<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 solve(int i, int j, string s1, string s2,vector<vector<int>> &dp){
        if(i == 0 or j == 0) return 0;
        if(dp[i][j] != -1) return dp[i][j];

        if(s1[i-1] == s2[j-1]){
            return dp[i][j] =  1 + solve(i-1,j-1,s1,s2,dp);
        }else{
            return dp[i][j] =  max(solve(i-1,j,s1,s2,dp),solve(i,j-1,s1,s2,dp));
        }
    }
    int longestCommonSubsequence(string s1, string s2) {
        if(s1 == s2) return s1.size();

        int n = s1.size(), m = s2.size();

        vector<vector<int>> dp(n+1,vector<int>(m+1,0));
    
        for(int i = 1; i<=n; ++i){
            for(int j = 1; j<=m; ++j){   
                if(s1[i-1] == s2[j-1]){
                    dp[i][j] =  1 + dp[i-1][j-1];
                }else{
                    dp[i][j] =  max(dp[i-1][j],dp[i][j-1]);
                }
            }
        }

        return dp[n][m];
    }
};
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