LeetCode 2323: Find Minimum Time to Finish All Jobs II Solution

Master LeetCode problem 2323 (Find Minimum Time to Finish All Jobs II), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2323. Find Minimum Time to Finish All Jobs II

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming with state compression. We will create a 2D dp array where dp[i][mask] represents the minimum total time to finish all jobs up to the i-th job with the assigned workers represented by the bitmask mask.

For each job, we iterate through all possible worker assignments (represented by the bitmask) and calculate the minimum total time based on the previous state. We update the dp array accordingly.

Finally, we return the minimum total time to finish all jobs with all workers assigned.

Time Complexity

The time complexity of this solution is O(n * 2^k), where n is the number of jobs and k is the number of workers.

Space Complexity

The space complexity of this solution is O(n * 2^k).

Solution Code

class Solution {
    public int minTime(int[] jobs, int k) {
        int n = jobs.length;
        int[][] dp = new int[n + 1][1 << k];
        
        for (int i = 0; i <= n; i++) {
            Arrays.fill(dp[i], Integer.MAX_VALUE / 2);
        }
        
        dp[0][0] = 0;
        
        for (int i = 1; i <= n; i++) {
            for (int mask = 0; mask < (1 << k); mask++) {
                for (int sub = mask; sub > 0; sub = (sub - 1) & mask) {
                    int workers = Integer.bitCount(sub);
                    dp[i][mask] = Math.min(dp[i][mask], Math.max(dp[i - 1][mask - sub], getSum(jobs, i - workers, i)));
                }
            }
        }
        
        return dp[n][(1 << k) - 1];
    }
    
    private int getSum(int[] jobs, int start, int end) {
        int sum = 0;
        for (int i = start; i < end; i++) {
            sum += jobs[i];
        }
        return sum;
    }
}

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