Interview Questions

Jeff H. - "Google Pay currently makes money from a small percentage of interchange from each transaction. They also make money with Mobile recharge sim card fees. Questions Do we have a timeframe for this? How could we 10x this? Let's think about Alphabet's goal and vision to make the world more accessible and useful. To 10x it, we will need a combination of improvements, new revenue sources, and some paradigm shifting thinking. The goal in this case is monetization, and a steep on at t"See full answer

Soraya B. - "A load balancer, web application servers and a large database. Database is the core of the application containing the songs and tags related to each song. Application server will provide the search interface to find songs, play them and search them in specific categories or channels."See full answer

Kevin Z. - "362880"See full answer

Fakhri A. - "static int countBinaryStrings(int n) { int a[] = new int[n]; int b[] = new int[n]; a[0] = b[0] = 1; for (int i = 1; i < n; i++) { a[i] = a[i - 1] + b[i - 1]; b[i] = a[i - 1]; } System.out.println(a[n - 1] + b[n - 1]); return a[n - 1] + b[n - 1]; }"See full answer

Mo I. - "m*n"See full answer

Padmanaban M. - "static int findLongestRepeatingSubSeq(String str) { int n = str.length(); int dp = new intn+1; for (int i=0; i<=n; i++) for (int j=0; j<=n; j++) dpi = 0; for (int i=1; i<=n; i++) { for (int j=1; j<=n; j++) { if (str.charAt(i-1)== str.charAt(j-1) && i != j) dpi =  1 + dpi-1; else dpi = Math.max(dpi, dpi-1); } } `return"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

Emma X. - "String commonStr(String str1, String str2) { int len1 = str1.length(); int len2 = str2.length(); if (len1 == 0 || len2 == 0) return ""; // let dpx reprsent the longest common str of 0...x int dp = new int len1 + 1; int maxLen = 0; int endIndex = 0; for (int i = 1; i <= len1; i++) { for (int j = 1; j <= len2; j++) { if (str1.charAt(i-1) == str2.charAt(j-1)) { dpi = dpi-1 + 1; "See full answer

Guilherme F. - "A much better solution than the one in the article, below: It looks like the ones writing articles here in Javascript do not understand the time/space complexity of javascript methods. shift, splice, sort, etc... In the solution article you have a shift and a sort being done inside a while, that is, the multiplication of Ns. My solution, below, iterates through the list once and then sorts it, separately. It´s O(N+Log(N)) class ListNode { constructor(val = 0, next = null) { th"See full answer

Ugo C. - "function merge(L1, L2) { let L3 = { data: null, next: null }; let prev = L3; while (L1 != null || L2 != null) { if (L1 == null) { prev.next = L2; L2 = L2.next; } else if (L2 == null) { prev.next = L1; L1 = L1.next; } else if (L1.data < L2.data) { prev.next = L1; L1 = L1.next; } else { prev.next = L2; L2 = L2.next; } prev = prev.next; } return L3.next; }"See full answer

Rishabh R. - "if decreasing arr, start from end and keep checking if next element increases by 1 or not. wherever not, put that value there."See full answer