What is the time complexity of Find Minimum Time to Finish All Jobs?

The optimal solution has a time complexity of O(k^n).

What is the space complexity of Find Minimum Time to Finish All Jobs?

The space complexity is O(n+k).

What data structures are used to solve Find Minimum Time to Finish All Jobs?

This problem uses Array, Dynamic Programming, Backtracking, Bit Manipulation, Bitmask. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Find Minimum Time to Finish All Jobs?

Insight 1: Sorting the jobs in descending order helps in pruning the search space during backtracking because assigning larger jobs earlier tends to lead to a higher maximum workload faster, making it easier to identify branches that exceed the current best answer. Insight 2: The problem can be solved using a backtracking algorithm, where we explore all possible assignments of jobs to workers. Backtracking allows us to explore different assignments while maintaining a running minimum of the maximum workload.

Which companies ask LeetCode 1723. Find Minimum Time to Finish All Jobs in interviews?

This problem is frequently asked at Lyft, PhonePe, Pinterest.

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Find Minimum Time to Finish All Jobs - LeetCode 1723 Solution

Find Minimum Time to Finish All Jobs - Complete Solution Guide

Find Minimum Time to Finish All Jobs is LeetCode problem 1723, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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: Inp

Detailed Explanation

The problem asks us to find the minimum possible maximum working time when assigning a set of jobs to a fixed number of workers. Each job must be assigned to exactly one worker, and the working time of a worker is the sum of the time it takes to complete all jobs assigned to them. The goal is to minimize the largest workload among all workers. We are given an array 'jobs' representing the time required for each job and an integer 'k' representing the number of workers.

Solution Approach

The solution uses a backtracking algorithm to explore all possible assignments of jobs to workers. It maintains an array `worker_loads` to track the current workload of each worker. The `dfs` function recursively assigns each job to a worker, updating the `worker_loads` array. Before exploring each possible assignment, it checks if the maximum workload exceeds the current best answer (`ans`). If it does, it prunes the branch. A crucial optimization is to sort the `jobs` array in descending order. Another important optimization occurs when a worker is currently idle (load = 0) because assigning a job to any idle worker in this scenario results in symmetrically equivalent search paths.

Step-by-Step Algorithm

Step 1: Sort the `jobs` array in descending order using `jobs.sort(reverse=True)` (Python) or similar sorting mechanisms in other languages. This helps improve the pruning of the search space.
Step 2: Initialize `worker_loads` array of size `k` with all elements set to 0. This array represents the current workload of each worker.
Step 3: Define a recursive `dfs` function that takes the current job index as input.
Step 4: Inside the `dfs` function, check if the maximum workload in `worker_loads` exceeds the current minimum maximum workload `ans`. If it does, return to prune the search branch.
Step 5: If all jobs have been assigned (base case: `job_index == n`), update `ans` with the maximum workload in `worker_loads` and return.
Step 6: Iterate through each worker (from 0 to `k-1`).
Step 7: Before assigning the current job to the current worker, check if the worker's workload plus the job's time exceeds the current minimum maximum workload `ans`. If it does, skip this assignment (continue to the next worker) to prune the search branch.
Step 8: Assign the current job to the current worker by adding the job's time to `worker_loads[i]`.
Step 9: Recursively call `dfs` with the next job index (`job_index + 1`).
Step 10: Backtrack by undoing the assignment: subtract the job's time from `worker_loads[i]` to revert to the previous state.
Step 11: Prune symmetric branches: If the current worker's load becomes 0 after undoing the assignment, break the inner loop. This is because assigning the current job to any subsequent worker with a load of 0 would lead to equivalent search paths.
Step 12: Call the `dfs` function initially with `job_index = 0`.
Step 13: Return the final `ans` value, which represents the minimum possible maximum working time.

Key Insights

Insight 1: Sorting the jobs in descending order helps in pruning the search space during backtracking because assigning larger jobs earlier tends to lead to a higher maximum workload faster, making it easier to identify branches that exceed the current best answer.
Insight 2: The problem can be solved using a backtracking algorithm, where we explore all possible assignments of jobs to workers. Backtracking allows us to explore different assignments while maintaining a running minimum of the maximum workload.
Insight 3: Pruning is crucial for optimizing the backtracking search. We prune branches that already exceed the current best minimum maximum workload and when assigning to an empty worker. When a worker has a load of 0 it's always best to assign the job to the free worker. This optimization is based on symmetry - if we have workers A and B with the same load, then the order we assign jobs to them doesn't matter. Once we see a zero load, the following branches are symmetrically equivalent.

Complexity Analysis

Time Complexity: O(k^n)

Space Complexity: O(n+k)

Topics

This problem involves: Array, Dynamic Programming, Backtracking, Bit Manipulation, Bitmask.

Companies

Asked at: Lyft, PhonePe, Pinterest.

Find Minimum Time to Finish All Jobs | LeetCode 1723

Hard

Array Dynamic Programming Backtracking Bit Manipulation Bitmask

Time: O(k^n)Space: O(n+k)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    private int ans = Integer.MAX_VALUE;

    public int minimumTimeRequired(int[] jobs, int k) {
        int n = jobs.length;
        this.ans = Integer.MAX_VALUE;

        Arrays.sort(jobs);
        reverseArray(jobs);

        int[] workerLoads = new int[k];

        dfs(jobs, k, workerLoads, 0);
        return this.ans;
    }

    private void dfs(int[] jobs, int k, int[] workerLoads, int jobIndex) {
        if (getMax(workerLoads) >= this.ans) {
            return;
        }

        int n = jobs.length;
        if (jobIndex == n) {
            this.ans = getMax(workerLoads);
            return;
        }

        int currentJob = jobs[jobIndex];

        for (int i = 0; i < k; i++) {
            if (workerLoads[i] + currentJob >= this.ans) {
                continue;
            }

            workerLoads[i] += currentJob;
            dfs(jobs, k, workerLoads, jobIndex + 1);
            workerLoads[i] -= currentJob;

            if (workerLoads[i] == 0) {
                break;
            }
        }
    }

    private int getMax(int[] arr) {
        int max = 0;
        for (int num : arr) {
            max = Math.max(max, num);
        }
        return max;
    }

    private void reverseArray(int[] arr) {
        int start = 0;
        int end = arr.length - 1;
        while (start < end) {
            int temp = arr[start];
            arr[start] = arr[end];
            arr[end] = temp;
            start++;
            end--;
        }
    }
}

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Problem Info

Description

Problem 1723: Find Minimum Time to Finish All Jobs - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Backtracking Bit Manipulation Bitmask

Companies

LyftPhonePePinterest

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:1723

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

13 steps

1
Step 1: Sort the `jobs` array in descending order using `jobs.sort(reverse=True)` (Python) or similar sorting mechanisms in other languages. This helps improve the pruning of the search space.
2
Step 2: Initialize `worker_loads` array of size `k` with all elements set to 0. This array represents the current workload of each worker.
3
Step 3: Define a recursive `dfs` function that takes the current job index as input.
4
Step 4: Inside the `dfs` function, check if the maximum workload in `worker_loads` exceeds the current minimum maximum workload `ans`. If it does, return to prune the search branch.
5
Step 5: If all jobs have been assigned (base case: `job_index == n`), update `ans` with the maximum workload in `worker_loads` and return.
6
Step 6: Iterate through each worker (from 0 to `k-1`).
7
Step 7: Before assigning the current job to the current worker, check if the worker's workload plus the job's time exceeds the current minimum maximum workload `ans`. If it does, skip this assignment (continue to the next worker) to prune the search branch.
8
Step 8: Assign the current job to the current worker by adding the job's time to `worker_loads[i]`.
9
Step 9: Recursively call `dfs` with the next job index (`job_index + 1`).
10
Step 10: Backtrack by undoing the assignment: subtract the job's time from `worker_loads[i]` to revert to the previous state.
11
Step 11: Prune symmetric branches: If the current worker's load becomes 0 after undoing the assignment, break the inner loop. This is because assigning the current job to any subsequent worker with a load of 0 would lead to equivalent search paths.
12
Step 12: Call the `dfs` function initially with `job_index = 0`.
13
Step 13: Return the final `ans` value, which represents the minimum possible maximum working time.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(k^n)

💾

Space Complexity

O(n+k)

📝

Why?

The time complexity arises from the exponential nature of the backtracking search. For each job, we have k choices of which worker to assign it to. The space complexity is linear because we have a recursion depth of n (number of jobs) and an array of size k (number of workers) to store worker loads.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Binary Search with Backtracking

Use binary search to find the minimum possible maximum working time. For a given 'mid' value, check if it's possible to assign all jobs to workers such that no worker's workload exceeds 'mid'. This feasibility check can be done using backtracking. This approach can slightly improve the time complexity in certain cases.

Time: O(n * k^n * log(sum(jobs))), where n is the number of jobs, k is the number of workers, and sum(jobs) is the sum of all job times (the search space for the binary search).Space: O(n + k), similar to the original solution.

Trade-offs: While the binary search adds a logarithmic factor, the backtracking part remains exponential. This approach is helpful when you have a tighter upper bound on possible 'mid' values but the base complexity is still exponential.

Dynamic Programming with Bitmasking

This is not ideal as it is slower. This approach uses a bitmask to represent the set of jobs that have been assigned and stores the minimum maximum workload for each possible subset of jobs. It can be helpful in situations where certain optimizations or constraints make it more suitable, but for the standard problem, the backtracking is preferred.

Time: O(3^n * k) approximately. There are 2^n subsets of jobs. For each one, we need to consider assigning that subset to each worker to find out whether the max job size exceeds target value. For assigning a task to worker `k`, we may need to compare workloads of the other `k-1` workersSpace: O(2^n) to store dp results for each possible set of jobs.

Trade-offs: Backtracking is preferred over Dynamic Programming for this problem as dynamic programming requires much more memory to store. However, dynamic programming may work if there were better-defined subproblems with good overlap.

✨

Key Insights

1Insight 1: Sorting the jobs in descending order helps in pruning the search space during backtracking because assigning larger jobs earlier tends to lead to a higher maximum workload faster, making it easier to identify branches that exceed the current best answer.
2Insight 2: The problem can be solved using a backtracking algorithm, where we explore all possible assignments of jobs to workers. Backtracking allows us to explore different assignments while maintaining a running minimum of the maximum workload.
3Insight 3: Pruning is crucial for optimizing the backtracking search. We prune branches that already exceed the current best minimum maximum workload and when assigning to an empty worker. When a worker has a load of 0 it's always best to assign the job to the free worker. This optimization is based on symmetry - if we have workers A and B with the same load, then the order we assign jobs to them doesn't matter. Once we see a zero load, the following branches are symmetrically equivalent.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: When k = jobs.length, the optimal solution is the maximum value in the jobs array. The algorithm correctly handles this by assigning each job to a separate worker, resulting in the maximum job time as the minimum maximum workload.
Edge case 2: When k = 1, the optimal solution is the sum of all job times. The algorithm correctly handles this by assigning all jobs to the single worker.
Edge case 3: When jobs array is empty, return 0. While the prompt specifies that 1 <= k <= jobs.length, it is good practice to consider this edge case. This solution would need a check `if jobs.length == 0: return 0;`

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to backtrack correctly, which leads to incorrect assignments and incorrect results. It's crucial to undo the changes made during the recursive call to explore other possibilities.

✗Common mistake 2: Not implementing pruning effectively, resulting in excessive exploration of the search space and exceeding time limits. The checks for exceeding the current minimum maximum workload and zero-load workers are crucial.

✗Common mistake 3: Not sorting jobs in descending order; this can lead to significantly slower execution times.

✗Common mistake 4: Not recognizing the symmetric nature of idle workers, thereby skipping the break statement in zero load optimization. This means that redundant branches will be explored, wasting processing time and potentially leading to slower execution.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Task scheduling in a multi-processor system, where the goal is to minimize the overall completion time by assigning tasks to processors in a balanced way. This algorithm, while simplified, reflects the challenges of efficient task allocation.

🏦Real-world scenario 2: Workload balancing in cloud computing, where virtual machines need to be assigned to physical servers in a way that minimizes the load on any single server. Similar backtracking or approximation techniques can be employed.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 12, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Find Minimum Time to Finish All Jobs"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?