LeetCode 1335: Minimum Difficulty of a Job Schedule Solution

Master LeetCode problem 1335 (Minimum Difficulty of a Job Schedule), a hard 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.

1335. Minimum Difficulty of a Job Schedule

Hard

Array Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We will define a 2D array dp where dp[i][j] represents the minimum difficulty to finish the first i jobs in j days. We will iterate over the jobs and days, updating the dp array based on the maximum difficulty for each day.

Initialize the dp array with size n+1 by d+1, where n is the number of jobs.
Set dp[0][0] = 0 since there is no difficulty if there are no jobs and days.
Iterate over the jobs and days:
- For each job i, update dp[i][j] by considering all previous jobs and days.
- Calculate the maximum difficulty for each day based on the current job and update dp[i][j].
The minimum difficulty to finish all jobs in d days is dp[n][d]. If it is INF, return -1.

Time Complexity: The time complexity of this solution is O(n^2 * d), where n is the number of jobs and d is the number of days.

Space Complexity: The space complexity of this solution is O(n * d).

Solution Code

class Solution {
    public int minDifficulty(int[] jobDifficulty, int d) {
        int n = jobDifficulty.length;
        if (n < d) return -1;
        
        int[][] dp = new int[n + 1][d + 1];
        for (int[] row : dp) {
            Arrays.fill(row, Integer.MAX_VALUE);
        }
        
        dp[0][0] = 0;
        
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= d; j++) {
                int maxDifficulty = 0;
                for (int k = i - 1; k >= j - 1; k--) {
                    maxDifficulty = Math.max(maxDifficulty, jobDifficulty[k]);
                    if (dp[k][j - 1] != Integer.MAX_VALUE) {
                        dp[i][j] = Math.min(dp[i][j], dp[k][j - 1] + maxDifficulty);
                    }
                }
            }
        }
        
        return dp[n][d] != Integer.MAX_VALUE ? dp[n][d] : -1;
    }
}

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