<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 i , int j){
         // base case 
         int n = s1.length();
         int m = s2.length();
         if(i>=n && j>=m) return 0;

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

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


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


          // if both are different
          int  pick = s1[i] + solve(s1,s2 , i+1,j);

          int notPick = s2[j] + solve(s1,s2,i, j+1);

          return min(pick,notPick);


    }

    // memoisation  ❤️❤️❤️❤️❤️❤️❤️

     int memo(string &s1, string &s2,int i , int j, vector<vector<int>>&dp){
         // base case 
         int n = s1.length();
         int m = s2.length();
         if(i>=n && j>=m) return 0;

          if(dp[i][j]!=-1)return dp[i][j];

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

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


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


          // if both are different
          int  pick = s1[i] + memo(s1,s2 , i+1,j,dp);

          int notPick = s2[j] + memo(s1,s2,i, j+1,dp);

          dp[i][j]=min(pick,notPick);

          return dp[i][j];


    }
     
    int minimumDeleteSum(string s1, string s2) {

        // return solve(s1,s2,0,0);

        // memoisation ❤️❤️❤️❤️❤️❤️
        int n = s1.length();
         int m = s2.length();
        vector<vector<int>>dp(n+10,vector<int>(m+10,-1));
         return memo(s1,s2,0,0,dp);

    }
};
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