<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 dp[501][501];
    int solve(string s1,string s2,int i,int j){
        int n1=s1.length();
        int n2=s2.length();
        if(i>=n1 && j>=n2)return 0;
        if(i>=n1)return dp[i][j]= s2[j]+solve(s1,s2,i,j+1);
        if(j>=n2)return dp[i][j]= s1[i]+solve(s1,s2,i+1,j);
        if(dp[i][j]!=-1)return dp[i][j];
        if(i>=n1)return n2-j;
        if(j>=n2)return n1-i;

        if(s1[i]==s2[j])return dp[i][j]= solve(s1,s2,i+1,j+1);

        else
            return dp[i][j]= min(s1[i]+solve(s1,s2,i+1,j),s2[j]+solve(s1,s2,i,j+1));
        
    }
    int minimumDeleteSum(string s1, string s2) {
        int n1=s1.length();
        int n2=s2.length();
        for (int i = n1 - 1; i >= 0; i--) {
            dp[i][n2] = dp[i + 1][n2] + s1[i];
        }
        for (int j = n2 - 1; j >= 0; j--) {
            dp[n1][j] = dp[n1][j + 1] + s2[j];
        }
        for(int i=n1-1;i>=0;i--){
            for(int j=n2-1;j>=0;j--){
                if(s1[i]==s2[j])dp[i][j]=dp[i+1][j+1];
                else{
                    dp[i][j]=min(dp[i+1][j]+s1[i],dp[i][j+1]+s2[j]);
                }
            }
        }
        return dp[0][0];
    }
};
9100