Data Structures & Algorithms Interview Questions

Rishabh R. - "if decreasing arr, start from end and keep checking if next element increases by 1 or not. wherever not, put that value there."See full answer

Shivani N. - "did well but messed up dequeue"See full answer

Nathan B. - " This is mostly correct and fairly fast. My code has a bug somewhere where it fails on cases like the last case, where there are negative number on both ends of the array and the sums . from collections import deque debug = True # False def prdbg(*x): global debug debug = True # False if debug: print(x) else: return def max_sum(arr, start, end): if type(arr) == type(''' "See full answer

Divya R. - "public Double calculateRatio(String source, String destination) { Double ratio=1.0; while(graph.containsKey(source) && !visited.contains(source)) { visited.add(source); Map valueMap=graph.get(source); if(valueMap.containsKey(destination)) { return ratio*=valueMap.get(destination); } Map.Entry firstEntry=valueMap.entrySet().iterator().next(); source=firstEntry.getKey(); ratio*=firstEntry.getValue(); System.out.println("Entered"); } return null; }"See full answer

Tiago R. - "function isPalindrome(s, start, end) { while (s[start] === s[end] && end >= start) { start++; end--; } return end <= start; } function longestPalindromicSubstring(s) { let longestPalindrome = ''; for (let i=0; i < s.length; i++) { let j = s.length-1; while (s[i] !== s[j] && i <= j) { j--; } if (s[i] === s[j]) { if (isPalindrome(s, i, j)) { const validPalindrome = s.substring(i, j+1"See full answer

Aziz V. - "ArrayList allows constant time access (O(1)) to elements using their index because it uses a dynamic array internally, whereas LinkedList requires traversal from the head node, resulting in linear time complexity (O(n))."See full answer

Анвар А. - "I'd translated this decimal number into binary form and if it looks like 1000 or 10 or 1 - one leading 1 and others are zeros - than return 1 else return 0 "See full answer

Ravi C. - "boolean isMatch(String s, String p) { int i=0; int j=0; int sID=-1; int prevM=-1; while(i<s.length()){ if(j<p.length() && (s.charAt(i)==p.charAt(j) || p.charAt(j)=='?')){ i++; j++; }else if(j<p.length() && p.charAt(j)=='*'){ sID=j; prevM=i; j++; }else if(sID!=-1){ j=sID+1; prevM++; i=prevM; }else{ return false; } } while(j<p.length() && p.charAt(j)=='*') j++; if(i!=s.length() || j!=p.leng"See full answer

Divyani .. - "Did the code in Python"See full answer

Chen J. - "Leetcode 347: Heap + Hashtable Follow up question: create heap with the length of K instead of N (more time complexity but less space )"See full answer

Saurabh S. - "Basic Approach As BST inorder traversal will result in a sequence of increasing order. Store that order in a vector and get the k-1 index to get the Kth smallest element, similarly access the N-K+1 th element will be the Kth largest element Time Complexity: O(n) Space Complexity O(n) Space Optimized Approach For Kth smallest , start inorder traversal, and keep a counter, decrement the counter when you access the node element. When the counter turns 0 that elementwill be the Kth smal"See full answer

Mark S. - "int a_array[10] = {3,6,4,7,2,1,9}; int index = 0; int index2 = 0; for ( index = 0; index < sizeof(a_array); index++ ) { int tmpindex = index + 1; if ( tmpindex <= sizeof(a_array) ) { for ( index2 = tmpindex; index2 < sizeof(a_array); index2++ ) { if ( aarray[index] <= aarray[index2] ) { print( "%d is the NGE of %d" array[index2], array[index]); break; "See full answer