Coding Interview Questions

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

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

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

Shubhangi S. - "sanity testing, black box testing , white box testing, smoke testing,performance testing"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

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

Jazeel A. - "public int CountNumberOfSpace(string s){ int n = s.Length; for(int i=0;i<n;i++){ if(s[i]==' ')count++; } return count; }"See full answer

Shivam P. - " First, sort the array in ascending order. This ensures that we can easily check the triangle inequality condition. Use a loop to iterate through the array. For each triplet of consecutive elements, check if they satisfy the triangle inequality condition a+b>ca+b>c. As soon as you find a valid tuple, return it. If no valid tuple is found, return null. This approach is efficient with a time complexity of O(nlog⁡n)O(nlogn) due to the sorting step, followed by a linear scan of the array"See full answer

Konstantin P. - "Count how many 1 and 2. calculate how many remaining items with '0'. Override existing data with knowledge of how many '0','1','2' in that order; TC=O(n), SC=O(1)"See full answer