<p>Given two strings <code>s1</code> and&nbsp;<code>s2</code>, return <em>the lowest <strong>ASCII</strong> sum of deleted characters to make two strings equal</em>.</p>

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

<pre>
<strong>Input:</strong> s1 = &quot;sea&quot;, s2 = &quot;eat&quot;
<strong>Output:</strong> 231
<strong>Explanation:</strong> Deleting &quot;s&quot; from &quot;sea&quot; adds the ASCII value of &quot;s&quot; (115) to the sum.
Deleting &quot;t&quot; from &quot;eat&quot; adds 116 to the sum.
At the end, both strings are equal, and 115 + 116 = 231 is the minimum sum possible to achieve this.
</pre>

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

<pre>
<strong>Input:</strong> s1 = &quot;delete&quot;, s2 = &quot;leet&quot;
<strong>Output:</strong> 403
<strong>Explanation:</strong> Deleting &quot;dee&quot; from &quot;delete&quot; to turn the string into &quot;let&quot;,
adds 100[d] + 101[e] + 101[e] to the sum.
Deleting &quot;e&quot; from &quot;leet&quot; adds 101[e] to the sum.
At the end, both strings are equal to &quot;let&quot;, and the answer is 100+101+101+101 = 403.
If instead we turned both strings into &quot;lee&quot; or &quot;eet&quot;, we would get answers of 433 or 417, which are higher.
</pre>

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

<ul>
	<li><code>1 &lt;= s1.length, s2.length &lt;= 1000</code></li>
	<li><code>s1</code> and <code>s2</code> consist of lowercase English letters.</li>
</ul>

import numpy as np

class Solution:
    def minimumDeleteSum(self, s1: str, s2: str) -> int:
        # need something like bottom up levenstein distance
        sol_mat = np.zeros((len(s1)+1,len(s2)+1), dtype=int)
        print(sol_mat)

        # set up col 0 (s1 is rows, s2 is cols)
        for i in range(1,len(s1)+1):
            sol_mat[i,0] = ord(s1[i-1])+sol_mat[i-1,0]
        
        # set up row 0
        for i in range(1,len(s2)+1):
            sol_mat[0,i] = ord(s2[i-1])+sol_mat[0,i-1]
        print(sol_mat)

        for i in range(1,len(s1)+1):
            for j in range(1,len(s2)+1):
                if s1[i-1] == s2[j-1]:
                    sol_mat[i,j] = sol_mat[i-1,j-1]
                else:
                    diag = sol_mat[i-1,j-1]+ord(s1[i-1])+ord(s2[j-1])
                    up = sol_mat[i-1,j] + ord(s1[i-1])
                    left = sol_mat[i,j-1] + ord(s2[j-1])
                    sol_mat[i,j] = min([diag,up,left])
        print(sol_mat)
        return sol_mat[-1,-1]

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