<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>

class Solution:
    def computeCost(self, s1: str, s2: str, i: int, j: int):
        if (i, j) in self.cache:
            return self.cache[i, j]
        if i < 0:
            return sum(ord(c) for c in s2[:j+1])
        if j < 0:
            return sum(ord(c) for c in s1[:i+1])
        if s1[i] == s2[j]:
            return self.computeCost(s1, s2, i - 1, j - 1)
        val = min(
            self.computeCost(s1, s2, i - 1, j    ) + ord(s1[i]), # delete s1[i]
            self.computeCost(s1, s2, i    , j - 1) + ord(s2[j]), # detele s2[j]
            self.computeCost(s1, s2, i - 1, j - 1) + ord(s1[i]) + ord(s2[j]), # delete both s1[i] and s2[j]
        )
        self.cache[i, j] = val
        return val

    def minimumDeleteSum(self, s1: str, s2: str) -> int:
        """
        
        For each character in both strings, consider what happens if we delete either:
            1. delete character from s1
            2. delete character from s2
            3. delete both characters from s1 and s2
            4. keep both

            Idea: consider s1 vs substring(s2) in a way to re-use results from previous s1 vs substring(s1).
        """
        if not s1:
            return sum(ord(c) for c in s2)
        if not s2:
            return sum(ord(c) for c in s1)
        self.cache = {}
        return self.computeCost(s1, s2, len(s1) - 1, len(s2) - 1)
37135