Mobile Engineer Interview Questions

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my solution was still O(n) - 3n to be exact. i "iterate" over the entire matrix 3 times - once to check for duplicates in the rows, once to check for duplicates in the columns, and once to check for duplicates within sub matrices. During each of those iterations I visit each element of the matrix exactly once and just check whether it's already in a dictionary of seen numbers which would be a constant time operation, so this is in total would be O(n). I can't think of how to check all those in just one iteration , if that's what you were saying you could do. (at least without using O(n) extra space.. mine only uses constant extra space) i would appreciate if you could elaborate Less

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I basically just checked for duplicate numbers in each row, column, and then each 3x3 inner matrix. each iteration, i reset a dictionary of "seen" numbers. "." characters wouldn't matter, if i saw a number 1-9 i would check if it was already in the dictionary, if it was return false. return true at the very end if never encountered repeat numbers in rows, columns, or inner squares. when I explained this approach to the interviewer and he seemed satisfied with it although it was not easy to tell what the hell he was thinking or saying. i was working in a shared text file (i think it was codepen?), and even though the interviewer could have typed too, he didn't (even when i had to ask him repeatedly to spell a word i couldn't make out).. Less

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1) Subtract each number from summation formula = N * (N+1) / 2 2) Hash table , zero out all entries when insert number -> 1 after all insert look for 0 entry 3) XOR all elems -> X XOR all # 1 to N -> Y XOR of X and Y -> missing # 15 ^ 12 ^ 15 = 12 Less

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Stupid Math this is called stalking not algorithm

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With 360 degrees in a circle and 12 hours throughout each cycle the distance between each number is 30 degrees. If the minute hand is pointing at the 3 and is fifteen minutes, fifteen minutes is 1/4 of the total amount of minutes on the clock, this puts the hour hand 30/4 degrees passed the 3 at 7.25 degrees. They'll enjoy you thinking analytically. Less

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The answer should be a right angle.

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can't u just parse char by char and use a bunch of shift operators and since ur just using charAt(i) to pick up a character u can use the same i and just raise it to the power of 10 and just or it with some bytes Less

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public static int parseInt(String stringToConvert) { int i =0,num=0; int isNeg = 1; int length = stringToConvert.length(); if(stringToConvert.charAt(0) =='-') { isNeg=-1; i=1; } while(i Less

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Hi your approach for both of the questions is perfect. Where did it go wrong? Why did they reject any ideas?? Less

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Use the input array itself as a buffer to record appearances

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It's actually a pretty simple dynamic programming problem you can solve using constant space, do it iteratively rather than recursively. Check it out of LeetCode. Less

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This can be calculated with a simple recursive Fibonacci

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int ans[] = new int[1]; //O(n) public int efficientDia(TreeNode root) { if(root == null) return 0; int left = efficientDia(root.left); int right = efficientDia(root.right); ans[0] = Math.max(ans[0], 1 + left+ right); return 1+ Math.max(left, right); } //O(n^2) public int getDiameter(TreeNode root) { if(root == null) return 0; int leftHeight = getHeight(root.left); int rightHeight = getHeight(root.right); if(ans[0] < 1 + leftHeight + rightHeight) { ans[0] = 1 + leftHeight + rightHeight; } return Math.max(getDiameter(root.left), getDiameter(root.right)); } private int getHeight(TreeNode root) { // TODO Auto-generated method stub if(root == null) return 0; return Math.max(getHeight(root.left), getHeight(root.right))+1; } Less

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Indians at all companies always ask tree questions, it makes them giggle inside. I know, because I'm half indian and have interviewed people... tee hee hee Less

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public void printNumbers () { for (int i = 1; i <= 100; ++i) { if (i % 15==0) { System.out.println("FooBizz"); } else if (i % 3 == 0) { System.out.println("Foo"); } else if (i % 5 == 0) { System.out.println("Bizz"); } else { System.out.println(i); } } } Less

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Answer in JS- Trying to remove the duplicate tickets for each person then compare the tickets to get the desired number k. And then get the number of such friends. Not comparing Jay's own tickets. const getFriendsCount = (N, M, k, ticketsArr) => { const uniqueTktsArr = ticketsArr.map(tktsArr => Array.from(new Set(tktsArr))) let friendsNum = 0 const jaysTickets = uniqueTktsArr[N] uniqueTktsArr.forEach((tktsArr, i) => { if (i === N) { return true } let match = 0 tktsArr.every(tkt => { if (jaysTickets.includes(tkt)) { if (++match === k) { friendsNum++ return false } } return true }) }) return friendsNum } Less