<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 solve(string s1,string s2,int n,int m){
        int dp[n+1][m+1];
        for(int i=0;i<n+1;i++){
            for(int j=0;j<m+1;j++){
                if(i==0 || j==0){
                    dp[i][j]=0;
                }
            }
        }
        for(int i=1;i<n+1;i++){
            for(int j=1;j<m+1;j++){
                if(s1[i-1]==s2[j-1]){
                    dp[i][j]=int(s1[i-1])+dp[i-1][j-1];
                }
                else{
                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
                }
            }
        }
        /*int i=n,j=m;
        string s="";
        while(i>=0 && j>=0){
            if(s1[i-1]==s2[j]){
                s+=s1[i-1];
                i--;
                j--;
            }
            else if(dp[i-1][j]>dp[i][j-1]){
                i--;
            }
            else{
                j--;
            }
        }
        cout<<s;
        return s;
        */
        return dp[n][m];
        
    }
    int ascisum(string s){
        int sum=0;
        for(int i=0;i<s.size();i++){
            sum+=int(s[i]);
        }
        return sum;
    }
    int minimumDeleteSum(string s1, string s2) {
        return ascisum(s1)+ascisum(s2)-2*solve(s1,s2,s1.size(),s2.size());
        /*reverse(s.begin(),s.end());
        int k=0;
        int sum=0;
        for(int i=0;i<s1.size();i++){
            if(s1[i]!=s[k]){
                int n=int(s1[i]);
                sum+=n;
            }
            else{
                k++;
            }
        }
        k=0;
        for(int i=0;i<s2.size();i++){
            if(s2[i]!=s[k]){
                int n=int(s2[i]);
                sum+=n;
            }
            else{
                k++;
            }
        }
        return sum;*/
    }
};
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