<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 minimumDeleteSum(self, s1: str, s2: str) -> int:
        self.sum_dict = {}
        ans = self.helper(s1, s2)
        #print(f'dict={self.sum_dict}')
        return ans
    
    def helper(self, s1: str, s2: str) -> int:
        if s1 == s2:
            return 0
        tuple = (s1, s2)
        if tuple in self.sum_dict:
            return self.sum_dict[tuple]
        if not s1 or not s2:
            s = s1 if s1 else s2
            delete_sum = sum(ord(c) for c in s)
            self.sum_dict[tuple] = delete_sum
            return delete_sum
        #find the 1st diff
        m, n = len(s1), len(s2)
        i, min_num = 0, min(m, n)
        while i < min_num and s1[i] == s2[i]:
            i += 1
        if i == min_num:
            delete_sum = self.helper(s1[min_num:], s2[min_num:])
        else:
            delete_sum = min(ord(s1[i]) + self.helper(s1[i+1:], s2[i:]), ord(s2[i]) + self.helper(s1[i:], s2[i+1:]))
            delete_sum = min(delete_sum, ord(s1[i]) + ord(s2[i]) + self.helper(s1[i+1:], s2[i+1:]))
        self.sum_dict[tuple] = delete_sum
        return delete_sum
51843