LeetCode #3683 — EASY

Earliest Time to Finish One Task

Build confidence with an intuition-first walkthrough focused on array fundamentals.

Solve on LeetCode

The Problem

Problem Statement

You are given a 2D integer array tasks where tasks[i] = [s_i, t_i].

Each [s_i, t_i] in tasks represents a task with start time s_i that takes t_i 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] = [s_i, t_i]
1 <= s_i, t_i <= 100

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[1,6],[2,3]]

Example 2

[[100,100],[100,100],[100,100]]

Step 02

Core Insight

What unlocks the optimal approach

Compute <code>finish[i] = s[i] + t[i]</code> for each task and take the minimum.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3683: Earliest Time to Finish One Task
class Solution {
    public int earliestTime(int[][] tasks) {
        int ans = 200;
        for (var task : tasks) {
            ans = Math.min(ans, task[0] + task[1]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3683: Earliest Time to Finish One Task
func earliestTime(tasks [][]int) int {
	ans := 200
	for _, task := range tasks {
		ans = min(ans, task[0]+task[1])
	}
	return ans
}

# Accepted solution for LeetCode #3683: Earliest Time to Finish One Task
class Solution:
    def earliestTime(self, tasks: List[List[int]]) -> int:
        return min(s + t for s, t in tasks)

// Accepted solution for LeetCode #3683: Earliest Time to Finish One Task
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3683: Earliest Time to Finish One Task
// class Solution {
//     public int earliestTime(int[][] tasks) {
//         int ans = 200;
//         for (var task : tasks) {
//             ans = Math.min(ans, task[0] + task[1]);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3683: Earliest Time to Finish One Task
function earliestTime(tasks: number[][]): number {
    return Math.min(...tasks.map(task => task[0] + task[1]));
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.