<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:
    def minDifficulty(self, jobDifficulty: List[int], d: int) -> int:
        # dp = [[0] * (len(jobDifficulty) + 1) for _ in range(d+1)]
        # difficulties = [0] * d
        # sum
        # # currently day at i-1 day, at j job
        # for i in range(1, d + 1):
        #     for j in range(i, len(jobDifficulty) +1):
        #         # prev += jobDifficulty[j]
        #         # dp[i-1][j-1] is the difficulty from previous day
        #         # dp[i][j-1] is the difficulty from last task
        #         dp[i][j] = dp[i-1][j-1] + max(dp[i][j-1], jobDifficulty[j-1])
        #         print(i, j, dp[i][j])
        # print(dp)
        # return min(dp[d-1])
        if len(jobDifficulty) < d:
            return -1
        memo = [[-1] * (d+1) for _ in range(len(jobDifficulty))]
        def min_diff(curr_start: int, num_day: int) -> int:
            if memo[curr_start][num_day] != -1:
                return memo[curr_start][num_day]
            # base case
            if num_day == 1:
                # do all remaining job in one day
                memo[curr_start][num_day] = max(jobDifficulty[curr_start:])
                return memo[curr_start][num_day]
            result = 1001 * d
            for j in range(curr_start+1, len(jobDifficulty)):
                result = min(result, min_diff(j, num_day-1) + max(jobDifficulty[curr_start:j]))
            memo[curr_start][num_day] = result
            return result
        return min_diff(0, d)
            
        
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