Problem statement Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. Solution Obviously the naive approach to this problem would be to subtract the divisor from the dividend until the dividend becomes less than the divisor, while keeping track of how many times . . . Read more
Problem statement Given a list, nums, of distinct numbers, return all possible unique permutations. Sample input
Sample output

[[1, 2, 1], [2, 1, 1], [1, 1, 2]] 
Solution In another post we showed how this problem can be solved iteratively. We can use the same technique but add a check to skip adding the number n to . . . Read more
Problem statement Given a list, nums, of distinct numbers, return all possible permutations. Sample input
Sample output

[ [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1] ] 
Solution In another post we showed how this problem can be solved using a recursive technique by swapping. We stumbled upon a nice iterative solution that we thought might be worth . . . Read more
Problem statement Given a list, nums, of distinct numbers, return all possible permutations. Sample input
Sample output

[ [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1] ] 
Solution We are using a technique called permutations by swapping. Full code
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

class Solution(object): def swap(self, nums, i, j): tmp = nums[i] nums[i] = nums[j] nums[j] = tmp def solve(self, nums, i, result): if i == len(nums)1: result.append(list(nums)) return for j in range(i, len(nums)): self.swap(nums, i, j) self.solve(nums, i+1, result) self.swap(nums, i, j) def permute(self, nums): """ :type nums: List[int] :rtype: List[List[int]] """ ans = [] self.solve(nums, 0, ans) return ans def main(): solution = Solution() print(solution.permute([1,2,3])) if __name__ == "__main__": main() 
Problem statement You are given an 2D matrix / array representing an image. Rotate matrix by 90 degrees (clockwise) inplace. Solution We will work through the matrix layers. The layers of a matrix are illustrated below in different colours: There are a total of layers. We then iterate of the . . . Read more
Problem statement A time series is a series of data points indexed in time order. They are commonly used in the financial world, especially in stock markets. In this challenge you are working with a time series of stock prices. You are given historical records where the stock at time . . . Read more
My main goal with this blog is to help fellow software engineers become better at what they do by sharing my experiences during my quest to become better myself. This does not just cover writing about algorithms and cool technologies, but also includes sharing experience in terms of career choices. This . . . Read more
Problem statement Given a string containing just the characters ( and ), find the length of the longest valid (wellformed) parentheses substring. Sample input
Sample output
Sample input 1
Sample output 1
Solution We will solve this problem using a Stack. We are going to iterate . . . Read more
Disclaimer: The experiences (e.g. BMW software engineer interview experience) mentioned in this blog were submitted by readers anonymously. It is by no means the intention of this blog to smear the companies mentioned. Rather the sole purpose is to give prospective candidates the opportunity to gain insight information on working . . . Read more
Problem statement Given an array of integers sorted in ascending order, find the starting and ending position of a given target value. If the target is not found in the array, return . Sample input
Sample output
Solution
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

class Solution(object): def search(self, nums, n, start, end, ans): if start > end: return mid = start + (end  start)//2 if n == nums[mid]: ans[0] = min(ans[0], mid) ans[1] = max(ans[1], mid) self.search(nums, n, start, mid1, ans) self.search(nums, n, mid+1, end, ans) def searchRange(self, nums, target): """ :type nums: List[int] :type target: int :rtype: List[int] """ ans = [len(nums), 1] self.search(nums, target, 0, len(nums)1, ans) if ans[0]==len(nums): return [1,1] return ans def main(): solution = Solution() print(solution.searchRange([5, 7, 7, 8, 8, 10], 8)) if __name__ == "__main__": main() 