<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) {
        return minDifficultyVersion(jobDifficulty, d);
    }

    int minDifficultyVersion(const vector<int>& jobDifficulty, int d) {
        const int n = static_cast<int>(jobDifficulty.size());

        if (n < d) {
            return -1;
        } else if (n == d) {
            int totalDifficulty = 0;
            for (int difficulty : jobDifficulty)
                totalDifficulty += difficulty;
            return totalDifficulty;
        }

        // dp[j]:
        //    the minimum difficulty OF
        //       the first (j+1) jobs, exactly scheduled in (i+1) days.
        vector<int> dp(n);
        dp[0] = jobDifficulty[0];

        // Initializing the dp array to store the maximum difficulty encountered so far.
        for (int i = 1; i < n; ++i) {
            dp[i] = max(jobDifficulty[i], dp[i - 1]);
        }

        vector<int> dpPrev(n);

        // Dynamic Programming to find the minimum difficulty.
        for (int i = 1; i < d; ++i) {
            dp.swap(dpPrev);
            for (int j = i; j < n; ++j) {
                int lastDayDifficulty = jobDifficulty[j];
                int tmpMin = lastDayDifficulty + dpPrev[j - 1];

                // Iterate to find the minimum difficulty for the current day.
                for (int t = j - 1; i - 1 < t; --t) {
                    lastDayDifficulty = max(lastDayDifficulty, jobDifficulty[t]);
                    tmpMin = min(tmpMin, lastDayDifficulty + dpPrev[t - 1]);
                }

                dp[j] = tmpMin;
            }
        }

        return dp.back();
    }
};

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