<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:
        def dp(i, k):
            if i >= len(houses):
                return 0

            if k == 0:
                return sys.maxsize

            if (i, k) in cache:
                return cache[(i, k)]

            arr = []
            med = -1
            j = i
            cost = 0
            ret = sys.maxsize

            while j < min(len(houses), len(houses) - (k - 1)):
                if med >= 0:
                    cost += (houses[j] - houses[arr[med]])

                arr.append(j)

                if len(arr) % 2 == 1:
                    med += 1
                    
                    if med > 0:
                        diff = houses[arr[med]] - houses[arr[med - 1]]
                        cost += (diff * (med))
                        cost -= (diff * (len(arr) - med))

                ret = min(ret, cost + dp(j + 1, k - 1))
                j += 1

            cache[(i, k)] = ret
            return ret

        if k == len(houses):
            return 0

        houses.sort()
        cache = {}
        return dp(0, k)
17600