carloscn
commented
1 year ago Analysis
pub fn hardest_worker(n: i32, logs: Vec<Vec<i32>>) -> i32
{
let mut ret:i32 = 0;
if logs.len() < 1 {
return ret;
}
ret = i32::MIN;
let mut index:usize = 0;
for i in 0..logs.len() {
let unit_time:i32;
if i == 0 {
unit_time = logs[i][1];
} else {
unit_time = logs[i][1] - logs[i - 1][1];
}
if unit_time > ret {
ret = unit_time;
index = i;
}
}
ret = logs[index][0];
return ret;
}
Description
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.
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.