<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>

/*
    Given 2 strings, return length of longest common subsequence
    Ex. text1 = "abcde", text2 = "ace" -> 3, "ace" is LCS

                j
            a   c   e
        a   3
        b       2     --> visualization of below, build DP bottom-up
    i   c       2
        d           1
        e           1

    Time: O(m x n)
    Space: O(m x n)
*/

class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int m = text1.size();
        int n = text2.size();
        
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        
        for (int i = m - 1; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                if (text1[i] == text2[j]) {
                    dp[i][j] = 1 + dp[i + 1][j + 1];
                } else {
                    dp[i][j] = max(dp[i + 1][j], dp[i][j + 1]);
                }
            }
        }
        
        return dp[0][0];
    }
};
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