<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.length();
        int m=s2.length();
        int arr1[n+1];
        arr1[0]=0;
        for(int i=1; i<=n; i++){
            arr1[i]=arr1[i-1]+s1[i-1];
        }
        int arr2[m+1];
        arr2[0]=0;
        for(int i=1; i<=m; i++){
            arr2[i]=arr2[i-1]+s2[i-1];
        }

        // int dp[n+1][m+1];
        // memset(dp, -1, sizeof(dp));

        // for(int i=0; i<=n; i++){
        //     for(int j=0; j<=m; j++){
        //         if(i==0 || j==0){
        //             dp[i][j] = arr1[i]+arr2[j];
        //         }else if(s1[i-1] == s2[j-1]){
        //             dp[i][j] = dp[i-1][j-1];
        //         }else{
        //             dp[i][j] = min({dp[i-1][j-1]+s1[i-1]+s2[j-1], dp[i-1][j]+s1[i-1], dp[i][j-1]+s2[j-1]});
        //         }
        //     }
        // }

        // return dp[n][m];

        int memo[n+1][m+1];
        memset(memo, -1, sizeof(memo));

        function<int(int, int)> fun = [&](int i, int j)->int{
            if(i==0 || j==0){
                return arr1[i]+arr2[j];
            }

            if(memo[i][j]!=-1){
                return memo[i][j];
            }

            if(s1[i-1] == s2[j-1]){
                return memo[i][j] = fun(i-1, j-1);
            }

            int a = s1[i-1]+s2[j-1]+fun(i-1, j-1);
            int b = s1[i-1] + fun(i-1, j);
            int c = s2[j-1] + fun(i, j-1);
            return memo[i][j] = min({a, b, c});
        };

        return fun(n, m);
    }
};
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