<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 m = s1.size();
        int n = s2.size();
        vector<vector<int>> dp(m + 1, vector<int> (n + 1, 0));

        for(int j = 1; j <= n; j++)
            dp[0][j] = s2[j - 1] + dp[0][j - 1];

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

        return dp[m][n];
    }
};

/*

if equal don't add ascii value

state[i][j] = state[i - 1][j - 1] if s1[i - 1] == s2[j - 1]
else min(s1[i - 1] + state[i - 1][j] + , s2[j - 1] + state[i][j - 1])


*/
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