SDE intern Mindtickle
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MindtickleCareerCompensationInterview

Mindtickle SDE Intern Interview Experience – Pune (35K/month)

Role: SDE Intern
Location: Pune
Stipend: 35K/month
Interview Duration: 1 hour

Process

Round 1: Technical Interview (1 hour)

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

code```

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):

#include <bits/stdc++.h>
using namespace std;

int maxPerformance(vector<int>& speed, vector<int>& efficiency, int k) {
  int n = speed.size();
  vector<pair<int,int>> emp;
  for (int i = 0; i < n; i++) {
      emp.push_back({efficiency[i], speed[i]});
  }

  sort(emp.rbegin(), emp.rend());

  priority_queue<int, vector<int>, greater<int>> 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/

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        vector<int> 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.

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