<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 functools
import math
class Solution:
    def minimumDeleteSum(self, s1: str, s2: str) -> int:
        m, n = len(s1), len(s2)
        dp = {}
        def dfs(i, j):
            if i <0 and j < 0:
                return 0
            if (i, j) in dp:
                return dp[(i, j)]
            if i < 0:
                dp[(i, j)]= sum([ord(k) for l, k in enumerate(s2) if l <= j ])
                return dp[(i, j)]
            if j < 0:
                dp[(i, j)] = sum([ord(k) for l, k in enumerate(s1) if l <= i ])
                return dp[(i, j)]
            if s1[i] == s2[j]:
                dp[(i, j)] = dfs(i-1, j-1)
            else:
                dp[(i, j)] = min(ord(s1[i])+dfs(i-1, j), ord(s2[j])+dfs(i, j-1))
            return dp[(i, j)]
        # m, n = len(s1)
        dp = [[math.inf for i in range(n+1)] for j in range(m+1)]
        dp[0][0] = 0
        ascii = {i:ord(i) for i in 'abcdefghijklmnopqrstuvwxyz'}
        for i in range(1, n+1):
            dp[0][i] = (ascii[s2[i-1]]+dp[0][i-1])
        
        for i in range(1, m+1):
            dp[i][0] = (ascii[s1[i-1]]+dp[i-1][0])

        for i in range(1, m+1):
            for j in range(1, n+1):
                if s1[i-1] == s2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min(ascii[s1[i-1]]+dp[i-1][j], ascii[s2[j-1]]+dp[i][j-1])
        # print(dp)
        return dp[m][n]

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