Easy
You are given a 2D integer array tasks
where tasks[i] = [si, ti]
.
Each [si, ti]
in tasks
represents a task with start time si
that takes ti
units of time to finish.
Return the earliest time at which at least one task is finished.
Example 1:
Input: tasks = [[1,6],[2,3]]
Output: 5
Explanation:
The first task starts at time t = 1
and finishes at time 1 + 6 = 7
. The second task finishes at time 2 + 3 = 5
. You can finish one task at time 5.
Example 2:
Input: tasks = [[100,100],[100,100],[100,100]]
Output: 200
Explanation:
All three tasks finish at time 100 + 100 = 200
.
Constraints:
1 <= tasks.length <= 100
tasks[i] = [si, ti]
1 <= si, ti <= 100
public class Solution {
public int earliestTime(int[][] tasks) {
int ans = 1000;
for (int[] task : tasks) {
int st = task[0];
int tm = task[1];
ans = Math.min(ans, st + tm);
}
return ans;
}
}