Uber SDE-2 Frontend Interview Experience: A Detailed Breakdown with Coding Questions and…

Overview

This document details the interview experience for a Software Engineer II (Frontend) position at Uber. The process consisted of five rounds, designed to evaluate the candidate's technical skills, problem-solving abilities, system design knowledge, and leadership qualities. This breakdown provides a comprehensive guide for prospective candidates.

Interview Rounds

The Uber SDE-2 Frontend interview process includes the following stages:

1. Online Assessment (35 mins)

   • Format: CodeSignal test with 2 coding questions and 8 multiple-choice questions (JavaScript, HTML, CSS).
   • Difficulty: Easy to Medium.
   • Coding Question 1: Playing with Digits (Easy) - Digit manipulation (Math).

     function sumOfDigits(num) {
       while (num >= 10) {
         let sum = 0;
         while (num > 0) {
           sum += num % 10;
           num = Math.floor(num / 10);
         }
         num = sum;
       }
       return num;
     }
     

     Explanation: The function calculates the sum of digits repeatedly until a single-digit number remains.
   • Coding Question 2: Longest Substring Without Repeating Characters (Medium) - HashMap and Sliding Window technique.

     function lengthOfLongestSubstring(s) {
       let map = new Map();
       let maxLen = 0;
       let start = 0;
     
       for (let end = 0; end < s.length; end++) {
         if (map.has(s[end])) {
           start = Math.max(map.get(s[end]) + 1, start);
         }
         map.set(s[end], end);
         maxLen = Math.max(maxLen, end - start + 1);
       }
       return maxLen;
     }
     

     Explanation: The algorithm uses a sliding window and a HashMap to find the longest substring without repeating characters.

2. DSA Coding Round (1 hr)

   • Format: Live coding round (1-2 Data Structures and Algorithms problems).
   • Difficulty: Medium to Hard.
   • Coding Problem: Minimum Time for Objects to Meet - Binary Search and Math.

     function minMeetingTime(positions, velocities) {
       let low = 0;
       let high = 1e9;
       
       while (high - low > 1e-6) {
         let mid = (low + high) / 2;
         let minPos = Number.MAX_VALUE;
         let maxPos = -Number.MAX_VALUE;
         for (let i = 0; i < positions.length; i++) {
           minPos = Math.min(minPos, positions[i] + velocities[i] * mid);
           maxPos = Math.max(maxPos, positions[i] - velocities[i] * mid);
         }
         if (minPos >= maxPos) {
           high = mid;
         } else {
           low = mid;
         }
       }
       return low;
     }
     

     Explanation: The code uses binary search to find the minimum time for all objects to meet at a single point.