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# Program to find The Employee That Worked on the Longest Task

Here, we will see Program to find The Employee That Worked on the Longest Task with code and algorithm.

You are given nÂ employees, each with a unique id fromÂ `0`Â toÂ `n - 1`. You are given a 2D integer arrayÂ `logs`Â whereÂ `logs[i] = [idi, leaveTimei]`Â where:

• `idi`Â is the id of the employee that worked on theÂ `ith`Â task
• `leaveTimei`Â is the time at which the employee finished theÂ `ith`Â task. All the valuesÂ `leaveTimei`Â areÂ unique.

Note that theÂ `ith`Â task starts the moment right after theÂ `(i - 1)th`Â task ends, and theÂ `0th`Â task starts at timeÂ `0`.

You have to returnÂ the id of the employee that worked the task with the longest time.Â If there is a tie between two or more employees, returnÂ theÂ smallestÂ id among them.

A customer’sÂ wealthÂ is the amount of money they have in all their bank accounts. The richest customer is the customer that has the maximumÂ wealth.

Example:

```1)
Input: n = 11, logs = [[1,4],[3,6],[1,10],[2,16]]
Output: 2
Explanation:
Task 0 started at 1 and ended at 4 with 3 units of times.
Task 1 started at 4 and ended at 6 with 2 units of times.
Task 2 started at 6 and ended at 10 with 4 units of times.
Task 3 started at 10 and ended at 16 with 6 units of times.
The task with the longest time is task 3 and the employee with id 2 is the one that worked on it, so we return 2.

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 employees that worked on it is 3, so we return 3.

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.```

## Program to find The Employee That Worked on the Longest Task code in C++

Code 1:

```#include <iostream>
#include <vector>

using namespace std;

int hardestWorker(int n, vector<vector<int>>& logs) {
int max = logs[0][1], id = logs[0][0];

for(int i = 1; i < logs.size(); i++) {
if((logs[i][1] - logs[i-1][1]) > max) {
max = logs[i][1] - logs[i-1][1];
id = logs[i][0];
} else if((logs[i][1] - logs[i-1][1]) == max) {
if(logs[i][0] < id) {
id = logs[i][0];
}
}
}
return id;
}

int main()
{
int n = 10;
vector<vector<int> > logs = {
{1, 4},
{3,6},
{1,10},
{2,16}
};
cout<<hardestWorker(n, logs);

return 0;
}```

Code 2:

```#include <iostream>
#include <vector>

using namespace std;

int hardestWorker(int n, vector<vector<int>>& logs) {
int start = 0;
int res = -1;
int time = -1;

for (vector<int> & log : logs) {
auto id = log[0];
auto end = log[1];
int duration = end-start;
if (duration > time) {
time = duration;
res = id;
} else if (duration == time && res > id) {
res = id;
}
start = end;
}
return res;
}

int main()
{
int n = 10;
vector<vector<int> > logs = {
{1, 4},
{3,6},
{1,10},
{2,16}
};
cout<<hardestWorker(n, logs);

return 0;
}```

Output:

`2`

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Posted in C++, Easy