Coding Interview Questions

Shivani N. - "did well but messed up dequeue"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

Kunal kumar S. - "simply check its size if the size if the size is greater than n then yes it has duplicate"See full answer

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

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

Vaibhav D. - "Make current as root. 2 while current is not null, if p and q are less than current, go left. If p and q are greater than current, go right. else return current. return null"See full answer

Natalie C. - "I walked through the code for a react.js based tic-tac-toe game written in typescript. The goal was to find ways to improve the code/ suggest improvements. I missed some areas like where state was being updated directly rather than using React's setState. There were issues around clear and maintainable logic, adherence to React best practices."See full answer

Anmol R. - "The example given is wrong. The 2nd test case should have answer 3, as we can get to 6 by using 3 coins of denomination 2."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

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

Susheel C. - "I wrote a function to determine if a given number is a power of 2 using logarithms."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