<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 {
public:
    int minimumDeleteSum(string s1, string s2) {
        int n=s1.size() , m=s2.size();
        vector<int> suffOne(n,0) , suffTwo(m,0);
        suffOne[n-1]=s1[n-1];
        suffTwo[m-1]=s2[m-1];
        for(int i=n-2;i>=0;i--) suffOne[i]=s1[i] + suffOne[i+1];
        for(int i=m-2;i>=0;i--) suffTwo[i]=s2[i] + suffTwo[i+1];
        int dp[n][m];
        memset(dp,-1,sizeof(dp));
        function<int(int,int)> f=[&](int i,int j)->int{
            if(i==n && j==m) return 0;
            if(i==n) return suffTwo[j];
            if(j==m) return suffOne[i];
            if(dp[i][j]!=-1) return dp[i][j];
            if(s1[i]==s2[j]){
                return dp[i][j]=f(i+1,j+1);
            }
            else{
                return dp[i][j]=min(s1[i] + f(i+1,j) , s2[j] + f(i,j+1));
            }
        };
        return f(0,0);
    }
};
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