Round 3 – Technical Interview
The third technical round further assessed my coding and problem-solving prowess. The questions asked in this round are listed below:
- Calculating the minimum number of steps necessary to traverse an array and reach its end.
Java
public class MinimumJumpsToEnd { public static int minJumpsToEnd(int[] arr) { int n = arr.length; if (n <= 1) { return 0; // 1 element or is empty, no jumps needed } int jumps = 1; int maxReach = arr[0]; // max reach from the first position int steps = arr[0]; // steps available from the first position for (int i = 1; i < n; i++) { if (i == n - 1) { return jumps; // reached the end } maxReach = Math.max(maxReach, i + arr[i]); // Update max reach steps--; if (steps == 0) { if (i >= maxReach) { return -1; // Can't proceed further } jumps++; steps = maxReach - i; } } return -1; // end not reachable } public static void main(String[] args) { int[] arr = {2, 3, 1, 1, 4}; System.out.println(minJumpsToEnd(arr)); } } |
Output
2
- Given two strings, find the length of their longest common subsequence. A subsequence is a sequence that appears in the same relative order in both strings, but not necessarily consecutively. For example, for the strings “ABCD” and “ACDF”, the longest common subsequence is “ACD”, with a length of 3.
Java
public class LongestCommonSubsequence { public static int longestCommonSubsequenceLength(String str1, String str2) { int m = str1.length(); int n = str2.length(); // a table to store the lengths of longest common // subsequences int[][] dp = new int[m + 1][n + 1]; // fill the table using dp for (int i = 1; i <= m; i++) { for (int j = 1; j <= n; j++) { if (str1.charAt(i - 1) == str2.charAt(j - 1)) { dp[i][j] = dp[i - 1][j - 1] + 1; } else { dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]); } } } return dp[m][n]; } public static void main(String[] args) { String str1 = "ABCD"; String str2 = "ACDF"; int result = longestCommonSubsequenceLength(str1, str2); System.out.println( "Length of longest common subsequence: " + result); } } |
Output
Length of longest common subsequence: 3
Loyalty Juggernaut Interview Experience for Product Engineer Role
I applied to Loyalty Juggernaut through the AngelList website and received a call from their HR shortly after to discuss interview scheduling. Given that I was in my 6th semester with the on-campus placement drive approaching, I decided to explore my options. Nonetheless, I proceeded with the interview rounds to gain valuable experience.