<p>Given the array <code>houses</code> where <code>houses[i]</code> is the location of the <code>i<sup>th</sup></code> house along a street and an integer <code>k</code>, allocate <code>k</code> mailboxes in the street.</p>

<p>Return <em>the <strong>minimum</strong> total distance between each house and its nearest mailbox</em>.</p>

<p>The test cases are generated so that the answer fits in a 32-bit integer.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/05/07/sample_11_1816.png" style="width: 454px; height: 154px;" />
<pre>
<strong>Input:</strong> houses = [1,4,8,10,20], k = 3
<strong>Output:</strong> 5
<strong>Explanation:</strong> Allocate mailboxes in position 3, 9 and 20.
Minimum total distance from each houses to nearest mailboxes is |3-1| + |4-3| + |9-8| + |10-9| + |20-20| = 5 
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/05/07/sample_2_1816.png" style="width: 433px; height: 154px;" />
<pre>
<strong>Input:</strong> houses = [2,3,5,12,18], k = 2
<strong>Output:</strong> 9
<strong>Explanation:</strong> Allocate mailboxes in position 3 and 14.
Minimum total distance from each houses to nearest mailboxes is |2-3| + |3-3| + |5-3| + |12-14| + |18-14| = 9.
</pre>

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

<ul>
	<li><code>1 &lt;= k &lt;= houses.length &lt;= 100</code></li>
	<li><code>1 &lt;= houses[i] &lt;= 10<sup>4</sup></code></li>
	<li>All the integers of <code>houses</code> are <strong>unique</strong>.</li>
</ul>

class Solution:
    def minDistance(self, houses: List[int], k: int) -> int:
        houses.sort()
        self.note = {}
        self.note2 = {}
        return self.dfs(houses, 0, k)
    
    def dfs(self, houses: List[int], left: int, k: int) -> int:
        if k == 1:
            return self.calc(houses, left, len(houses) - 1)
        
        key = (k, left)
        if key in self.note:
            return self.note[key]
        
        result = sys.maxsize
        for right in range(left + 1, len(houses)):
            temp = self.calc(houses, left, right - 1)
            result = min(result, temp + self.dfs(houses, right, k - 1))
            
        self.note[key] = result
        return result
    
    def calc(self, houses: List[int], left: int, right: int) -> int:
        key = (left, right)
        if key in self.note2:
            return self.note2[key]

        result = 0
        while left < right:
            result += houses[right] - houses[left]
            left += 1
            right -= 1

        self.note2[key] = result        
        return result
17500