<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.length(), n = s2.length();

        int dp[m+1][n+1][2];
        for(int i = 0; i <= m; i++)
            for(int j = 0; j <= n; j++)
                dp[i][j][0] = dp[i][j][1] = 0;
    
        for(int i = 1; i <= m; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                if(s1[i-1] == s2[j-1])
                {
                    dp[i][j][0] = 1 + dp[i-1][j-1][0];
                    dp[i][j][1] = dp[i-1][j-1][1] + (s1[i-1]);
                }
                else
                {
                    if( dp[i-1][j][0] > dp[i][j-1][0] )
                    {
                        dp[i][j][0] = dp[i-1][j][0];
                        dp[i][j][1] = dp[i-1][j][1];
                    }
                    else if( dp[i-1][j][0] < dp[i][j-1][0] )
                    {
                        dp[i][j][0] = dp[i][j-1][0];
                        dp[i][j][1] = dp[i][j-1][1];
                    }
                    else
                    {
                        dp[i][j][0] = dp[i-1][j][0];
                        dp[i][j][1] = max(dp[i][j-1][1] , dp[i-1][j][1]);
                    }
                }
            }
        }

        int res = 0;
        int sum = 0;
        for(int i = 0; i < m; i++)
            sum = sum + (s1[i]);
        res = (sum - dp[m][n][1]);

        sum = 0;
        for(int i = 0; i < n; i++)
            sum = sum + (s2[i]);
        res += (sum - dp[m][n][1]);

        return res;

    }
};
9000