leetcode 1335 - Minimum Difficulty of a Job Schedule

https://leetcode.com/problems/minimum-difficulty-of-a-job-schedule/
https://zxi.mytechroad.com/blog/dynamic-programming/leetcode-1335-minimum-difficulty-of-a-job-schedule/

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

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class Solution {
public:
int minDifficulty(vector<int>& jobDifficulty, int d) {
int n = jobDifficulty.size();

if (d > n) return -1;

vector<vector<int>> dp(n+1, vector<int>(d+1, INT_MAX/2));

dp[0][0] = 0;

for (int i = 1; i <= n; i++) {
for (int j = 1; j <= d; j++) {
int md = 0;
for (int k = i - 1; k >= j - 1; k--) {
md = max(md, jobDifficulty[k]);

dp[i][j] = min(dp[i][j], dp[k][j-1] + md);
}
}
}

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