<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 recursion(index1,index2):
            if index1 < 0:
                # left over in s2
                ans = 0
                for i in range(0,index2+1):
                    ans += ord(s2[i])
                return ans

            if index2 < 0:
                ans = 0
                for i in range(0,index1+1):
                    ans += ord(s1[i])
                return ans

            if (index1,index2) in dp:
                return dp[(index1,index2)]
            take = 10**9
            not_take = 10**9

            if s1[index1] == s2[index2]:
                not_take = 0 + recursion(index1-1,index2-1)

            else:
                take = min(ord(s1[index1]) + recursion(index1-1,index2),
                           ord(s2[index2]) + recursion(index1,index2-1))

            dp[(index1,index2)] =  min(take,not_take)
            return dp[(index1,index2)]

        m = len(s1)
        n = len(s2)
        return recursion(m-1,n-1)
135185