LeetCode #1723 — HARD

Find Minimum Time to Finish All Jobs

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an integer array jobs, where jobs[i] is the amount of time it takes to complete the i^th job.

There are k workers that you can assign jobs to. Each job should be assigned to exactly one worker. The working time of a worker is the sum of the time it takes to complete all jobs assigned to them. Your goal is to devise an optimal assignment such that the maximum working time of any worker is minimized.

Return the minimum possible maximum working time of any assignment.

Example 1:

Input: jobs = [3,2,3], k = 3
Output: 3
Explanation: By assigning each person one job, the maximum time is 3.

Example 2:

Input: jobs = [1,2,4,7,8], k = 2
Output: 11
Explanation: Assign the jobs the following way:
Worker 1: 1, 2, 8 (working time = 1 + 2 + 8 = 11)
Worker 2: 4, 7 (working time = 4 + 7 = 11)
The maximum working time is 11.

Constraints:

1 <= k <= jobs.length <= 12
1 <= jobs[i] <= 10⁷

Step 01

Brute Force Baseline

Problem summary: You are given an integer array jobs, where jobs[i] is the amount of time it takes to complete the ith job. There are k workers that you can assign jobs to. Each job should be assigned to exactly one worker. The working time of a worker is the sum of the time it takes to complete all jobs assigned to them. Your goal is to devise an optimal assignment such that the maximum working time of any worker is minimized. Return the minimum possible maximum working time of any assignment.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming · Backtracking · Bit Manipulation

Example 1

[3,2,3]
3

Example 2

[1,2,4,7,8]
2

Core Insight

What unlocks the optimal approach

We can select a subset of tasks and assign it to a worker then solve the subproblem on the remaining tasks

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

Largest constraint values

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 #1723: Find Minimum Time to Finish All Jobs
class Solution {
    private int[] cnt;
    private int ans;
    private int[] jobs;
    private int k;

    public int minimumTimeRequired(int[] jobs, int k) {
        this.k = k;
        Arrays.sort(jobs);
        for (int i = 0, j = jobs.length - 1; i < j; ++i, --j) {
            int t = jobs[i];
            jobs[i] = jobs[j];
            jobs[j] = t;
        }
        this.jobs = jobs;
        cnt = new int[k];
        ans = 0x3f3f3f3f;
        dfs(0);
        return ans;
    }

    private void dfs(int i) {
        if (i == jobs.length) {
            int mx = 0;
            for (int v : cnt) {
                mx = Math.max(mx, v);
            }
            ans = Math.min(ans, mx);
            return;
        }
        for (int j = 0; j < k; ++j) {
            if (cnt[j] + jobs[i] >= ans) {
                continue;
            }
            cnt[j] += jobs[i];
            dfs(i + 1);
            cnt[j] -= jobs[i];
            if (cnt[j] == 0) {
                break;
            }
        }
    }
}

// Accepted solution for LeetCode #1723: Find Minimum Time to Finish All Jobs
func minimumTimeRequired(jobs []int, k int) int {
	cnt := make([]int, k)
	ans := 0x3f3f3f3f
	sort.Slice(jobs, func(i, j int) bool {
		return jobs[i] > jobs[j]
	})
	var dfs func(int)
	dfs = func(i int) {
		if i == len(jobs) {
			mx := slices.Max(cnt)
			ans = min(ans, mx)
			return
		}
		for j := 0; j < k; j++ {
			if cnt[j]+jobs[i] >= ans {
				continue
			}
			cnt[j] += jobs[i]
			dfs(i + 1)
			cnt[j] -= jobs[i]
			if cnt[j] == 0 {
				break
			}
		}
	}
	dfs(0)
	return ans
}

# Accepted solution for LeetCode #1723: Find Minimum Time to Finish All Jobs
class Solution:
    def minimumTimeRequired(self, jobs: List[int], k: int) -> int:
        def dfs(i):
            nonlocal ans
            if i == len(jobs):
                ans = min(ans, max(cnt))
                return
            for j in range(k):
                if cnt[j] + jobs[i] >= ans:
                    continue
                cnt[j] += jobs[i]
                dfs(i + 1)
                cnt[j] -= jobs[i]
                if cnt[j] == 0:
                    break

        cnt = [0] * k
        jobs.sort(reverse=True)
        ans = inf
        dfs(0)
        return ans

// Accepted solution for LeetCode #1723: Find Minimum Time to Finish All Jobs
/**
 * [1723] Find Minimum Time to Finish All Jobs
 *
 * You are given an integer array jobs, where jobs[i] is the amount of time it takes to complete the i^th job.
 * There are k workers that you can assign jobs to. Each job should be assigned to exactly one worker. The working time of a worker is the sum of the time it takes to complete all jobs assigned to them. Your goal is to devise an optimal assignment such that the maximum working time of any worker is minimized.
 * Return the minimum possible maximum working time of any assignment.
 *  
 * Example 1:
 *
 * Input: jobs = [3,2,3], k = 3
 * Output: 3
 * Explanation: By assigning each person one job, the maximum time is 3.
 *
 * Example 2:
 *
 * Input: jobs = [1,2,4,7,8], k = 2
 * Output: 11
 * Explanation: Assign the jobs the following way:
 * Worker 1: 1, 2, 8 (working time = 1 + 2 + 8 = 11)
 * Worker 2: 4, 7 (working time = 4 + 7 = 11)
 * The maximum working time is 11.
 *  
 * Constraints:
 *
 * 	1 <= k <= jobs.length <= 12
 * 	1 <= jobs[i] <= 10^7
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/find-minimum-time-to-finish-all-jobs/
// discuss: https://leetcode.com/problems/find-minimum-time-to-finish-all-jobs/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    // Credit: https://leetcode.com/problems/find-minimum-time-to-finish-all-jobs/solutions/3199728/just-a-runnable-solution/
    pub fn minimum_time_required(jobs: Vec<i32>, k: i32) -> i32 {
        let mut jobs = jobs;

        jobs.sort_unstable();

        let mut left = jobs.iter().sum::<i32>() / k;
        let mut right = jobs.iter().sum::<i32>();

        while left < right {
            let mid = (left + right) / 2;
            if Self::check(&jobs, k, mid) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }

        left
    }

    fn check(jobs: &[i32], k: i32, limit: i32) -> bool {
        let mut workloads = vec![0; k as usize];
        Self::dfs_helper(jobs, &mut workloads, 0, limit)
    }

    fn dfs_helper(jobs: &[i32], workloads: &mut [i32], i: usize, limit: i32) -> bool {
        if i >= jobs.len() {
            return true;
        }
        let cur = jobs[i];
        for j in 0..workloads.len() {
            if workloads[j] + cur <= limit {
                workloads[j] += cur;
                if Self::dfs_helper(jobs, workloads, i + 1, limit) {
                    return true;
                }
                workloads[j] -= cur;
            }
            if workloads[j] == 0 || workloads[j] + cur == limit {
                break;
            }
        }
        false
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_1723_example_1() {
        let jobs = vec![3, 2, 3];
        let k = 3;

        let result = 3;

        assert_eq!(Solution::minimum_time_required(jobs, k), result);
    }

    #[test]
    fn test_1723_example_2() {
        let jobs = vec![1, 2, 4, 7, 8];
        let k = 2;

        let result = 11;

        assert_eq!(Solution::minimum_time_required(jobs, k), result);
    }
}

// Accepted solution for LeetCode #1723: Find Minimum Time to Finish All Jobs
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1723: Find Minimum Time to Finish All Jobs
// class Solution {
//     private int[] cnt;
//     private int ans;
//     private int[] jobs;
//     private int k;
// 
//     public int minimumTimeRequired(int[] jobs, int k) {
//         this.k = k;
//         Arrays.sort(jobs);
//         for (int i = 0, j = jobs.length - 1; i < j; ++i, --j) {
//             int t = jobs[i];
//             jobs[i] = jobs[j];
//             jobs[j] = t;
//         }
//         this.jobs = jobs;
//         cnt = new int[k];
//         ans = 0x3f3f3f3f;
//         dfs(0);
//         return ans;
//     }
// 
//     private void dfs(int i) {
//         if (i == jobs.length) {
//             int mx = 0;
//             for (int v : cnt) {
//                 mx = Math.max(mx, v);
//             }
//             ans = Math.min(ans, mx);
//             return;
//         }
//         for (int j = 0; j < k; ++j) {
//             if (cnt[j] + jobs[i] >= ans) {
//                 continue;
//             }
//             cnt[j] += jobs[i];
//             dfs(i + 1);
//             cnt[j] -= jobs[i];
//             if (cnt[j] == 0) {
//                 break;
//             }
//         }
//     }
// }

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 × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.