<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 lcs(string s1,string s2){
        int n=s1.length();
        int m=s2.length();
        int t[n+1][m+1];
        
        for(int i=0;i<=n;i++){
            for(int j=0;j<=m;j++){
                if(i==0 ||j==0){
                    t[i][j]=0;
                    continue;
                }
                if(s1[i-1]==s2[j-1]){
                    t[i][j]=(int)s1[i-1]+t[i-1][j-1];
                }
                else{
                    t[i][j]=max(t[i-1][j],t[i][j-1]);
                }
            }
        }
        return t[n][m];
    }
    int minimumDeleteSum(string s1, string s2) {
        int sum=0;
        for(int i=0;i<s1.length();i++){
            sum+=(int)s1[i];
        }
        for(int i=0;i<s2.length();i++){
            sum+=(int)s2[i];
        }
        return sum-2*(lcs(s1,s2));
    }
};
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