<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>

#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC optimize("no-math-errno,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4")

static const auto init = [](){
    cin.tie(nullptr);
    cout.tie(nullptr);
    ios_base::sync_with_stdio(false);

    return -1;
}();

class Solution {
public:
    static int longestCommonSubsequence(const string &text1, const string &text2) {
        short int m = (int)text1.size();
        short int n = (int)text2.size();
        short int dp[n + 1];

        //dp[j] top
        //dp[j - 1] topLeft

        memset(dp, 0, sizeof dp);

        for (short int i = 1; i <= m; ++i) {
            short int topLeft = 0;
            for (short int j = 1; j <= n; ++j) {
                short int nextTopLeft = dp[j];
                if (text1[i - 1] == text2[j - 1]) {
                    dp[j] = 1 + topLeft;
                } else {
                    dp[j] = max(dp[j], dp[j - 1]);
                }
                topLeft = nextTopLeft;
            }
        }
        return (int)dp[n];
    }
};

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