<p>You want to schedule a list of jobs in <code>d</code> days. Jobs are dependent (i.e To work on the <code>i<sup>th</sup></code> job, you have to finish all the jobs <code>j</code> where <code>0 &lt;= j &lt; i</code>).</p>

<p>You have to finish <strong>at least</strong> one task every day. The difficulty of a job schedule is the sum of difficulties of each day of the <code>d</code> days. The difficulty of a day is the maximum difficulty of a job done on that day.</p>

<p>You are given an integer array <code>jobDifficulty</code> and an integer <code>d</code>. The difficulty of the <code>i<sup>th</sup></code> job is <code>jobDifficulty[i]</code>.</p>

<p>Return <em>the minimum difficulty of a job schedule</em>. If you cannot find a schedule for the jobs return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/01/16/untitled.png" style="width: 365px; height: 370px;" />
<pre>
<strong>Input:</strong> jobDifficulty = [6,5,4,3,2,1], d = 2
<strong>Output:</strong> 7
<strong>Explanation:</strong> 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 
</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p><strong class="example">Example 3:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= jobDifficulty.length &lt;= 300</code></li>
	<li><code>0 &lt;= jobDifficulty[i] &lt;= 1000</code></li>
	<li><code>1 &lt;= d &lt;= 10</code></li>
</ul>

class Solution {
public:
    int solve(int ind, int days, vector<int> jd, int n,
              vector<vector<int>>& dp) {
        if (days == 1) {
            int maxi = jd[ind];
            for (int i = ind; i < n; i++)
                maxi = max(maxi, jd[i]);

            return maxi;
        }
        if (dp[ind][days] != -1)
            return dp[ind][days];

        int maxi = jd[ind];
        int finalResult = INT_MAX;

        for (int i = ind; i <= n - days; i++) {
            maxi = max(maxi, jd[i]);
            int result = maxi + solve(i + 1, days - 1, jd, n, dp);
            finalResult = min(result, finalResult);
        }
        return dp[ind][days] = finalResult;
    }
    int minDifficulty(vector<int>& jd, int d) {
        int n = jd.size();
        if (d > n)
            return -1;
        vector<vector<int>> t(301, vector<int>(11, 0));
        for (int i = 0; i < n; i++) {
            t[i][1] = *max_element(begin(jd) + i, end(jd));
        }
         for (int days = 2; days <= d; days++) {
            for (int i = 0; i < n ; i++) {
                int maxDifficulty = INT_MIN;
                int result        = INT_MAX;

                for (int j = i; j <= n - days; j++) {
                    maxDifficulty   = max(maxDifficulty, jd[j]);
                    result          = min(result, maxDifficulty + t[j + 1][days - 1]);
                }

                t[i][days] = result;
            }
        }

        return t[0][d];
    }
};
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