Find Minimum Time to Finish All Jobs Interview Question for Amazon

Optimal Assignment of Jobs to Workers

This problem is about finding the optimal way to assign jobs to workers such that the maximum working time of any worker is minimized. This can be solved using binary search combined with a backtracking approach to check if a given maximum working time is feasible.

1. Naive (Brute Force) Solution

One brute-force approach would be to try all possible assignments of jobs to workers and calculate the maximum working time for each assignment. This can be done by generating all possible partitions of the jobs array into k subsets and then calculating the maximum sum of each subset. The minimum of these maximum sums would be the answer.

This approach is highly inefficient due to the exponential number of possible assignments. Generating all possible partitions is computationally expensive, making it impractical for larger input sizes.

2. Optimal Solution (Binary Search + Backtracking)

The optimal solution involves using binary search to find the minimum possible maximum working time. We can define a search space where the lower bound is the maximum job length, and the upper bound is the sum of all job lengths. For each value (potential maximum working time) in this space, we use a backtracking algorithm to determine if it's possible to assign all jobs to k workers such that no worker exceeds this maximum working time.

Algorithm:

Define Search Space: The lower bound low is the maximum value in the jobs array. The upper bound high is the sum of all values in the jobs array.
Binary Search: Perform binary search within the range [low, high].
Feasibility Check (Backtracking): For each mid value during binary search, check if it's possible to assign jobs such that no worker's workload exceeds mid using a backtracking algorithm.
Update Search Space:
- If the assignment is feasible, it means we can potentially find a smaller maximum working time. Update high = mid - 1.
- If the assignment is not feasible, it means the maximum working time needs to be larger. Update low = mid + 1.
Return Result: When the binary search completes (low > high), low will be the minimum possible maximum working time.

Code Implementation (Java):

public class AssignJobs {

    public int minimizeMax(int[] jobs, int k) {
        int low = 0, high = 0;
        for (int job : jobs) {
            low = Math.max(low, job);
            high += job;
        }

        int ans = high;
        while (low <= high) {
            int mid = (low + high) / 2;
            if (isPossible(jobs, k, mid)) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return ans;
    }

    private boolean isPossible(int[] jobs, int k, int maxLoad) {
        int[] workers = new int[k];
        return backtrack(jobs, workers, 0, maxLoad);
    }

    private boolean backtrack(int[] jobs, int[] workers, int jobIndex, int maxLoad) {
        if (jobIndex == jobs.length) {
            return true;
        }

        for (int i = 0; i < workers.length; i++) {
            if (workers[i] + jobs[jobIndex] <= maxLoad) {
                workers[i] += jobs[jobIndex];
                if (backtrack(jobs, workers, jobIndex + 1, maxLoad)) {
                    return true;
                }
                workers[i] -= jobs[jobIndex]; // Backtrack
            }
            // Optimization: If the worker is empty, assigning the job to another empty worker
            // won't lead to a different result. Prevents duplicate assignments.
            if (workers[i] == 0) break;
        }
        return false;
    }

    public static void main(String[] args) {
        AssignJobs solution = new AssignJobs();

        // Example 1
        int[] jobs1 = {3, 2, 3};
        int k1 = 3;
        System.out.println("Example 1: " + solution.minimizeMax(jobs1, k1)); // Output: 3

        // Example 2
        int[] jobs2 = {1, 2, 4, 7, 8};
        int k2 = 2;
        System.out.println("Example 2: " + solution.minimizeMax(jobs2, k2)); // Output: 11

        // Example 3
        int[] jobs3 = {5, 5, 5, 4, 6};
        int k3 = 4;
        System.out.println("Example 3: " + solution.minimizeMax(jobs3, k3)); // Output: 6
    }
}

3. Big(O) Run-time Analysis

Binary Search: The binary search runs in O(log(Sum - Max)) where Sum is the sum of all job lengths and Max is the maximum job length. This provides the logarithmic factor of the overall time complexity.
Backtracking: In the worst case, the backtracking algorithm explores all possible assignments. The number of possible assignments for each job is k, and we have n jobs. Thus, in the worst case, the backtracking might take O(k^n) time. However, the optimizations and pruning help reduce the average time.
Overall: The overall time complexity is O(log(Sum - Max) * k^n) in the worst case, where n is the number of jobs and k is the number of workers. Because n is small (n <= 12), this is generally acceptable. The efficiency primarily depends on how effectively the backtracking is optimized.

4. Big(O) Space Usage Analysis

Workers Array: The workers array in the isPossible function requires O(k) space to track the current load of each worker.
Recursion Stack: The backtracking algorithm uses a call stack. In the worst case, the depth of the recursion can be n (number of jobs).
Overall: The overall space complexity is O(k + n), where k is the number of workers, and n is the number of jobs. The space is mainly used for storing the workers array and the recursion call stack.

5. Edge Cases and Handling

k = 1: If there is only one worker, the minimum maximum working time is simply the sum of all job lengths. The binary search will still work correctly, but it is an important edge case to consider.
k >= n: If the number of workers is greater than or equal to the number of jobs, we can assign each job to a separate worker. In this case, the minimum maximum working time is the maximum job length. This is naturally handled by the binary search because the lower bound is set to the maximum job length.
Empty Jobs Array: If the jobs array is empty, the minimum maximum working time is 0. This should be handled as a base case. The provided code does not explicitly handle an empty jobs array, but it will behave reasonably as the low variable will remain 0, and the while loop will not execute, directly returning ans = 0.
Large Job Lengths: If the job lengths are very large, the sum of all job lengths might cause an integer overflow. To prevent this, using long data type would be better when computing the sum and performing binary search.