<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(int i ,int j , string &s1 , string &s2 , vector<vector<int>> &dp){
        if (i < 0) {
            // If s1 is empty, sum ASCII values of remaining characters in s2
            int sum = 0;
            for (; j >= 0; --j) {
                sum += s2[j];
            }
            return sum;
        }
        if (j < 0) {
            // If s2 is empty, sum ASCII values of remaining characters in s1
            int sum = 0;
            for (; i >= 0; --i) {
                sum += s1[i];
            }
            return sum;
        }
        if(dp[i][j]!=-1) return dp[i][j];
        if(s1[i]==s2[j]) return dp[i][j] = solve(i-1,j-1,s1,s2,dp);

        return dp[i][j] = min(s2[j]+solve(i,j-1,s1,s2,dp) , s1[i]+solve(i-1,j,s1,s2,dp));
    }
    int minimumDeleteSum(string s1, string s2) {
        //inserting in string 1 is equal to delete in string 2 
        int n = s1.size(),m=s2.size();
        vector<vector<int>> dp(n,vector<int>(m,-1));
        return solve(n-1,m-1,s1,s2,dp);
    }
};
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