# 1335.Minimum-Difficulty-of-a-Job-Schedule

## 1335. Minimum Difficulty of a Job Schedule

## 题目地址

[https://leetcode.com/problems/minimum-difficulty-of-a-job-schedule/](https://leetcode.com/problems/minimum-difficulty-of-a-job-schedule/)

## 题目描述

```
You want to schedule a list of jobs in d days. Jobs are dependent (i.e To work on the i-th job, you have to finish all the jobs j where 0 <= j < i).

You have to finish at least one task every day. The difficulty of a job schedule is the sum of difficulties of each day of the d days. The difficulty of a day is the maximum difficulty of a job done in that day.

Given an array of integers jobDifficulty and an integer d. The difficulty of the i-th job is jobDifficulty[i].

Return the minimum difficulty of a job schedule. If you cannot find a schedule for the jobs return -1.


Example 1:
Input: jobDifficulty = [6,5,4,3,2,1], d = 2
Output: 7
Explanation: First day you can finish the first 5 jobs, total difficulty = 6.
Second day you can finish the last job, total difficulty = 1.
The difficulty of the schedule = 6 + 1 = 7 

Example 2:
Input: jobDifficulty = [9,9,9], d = 4
Output: -1
Explanation: If you finish a job per day you will still have a free day. you cannot find a schedule for the given jobs.

Example 3:
Input: jobDifficulty = [1,1,1], d = 3
Output: 3
Explanation: The schedule is one job per day. total difficulty will be 3.

Example 4:
Input: jobDifficulty = [7,1,7,1,7,1], d = 3
Output: 15

Example 5:
Input: jobDifficulty = [11,111,22,222,33,333,44,444], d = 6
Output: 843

Constraints:
1 <= jobDifficulty.length <= 300
0 <= jobDifficulty[i] <= 1000
1 <= d <= 10
```

## 代码

### Approach #1 DFS

Time: O(nnd) && Space: O(nd)

```java
class Solution {
    public int minDifficulty(int[] jobDifficulty, int d) {
        int n = jobDifficulty.length;
        if (d > n) {
            return -1;
        }

        int[][] memo = new int[d + 1][n];

        for (int i = 0; i <= d; i++) {
            Arrays.fill(memo[i], -1);
        }

        return dfs(d, 0, jobDifficulty, memo);
    }

    private int dfs(int day, int index, int[] jobs, int[][] memo) {
        final int n = jobs.length;
        if (day == 0 && index == n) {
            return 0;
        }
        if (day == 0 || index == n) {
            return Integer.MAX_VALUE;
        }

        if (memo[day][index] != -1) {
            return memo[day][index];
        }

        int curMax = jobs[index];
        int min = Integer.MAX_VALUE;
        for (int i = index; i < n; i++) {
            curMax = Math.max(jobs[i], curMax);
            int temp = dfs(day - 1, i + 1, jobs, memo);
            if (temp != Integer.MAX_VALUE) {
                min = Math.min(min, temp + curMax);
            }
        }

        memo[day][index] = min;

        return min;
    }
}
```

### Approach #2 DP

Time: O(nnd) && Space: O(nd)

```java
class Solution {
  public int minDifficulty(int[] jobDifficulty, int d) {
    int n = jobDifficulty.length;
    if (d > n)  return -1;
    int[][] dp = new int[d + 1][n + 1];

      // 初始化条件很重要
    for (int i = 0; i <= d; i++) {
       Arrays.fill(dp[i], Integer.MAX_VALUE);
    }
    dp[0][0] = 0;
    for (int day = 1; day <= d; day++) {
        for (int task = day; task <= n; task++) {
            int localMax = jobDifficulty[task-1];
            for (int s = task; s >= day; s--) {
                localMax = Math.max(localMax, jobDifficulty[s-1]);

                if (dp[day-1][s-1] != Integer.MAX_VALUE) { // 注意越界
                    dp[day][task] = Math.min(dp[day][task], dp[day-1][s-1] + localMax);
                }

            }
        }
    }

    return dp[d][n];
  }
}
```
