2432. The Employee That Worked on the Longest Task LeetCode Solution

In this guide, you will get 2432. The Employee That Worked on the Longest Task LeetCode Solution with the best time and space complexity. The solution to The Employee That Worked on the Longest Task problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

  1. Problem Statement
  2. Complexity Analysis
  3. The Employee That Worked on the Longest Task solution in C++
  4. The Employee That Worked on the Longest Task solution in Java
  5. The Employee That Worked on the Longest Task solution in Python
  6. Additional Resources
2432. The Employee That Worked on the Longest Task LeetCode Solution image

Problem Statement of The Employee That Worked on the Longest Task

There are n employees, each with a unique id from 0 to n – 1.
You are given a 2D integer array logs where logs[i] = [idi, leaveTimei] where:

idi is the id of the employee that worked on the ith task, and
leaveTimei is the time at which the employee finished the ith task. All the values leaveTimei are unique.

Note that the ith task starts the moment right after the (i – 1)th task ends, and the 0th task starts at time 0.
Return the id of the employee that worked the task with the longest time. If there is a tie between two or more employees, return the smallest id among them.

Example 1:

Input: n = 10, logs = [[0,3],[2,5],[0,9],[1,15]]
Output: 1
Explanation:
Task 0 started at 0 and ended at 3 with 3 units of times.
Task 1 started at 3 and ended at 5 with 2 units of times.
Task 2 started at 5 and ended at 9 with 4 units of times.
Task 3 started at 9 and ended at 15 with 6 units of times.
The task with the longest time is task 3 and the employee with id 1 is the one that worked on it, so we return 1.

See also  491. Non-decreasing Subsequences LeetCode Solution

Example 2:

Input: n = 26, logs = [[1,1],[3,7],[2,12],[7,17]]
Output: 3
Explanation:
Task 0 started at 0 and ended at 1 with 1 unit of times.
Task 1 started at 1 and ended at 7 with 6 units of times.
Task 2 started at 7 and ended at 12 with 5 units of times.
Task 3 started at 12 and ended at 17 with 5 units of times.
The tasks with the longest time is task 1. The employee that worked on it is 3, so we return 3.

Example 3:

Input: n = 2, logs = [[0,10],[1,20]]
Output: 0
Explanation:
Task 0 started at 0 and ended at 10 with 10 units of times.
Task 1 started at 10 and ended at 20 with 10 units of times.
The tasks with the longest time are tasks 0 and 1. The employees that worked on them are 0 and 1, so we return the smallest id 0.

Constraints:

2 <= n <= 500
1 <= logs.length <= 500
logs[i].length == 2
0 <= idi <= n – 1
1 <= leaveTimei <= 500
idi != idi+1
leaveTimei are sorted in a strictly increasing order.

Complexity Analysis

  • Time Complexity: O(n)
  • Space Complexity: O(1)

2432. The Employee That Worked on the Longest Task LeetCode Solution in C++

class Solution {
 public:
  int hardestWorker(int n, vector<vector<int>>& logs) {
    int ans = logs[0][0];
    int maxWorkingTime = logs[0][1];

    for (int i = 1; i < logs.size(); ++i) {
      const int id = logs[i][0];
      const int workingTime = logs[i][1] - logs[i - 1][1];
      if (workingTime > maxWorkingTime) {
        ans = id;
        maxWorkingTime = workingTime;
      } else if (workingTime == maxWorkingTime) {
        ans = min(ans, id);
      }
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

2432. The Employee That Worked on the Longest Task LeetCode Solution in Java

class Solution {
  public int hardestWorker(int n, int[][] logs) {
    int ans = logs[0][0];
    int maxWorkingTime = logs[0][1];

    for (int i = 1; i < logs.length; ++i) {
      final int id = logs[i][0];
      final int workingTime = logs[i][1] - logs[i - 1][1];
      if (workingTime > maxWorkingTime) {
        ans = id;
        maxWorkingTime = workingTime;
      } else if (workingTime == maxWorkingTime) {
        ans = Math.min(ans, id);
      }
    }

    return ans;
  }
}
// code provided by PROGIEZ

2432. The Employee That Worked on the Longest Task LeetCode Solution in Python

class Solution:
  def hardestWorker(self, n: int, logs: list[list[int]]) -> int:
    ans = logs[0][0]
    maxWorkingTime = logs[0][1]

    for (_, prevLeaveTime), (id, leaveTime) in zip(logs, logs[1:]):
      workingTime = leaveTime - prevLeaveTime
      if workingTime > maxWorkingTime:
        ans = id
        maxWorkingTime = workingTime
      elif workingTime == maxWorkingTime:
        ans = min(ans, id)

    return ans
# code by PROGIEZ

Additional Resources

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