Coding Interview Questions

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

The_ A. - "The best and average of both the algorithms is same which is O(nlog(n)), but the worst time complexity of QuickSort is O(n^2){a case where all the elements are sorted in opposite to the fashion/order you want} while the worst TC for merge sort remains the same O(nlog(n)). and The SC for QS=O(logn) and MS=O(n)."See full answer

Divyani .. - "Did the code in Python"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

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

Prachi G. - "The solution produces the same result as the 'prescribed solution' yet it does not get accepted In the test results section transcript['year'] = transcript['year'].astype(str) df = pd.pivottable(data = transcript, index = 'studentid', columns = 'year', values = 'yearlygpa', aggfunc = 'mean').resetindex() df = df[(df['2021'] < df['2022']) & (df['2022'] < df['2023'])] df['average_gpa'] = df[['2021', '2022', '2023']].mean(axis=1).round(2) return df "See full answer

Sean L. - " debug your code below departments = pd.DataFrame({ 'id': [1, 2, 3, 4, 5], 'name': ['Reporting', 'Engineering', 'Marketing', 'Biz Dev', 'Silly Walks'] }) employees = pd.DataFrame({ 'id': [1, 2, 3, 4, 5, 6], 'first_name': ['John', 'Ava', 'Cailin', 'Mike', 'Ian', 'John'], 'last_name': ['Smith', 'Muffinson', 'Ninson', 'Peterson', 'Peterson', 'Mills'], 'salary': [20000, 10000, 30000, 20000, 80000, 50000], 'department_id': [1, 5, 2, 2, 2, 3] }) projects = p"See full answer

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

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