Software Engineer Coding Interview Questions

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

Mourya C. - "I try to solve this initially using quick select where will take a pivot element and position the remaining elements and check if the current index is answer or not and continue the same but it requires o(n*n), but interviewee is expecting the best from me, so at the end i tried solving using heaps where will check the difference between k and n-k to use min or max heap after that we will heap the array, and will keep popping the element k-1 and return the peek one which leads to answer."See full answer

Iris F. - "#include #include #include using namespace std; void printComs(int prev, int start, int end, int target) { if (start >= end) return; while (start target) { end--; } else { st"See full answer

Rasika D. - "A doubly linked list is like a train where the engine is its head, and the compartments doors are interconnected, you can travel though them. The last compartment is the tail and is not connected with anything. Following is a visual: Head C1 C2 C3....Cn Tail Pseudocode: **class Node: Data pointer next pointer prev End class DLL = NULL //pointer to list insertion: if DLL is NULL: head = new Node() DLL = head DLL.prev = NU"See full answer

Susmita S. - "Function that transforms each elements in a collection and returns new collection with transformed elements"See full answer

Shar N. - "standard answer for this."See full answer

Aniket G. - "public class BoggleBoard { public static List findWords(char board, Set dictionary) { int rows = board.length; int cols = board[0].length; boolean visited = new booleanrows; int directions = {{1,0}, {-1,0}, {0,1}, {0,-1}}; List result = new ArrayList(); for(int i=0; i<rows; i++) { for(int j=0; j<cols; j++) { dfs(board, visited, i, j, dictionary, "", result, dire"See full answer

Mohith J. - "Recursion: 0 if NULL, else 1+max(height(left), height(right))"See full answer

Gopal D. - "(Like a Rummy Game) There are 3 colors of tiles. Each tile has a number 1-9 on it. So the 27-tile set makes a deck. There are 4 decks. (Total = 108 tiles) Tile Colors = {Red, Black, Green} The tiles could be grouped together in patterns Types of patterns : The three tiles are identical (R2, R2, R2) The three tiles are of same color and sequential (R2, R3, R4). Sequence cannot be overlapping (R8, R9, R1 is not a pattern) Help : Tile Notations - R2 denotes Red tile having num 2 A player"See full answer

Anonymous Emu - "let str = 'this is a test of programs'; let obj={}; for (let s of str ) obj[s]?(obj[s]=obj[s]+1):(obj[s]=1) console.log(JSON.stringify(obj))"See full answer