<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 minDifficulty(vector<int>& jobDifficulty, int d) {
        int n = jobDifficulty.size();
        
        // If we can't split jobs into exactly d days, return -1.
        if (d > n) return -1;
        
        // DP table: dp[i][cut] is the minimum difficulty to schedule jobs 0..i with 'cut' cuts.
        vector<vector<int>> dp(n, vector<int>(d, INT_MAX));
        
        // Base case: When we make d-1 cuts, the remaining jobs after the last cut are assigned to the last day.
        dp[0][0] = jobDifficulty[0];
        
        // Preprocessing to fill the first column (cut == 0, meaning no cuts have been made yet)
        for (int i = 1; i < n; ++i) {
            dp[i][0] = max(dp[i-1][0], jobDifficulty[i]);
        }
        
        // Fill the DP table for 1 to d-1 cuts
        for (int cut = 1; cut < d; ++cut) {
            // Loop through all job indices where we can place the cut
            for (int i = cut; i < n; ++i) {
                int maxDifficulty = 0;
                
                // Try placing the cut at different positions
                for (int j = i; j >= cut; --j) {
                    maxDifficulty = max(maxDifficulty, jobDifficulty[j]);
                    dp[i][cut] = min(dp[i][cut], dp[j-1][cut-1] + maxDifficulty);
                }
            }
        }
        
        // The answer will be in dp[n-1][d-1], the minimum difficulty for all jobs split into d days
        return dp[n-1][d-1];
    }
};

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