<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:
        dp={}
        def helper(i,j):
            if i<0 or j<0:
                return ''
            if (i,j) in dp:
                return dp[(i,j)]
            if s1[i]==s2[j]:
                dp[(i,j)]=helper(i-1,j-1)+s1[i]
            else:
                op1=helper(i-1,j)
                op2=helper(i,j-1)
                
                if len(op1)==len(op2):
                    asci1,asci2=0,0
                    asci1=sum(ord(ch) for ch in op1)
                    asci2=sum(ord(ch) for ch in op2)
                    dp[(i,j)]=(op1 if asci1>asci2 else op2)
                else:
                    dp[(i,j)]=(op1 if len(op1)>len(op2) else op2)
            return dp[(i,j)]
        lcs=helper(len(s1)-1,len(s2)-1)

        lcsAsci=sum(ord(ch) for ch in lcs)
        st1=sum(ord(ch) for ch in s1)
        st2=sum(ord(ch) for ch in s2)

        return (st1-lcsAsci)+(st2-lcsAsci)

        
66550