) for x in str(S)]
print(res)
if res.count(1) < 2 or res.count(0) < 1:
print("its has no binary gap")
else:
positionof1 = [i for i,x in enumerate(res) if x==1]
print(positionof1)
differnce = [abs(j-i) for i,j in zip(positionof1, positionof1[1:])]
differnce[:] = [differnce - 1 for differnce in differnce]
differnce.sort()
print(differnce[-1])