<p>There is an <code>m x n</code> cake that needs to be cut into <code>1 x 1</code> pieces.</p>

<p>You are given integers <code>m</code>, <code>n</code>, and two arrays:</p>

<ul>
	<li><code>horizontalCut</code> of size <code>m - 1</code>, where <code>horizontalCut[i]</code> represents the cost to cut along the horizontal line <code>i</code>.</li>
	<li><code>verticalCut</code> of size <code>n - 1</code>, where <code>verticalCut[j]</code> represents the cost to cut along the vertical line <code>j</code>.</li>
</ul>

<p>In one operation, you can choose any piece of cake that is not yet a <code>1 x 1</code> square and perform one of the following cuts:</p>

<ol>
	<li>Cut along a horizontal line <code>i</code> at a cost of <code>horizontalCut[i]</code>.</li>
	<li>Cut along a vertical line <code>j</code> at a cost of <code>verticalCut[j]</code>.</li>
</ol>

<p>After the cut, the piece of cake is divided into two distinct pieces.</p>

<p>The cost of a cut depends only on the initial cost of the line and does not change.</p>

<p>Return the <strong>minimum</strong> total cost to cut the entire cake into <code>1 x 1</code> pieces.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">m = 3, n = 2, horizontalCut = [1,3], verticalCut = [5]</span></p>

<p><strong>Output:</strong> <span class="example-io">13</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/04/ezgifcom-animated-gif-maker-1.gif" style="width: 280px; height: 320px;" /></p>

<ul>
	<li>Perform a cut on the vertical line 0 with cost 5, current total cost is 5.</li>
	<li>Perform a cut on the horizontal line 0 on <code>3 x 1</code> subgrid with cost 1.</li>
	<li>Perform a cut on the horizontal line 0 on <code>3 x 1</code> subgrid with cost 1.</li>
	<li>Perform a cut on the horizontal line 1 on <code>2 x 1</code> subgrid with cost 3.</li>
	<li>Perform a cut on the horizontal line 1 on <code>2 x 1</code> subgrid with cost 3.</li>
</ul>

<p>The total cost is <code>5 + 1 + 1 + 3 + 3 = 13</code>.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">m = 2, n = 2, horizontalCut = [7], verticalCut = [4]</span></p>

<p><strong>Output:</strong> <span class="example-io">15</span></p>

<p><strong>Explanation:</strong></p>

<ul>
	<li>Perform a cut on the horizontal line 0 with cost 7.</li>
	<li>Perform a cut on the vertical line 0 on <code>1 x 2</code> subgrid with cost 4.</li>
	<li>Perform a cut on the vertical line 0 on <code>1 x 2</code> subgrid with cost 4.</li>
</ul>

<p>The total cost is <code>7 + 4 + 4 = 15</code>.</p>
</div>

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

<ul>
	<li><code>1 &lt;= m, n &lt;= 20</code></li>
	<li><code>horizontalCut.length == m - 1</code></li>
	<li><code>verticalCut.length == n - 1</code></li>
	<li><code>1 &lt;= horizontalCut[i], verticalCut[i] &lt;= 10<sup>3</sup></code></li>
</ul>

class Solution:
    def minimumCost(self, m: int, n: int, horizontalCut: List[int], verticalCut: List[int]) -> int:
        # MINCOST(i,j,k,l) is the minimum cost to cut a rectangular section part of cake from (i,j) to (k,l)
        # into 1X1 blocks
        # target is MINCOST(0,0,m,n)
        # MINCOST(i,j,i+1,j+1) is 0 for all i,j
        # MINCOST(i,j,k,l) = MIN(MIN[for i < x < k](horizontalCut[x-1] + MINCOST(i,j,x,l) + MINCOST(x,j,k,l)), MIN[for j < x < l](verticalCut[x-1] + MINCOST(i,j,k,x) + MINCOST(i,x,k,l)))
        # rather than building 4D DP table I use memoization with recursion.
        # max recursion depth is 40
        calculated_val = {}
        def min_cost(i,j,k,l):
            if (i,j,k,l) in calculated_val:
                return calculated_val[(i,j,k,l)]
            elif k == i+1 and l == j+1:
                calculated_val[(i,j,k,l)] = 0
                return calculated_val[(i,j,k,l)]
            else:
                min_v = 1000000000000000
                for x in range(i+1,k):
                    min_s = horizontalCut[x-1] + min_cost(i,j,x,l) + min_cost(x,j,k,l)
                    if min_s < min_v:
                        min_v = min_s
                for x in range(j+1,l):
                    min_s = verticalCut[x-1] + min_cost(i,j,k,x) + min_cost(i,x,k,l)
                    if min_s < min_v:
                        min_v = min_s
                calculated_val[(i,j,k,l)] = min_v
                return calculated_val[(i,j,k,l)]

        return min_cost(0,0,m,n)




         
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