<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 {
    string a, b ;
    vector<vector<int>> dp ;
    int fun(int i, int j) {
        if(dp[i+1][j+1] != -1) return dp[i+1][j+1];
        if(i < 0) {
            int ans = 0 ;
            for(int k=0;k<=j;k++)
                ans += b[k] ;
            return dp[i+1][j+1] = ans ;
        }
        else if(j<0){
            int ans = 0 ;
            for(int k=0;k<=i;k++)
                ans += a[k] ;
            return dp[i+1][j+1] = ans ;
        }
        if(a[i] != b[j]) {
            return dp[i+1][j+1] = min(
                a[i] + fun(i-1, j),
                b[j] + fun(i, j-1)
            );
        }
        return dp[i+1][j+1] = fun(i-1, j-1);
    }
public:
    int minimumDeleteSum(string s1, string s2) {
        a = s1 ;
        b = s2 ;
        int n, m ;
        n = s1.size() ;
        m = s2.size() ;
        dp = vector<vector<int>>(n+1, vector<int>(m+1,-1));
        return fun(n-1, m-1);
    }
};
17000