<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:
        ######## recursion with memo
        # memo = {}
        # def helper(i, j):
        #     if i == len(s1):
        #         return sum([ord(char) for char in s2[j:]])
        #     elif j == len(s2):
        #         return sum([ord(char) for char in s1[i:]])
        #     else:
        #         if (i,j) in memo:
        #             return memo[i,j]
        #         if s1[i] == s2[j]:
        #             result = helper(i+1, j+1)
        #         else:
        #             result = min(ord(s2[j]) + helper(i, j+1), ord(s1[i]) + helper(i+1, j))
        #         memo[i, j]  = result
        #         return result
        # return helper(0, 0)

        ####### DP approach
        DP = np.zeros((1+len(s1), 1+len(s2)), dtype=np.int64)
        # DP[i,j] represents the min ASCII sum for the two substrings s1[:i] and s2[:j]
        DP[0,1:] = np.cumsum([ord(char) for char in s2])
        DP[1:,0] = np.cumsum([ord(char) for char in s1])
        for i in range(1, len(s1)+1):
            for j in range(1, len(s2)+1):
                # upperleft
                route1 = (ord(s1[i-1]) + ord(s2[j-1]))*(s1[i-1] != s2[j-1]) + DP[i-1,j-1]
                # left
                route2 = ord(s2[j-1]) + DP[i,j-1]
                # upper
                route3 = ord(s1[i-1]) + DP[i-1,j]
                DP[i,j] = min(min(route1, route2), route3)
        #print(DP)
        return DP[-1,-1] 
29781