Software Engineer Coding Interview Questions

Alex M. - "A red-black tree is a self-balancing binary search tree. The motivation for this is that the benefits of O(logN) search, insertion, and deletion that a binary tree provides us will disappear if we let the tree get too "imbalanced" (e.g. there are too many nodes on one side of the tree or some branches have a depth that is way out of proportion to the average branch depth). This imbalance will occur if we don't adjust the tree after inserting or deleting nodes, hence our need for self-balancing c"See full answer

Bhaskar B. - "This could be done using two-pointer approach assuming array is sorted: left and right pointers. We need track two sums (left and right) as we move pointers. For moving pointers we will move left to right by 1 (increment) when right sum is greater. We will move right pointer to left by 1 (decrement) when left sum is greater. at some point we will either get the sum same and that's when we exit from the loop. 0-left will be one array and right-(n-1) will be another array. We are not going to mo"See full answer

Nikhil R. - "public class sample { public int [] merge(int [] a, int [] b) { if(a == null || a.length == 0 || b == null || b.length == 0) return null; int i = 0, j = 0, index = -1; int [] merged = new int[a.length + b.length]; while (i < a.length && j < b.length) { if(a[i] < b[i]) merged[++index] = a[i++]; else merged[++index] = b[j++]; } while (i < a.length) { merged[++index] = a[i++]; } "See full answer

Rick E. - " O(n) time, O(1) space from typing import List def maxsubarraysum(nums: List[int]) -> int: if len(nums) == 0: return 0 maxsum = currsum = nums[0] for i in range(1, len(nums)): currsum = max(currsum + nums[i], nums[i]) maxsum = max(currsum, max_sum) return max_sum debug your code below print(maxsubarraysum([-1, 2, -3, 4])) `"See full answer

Nasit S. - "Depend on the array size and number of 0's theere."See full answer

Ashley M. - "sliding window"See full answer

B. T. - "Example: bucket A: 3 liters capacity bucket B: 5 liters capacity goal: 4 liters You are asked to print the logical sequence to get to the 4 liters of water in one bucket. Follow up: How would you solve the problem if you have more than 2 buckets of water?"See full answer

Prashanth A. - "We are asked to calculate Sum(over value) for time in (t - window_size, t) where key in (key criteria). To develop a function to set this up. Let w be the window size. I would have an observer of some kind note the key-value, and for the first w windows just add the value to a temporary variable in memory if the key meets the key criteria. Then it would delete the oldest value and add the new value if the new key meets the criteria. At each step after "w", we would take the sum / w and store"See full answer

Tiago R. - "class TrieNode { constructor() { this.children = {}; this.isEndOfWord = false; } } class Trie { constructor() { this.root = new TrieNode(); } insert(word) { let node = this.root; for (const char of word) { if (!node.children[char]) { node.children[char] = new TrieNode(); } node = node.children[char]; } node.isEndOfWord = true; } search(word) { l"See full answer

Frederick A. - "This problem can be solved with two approaches Iterative approach Recursive approach Quite easy to think about the iterative approach, you can make use of a while loop in that case. But what if you want to make use of previously computed values? That case going for the recursive solution is quite useful. class Collatz: def init(self) -> None: self.cache = {} self.steps = 0 def steps_from(self, n) -> int: # base case if n == 1: "See full answer

Jalpa S. - "def containSubString(mainString, SubString): s1 = "hello world" # main String s2 = "hello" s3 = "world" s4 = "Nothing" answer1 = containSubString(s1, s2) answer2 = containSubString(s1, s3) answer3 = containSubString(s1, s4) print(answer1 , answer2, answer) "See full answer

Sonia M. - "virtual function is a member function declared with virtual keyword in a base class. It enables derived classes to redefine this function with their own specific implementations."See full answer

Tiago R. - "function permute(nums) { if (nums.length <= 1) { return [nums]; } const prevPermutations = permute(nums.slice(0, nums.length-1)); const currentNum = nums[nums.length-1]; const permutations = new Set(); for (let prev of prevPermutations) { for (let i=0; i < prev.length; i++) { permutations.add([...prev.slice(0, i), currentNum, ...prev.slice(i)]); } permutations.add([...prev, currentNum]); } return [...permutations]"See full answer

Nils G. - "Bitshift the number to the right and keep track of the 1's you encounter. If you bitshift it completely and only encounter one 1, it is a power of two."See full answer