Data Structures & Algorithms Interview Questions

Ahmed A. - "Good Question, but I would've marked this as medium not hard difficulty, since it's just a straightforward traversal."See full answer

Md kamrul H. - "public class HashMap { public class Element { T key; V value; Element(T k, V v) { this.key = k; this.value = v; } } private static final int DEFAULT_CAPACITY = 16; private static final float LOAD_FACTOR = 0.75f; private LinkedList[] table = new LinkedList[DEFAULT_CAPACITY]; private int size = 0; private int threshold = (int) (DEFAULTCAPACITY * LOADFACTOR); public void put(T k"See full answer

Rohit B. - "class Solution: def missingNumber(self, nums: list[int]) -> int: Sorting approach n = len(nums) s = n*(n+1)//2 r = s - sum(nums) return self.r l = [3,0,1] print(missingNumber(l))"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

Ravi teja N. - "I solved it using recursion and then memoization. Used Dynamic programming approach"See full answer

Divya R. - "static boolean sudokuSolve(char board) { return sudokuSolve(board, 0, 0); } static boolean sudokuSolve(char board, int r, int c) { if(c>=board[0].length) { r=r+1; c=0; } if(r>=board.length) return true; if(boardr=='.') { for(int num=1; num<=9; num++) { boardr=(char)('0' + num); if(isValidPosition(board, r, c)) { if(sudokuSolve(board, r, c+1)) return true; } boardr='.'; } } else { return sudokuSolve(board, r, c+1); } return false; } static boolean isValidPosition(char b"See full answer

Sravanthi M. - "public static char getRepeatingCharacterInGivenString(String str){ char result = '0'; HashSet set = new HashSet(); for(int i=0;i<str.length();i++){ char c = str.charAt(i); if(!set.contains(c)){ set.add(c); } else{ result= c; break; } } return result; }"See full answer

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

Anonymous Roadrunner - "from typing import List def traprainwater(height: List[int]) -> int: if not height: return 0 l, r = 0, len(height) - 1 leftMax, rightMax = height[l], height[r] res = 0 while l < r: if leftMax < rightMax: l += 1 leftMax = max(leftMax, height[l]) res += leftMax - height[l] else: r -= 1 rightMax = max(rightMax, height[r]) "See full answer

Gaddipati V. - "NA"See full answer

Aelaf G. - "This depends on the list of documents and the length of the documents. My implementation will use Trie with node containing the following: class TrieNode { is_end: boolean, instances: { docid → [wordpositions] }, children: array[26] } Look up for a word will give result instances{docid:wordposition...} dictionary (which can be further improved by methods like max instance on a document....you name it...) Trie space is proportional to the total characters in"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

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

Parth S. - "Assuming that trades will have information like trade_type buy or sell trade_price with these tuples, one can iterate over each trade while maintaining a stack which maintains all the open buy trades. If we encounter a sell trade then we pop one element make it a buy/sell pair and calculate the profit/loss for that pair. Moreover, keep adding pair-wise profit/loss to calculate overall profit as we continue iterating over trades. At the end print pairs and their profit/loss along with"See full answer

Richard W. - "static HashMap flattenDictionary(HashMap dict) { HashMap flatDict = new HashMap(); flattenDictionaryHelper("", dict, flatDict); return flatDict; } static void flattenDictionaryHelper(String initalKey, HashMap dict, HashMap flatDict) { for(Map.Entry entry : dict.entrySet()) { String key = entry.getKey(); Object value = entry.getValue(); String ne"See full answer