I had a single technical interview where I was asked two problems: Q1. Maximum Performance of a Team Given: - `speed[]` array - `efficiency[]` array - `K` = maximum team size Performance formula: Performance = (sum of speeds in the team) × (minimum efficiency in the team) Examples: - Example 1 speed = [2, 3, 4] efficiency = [1, 2, 2] K = 2 Output: 14 - Example 2 speed = [5, 8, 2, 9] efficiency = [1, 9, 2, 4] K = 2 Output: 68 - Example 3 speed = [1, 2, 3, 4, 5] efficiency = [5, 4, 3, 2, 1] K = 3 Output: 18 My Approach: - Sort employees by efficiency (descending) - Use a min-heap to maintain top K speeds - Calculate performance for each combination - Time Complexity: O(n log n) Code (similar logic): ```cpp #include using namespace std; int maxPerformance(vector& speed, vector& efficiency, int k) { int n = speed.size(); vector> emp; for (int i = 0; i < n; i++) { emp.push_back({efficiency[i], speed[i]}); } sort(emp.rbegin(), emp.rend()); priority_queue, greater> minHeap; long long sumSpeeds = 0, maxPerf = 0; for (auto &e : emp) { int eff = e.first; int spd = e.second; minHeap.push(spd); sumSpeeds += spd; if (minHeap.size() > k) { sumSpeeds -= minHeap.top(); minHeap.pop(); } maxPerf = max(maxPerf, sumSpeeds * 1LL * eff); } return (int)maxPerf; } ``` Q2. Longest Increasing Subsequence (LIS) https://leetcode.com/problems/longest-increasing-subsequence/description/ ```cpp class Solution { public: int lengthOfLIS(vector& nums) { int n = nums.size(); vector dp(n, 1); for (int i = 1; i < n; i++) { for (int j = 0; j < i; j++) { if (nums[i] > nums[j]) dp[i] = max(dp[i], 1 + dp[j]); } } int ans = INT_MIN; for (int i = 0; i < n; i++) ans = max(ans, dp[i]); return ans; } }; ``` After the Interview A week later, they asked if I wanted to join. By then, I had secured an offer from BrowserStack, which I felt had better work culture and reputation. So, I politely declined to proceed further.