<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 {
private:
    int solve(string& s1, string& s2, int i, int j, vector<vector<int>>& dp){
        int n1=s1.length(), n2=s2.length();

        if(i == n1 && j == n2)
            return 0;
        
        // s1 is exhausted; delete the
        // remaining (n2 - j) chars from s2
        if(i == n1)
        {
            int sum = 0;
            for(int k=j; k<n2; k++)
                sum += int(s2[k]);

            return sum;
        }
        
        // s2 is exhausted; delete the
        // remaining (n1 - i) chars from s1
        if(j == n2)
        {
            int sum = 0;
            for(int k=i; k<n1; k++)
                sum += int(s1[k]);
                
            return sum;
        }

        if(dp[i][j] != -1)
            return dp[i][j];

        if(s1[i] == s2[j])
        {
            dp[i][j] = solve(s1, s2, i+1, j+1, dp);
            return dp[i][j];
        }
        
        int deleteFromS1 = int(s1[i]) + solve(s1, s2, i+1, j, dp);
        int deleteFromS2 = int(s2[j]) + solve(s1, s2, i, j+1, dp);

        dp[i][j] = min(deleteFromS1, deleteFromS2);
        return dp[i][j]; 
    }

public:
    int minimumDeleteSum(string s1, string s2) {
        int n1=s1.length(), n2=s2.length();
        vector<vector<int>> dp(n1, vector<int>(n2, -1));

        return solve(s1, s2, 0, 0, dp);
    }
};
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