Coding Interview Questions

Chao peter Y. - "The question is incomplete --- the code only passes if you return the data frame sorted by BOTH department name AND rank. While in the problem description, it mentions to only rank by department name: "The results should be ordered by department name." Not a big difference I know, but students shouldn't need to look into the solution to get the necessary knowledge to answer the question."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

Ankur P. - "I will Use Bit manipulation for solving this question suppose if n is a power of 2 so it should satisfy (n&0)==0 else (n&0)==1"See full answer

Praful B. - "Using a join: This is the most common way to do a lookup."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

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

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

Rishabh N. - "I first asked few clarifying questions like the return array may need not contain the list of building in the same order, to which the interviewer agreed. Then I came up with an approach where we iterate the array from right to left and keep a max variable which will keep the value of the current max. When we find an item which is greater than max we update the max and add this element into our solution. The interviewer agreed for the approach. I discussed few corner scenarios with the interview"See full answer

Sai R. - "def countuniqueoutfits(totalpants: int, uniquepants: int, totalshirts: int, uniqueshirts: int, totalhats: int, uniquehats: int) -> int: """ Number of unique outfits can simply be defined by (uniquepantschoose1uniqueshirtschoose1uniquehatschoose_1) (uniquepantschoose1*uniqueshirtschoose1) # Not wearing a hat nchoosek is n """ res = (uniquepants*uniqueshirtsuniquehats) + (uniquepantsunique_shirts) return res print(countuniqueoutfits(2, 1, 1, 1, 3, 2))"See full answer