<p>Given two strings <code>text1</code> and <code>text2</code>, return <em>the length of their longest <strong>common subsequence</strong>. </em>If there is no <strong>common subsequence</strong>, return <code>0</code>.</p>

<p>A <strong>subsequence</strong> of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.</p>

<ul>
	<li>For example, <code>&quot;ace&quot;</code> is a subsequence of <code>&quot;abcde&quot;</code>.</li>
</ul>

<p>A <strong>common subsequence</strong> of two strings is a subsequence that is common to both strings.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> text1 = &quot;abcde&quot;, text2 = &quot;ace&quot; 
<strong>Output:</strong> 3  
<strong>Explanation:</strong> The longest common subsequence is &quot;ace&quot; and its length is 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> text1 = &quot;abc&quot;, text2 = &quot;abc&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> The longest common subsequence is &quot;abc&quot; and its length is 3.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> text1 = &quot;abc&quot;, text2 = &quot;def&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> There is no such common subsequence, so the result is 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= text1.length, text2.length &lt;= 1000</code></li>
	<li><code>text1</code> and <code>text2</code> consist of only lowercase English characters.</li>
</ul>

// class Solution {
// public:
//     int dp[1001][1001];
//     // int find(string &s,string &t,int n,int m,int i,int j){
//     //     if(i>=n || j>=m){
//     //         return 0;
//     //     }
//     //     if(dp[i][j]!=-1){
//     //         return dp[i][j];
//     //     }
//     //     int a=0,b=0,c=0;
//     //     if(s[i]==t[j]){
//     //         a=1+find(s,t,n,m,i+1,j+1);
//     //     }
//     //     else{
//     //         b=find(s,t,n,m,i+1,j);
//     //         c=find(s,t,n,m,i,j+1);
//     //     }
//     //     return dp[i][j]=max({a,b,c});
//     // }
//     // int longestCommonSubsequence(string s, string t) {
//     //     int n=s.length();
//     //     int m=t.length();
//     //     memset(dp,-1,sizeof(dp));
//     //     return find(s,t,n,m,0,0);
//     // }

//     int longestCommonSubsequence(string s, string t) {
//         int n = s.length();
//         int m = t.length();
//         memset(dp, 0, sizeof(dp));
//         for (int i = n - 1; i >= 0; i--) {
//             for (int j = m - 1; j >= 0; j--) {
//                 int a=0,b=0,c=0;
//                 if (s[i] == t[j]) {
//                     a = 1 + dp[i + 1][j + 1];
//                 } 
//                 else {
//                     b = dp[i + 1][j];
//                     c = dp[i][j + 1];
//                 }
//                 dp[i][j] = max({a, b, c});
//             }
//         }
//         return dp[0][0];
//     }
// };

class Solution{
    public:
    int dp[1001][1001];
    int find(string &s,string &t,int n,int m,int i,int j){
        if(i<=0 || j<=0){
            return 0;
        }
        if(dp[i][j]!=-1){
            return dp[i][j];
        }
        int a=0,b=0,c=0;
        if(s[i-1]==t[j-1]){
            a=1+find(s,t,n,m,i-1,j-1);
        }
        else{
            b=0+find(s,t,n,m,i-1,j);
            c=0+find(s,t,n,m,i,j-1);
        }
        return dp[i][j]=max({a,b,c});
    }

    int longestCommonSubsequence(string s, string t) {
        int n = s.length();
        int m = t.length();
        memset(dp, -1, sizeof(dp));
        return find(s,t,n,m,n,m);
    }
};
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