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+---
+id: longest-common-subsequence
+title: Longest Common Subsequence
+sidebar_label: "Longest Common Subsequence"  
+sidebar_position: 4  
+description: Longest Common Subsequence (LCS) is a dynamic programming technique that finds the longest subsequence common to two strings. It’s widely used in text comparison, bioinformatics, and file differencing tools. 
+tags: [DP, Longest common subsequence ]  
+---
+# Longest Common Subsequence (LCS)
+
+## Introduction
+
+The **Longest Common Subsequence (LCS)** is a fundamental problem in computer science, particularly in the field of dynamic programming. It helps in finding the longest subsequence that is common to two sequences (strings or arrays). A **subsequence** is a sequence that appears in the same relative order but not necessarily contiguously within the original sequence.
+
+### Example
+
+For example, consider the two strings:
+- String X: `ABCBDAB`
+- String Y: `BDCAB`
+
+The longest common subsequence (LCS) between these two strings is `"BCAB"`, which has a length of 4.
+
+## Problem Definition
+
+Given two sequences (X and Y), the goal is to find the length of the longest subsequence that is common to both sequences.
+
+## Dynamic Programming Approach
+
+To solve the LCS problem, we can utilize dynamic programming by creating a 2D table where `dp[i][j]` represents the length of the LCS of the first `i` characters of string X and the first `j` characters of string Y.
+
+### Recurrence Relation
+
+1. If the characters of X and Y match (i.e., `X[i-1] == Y[j-1]`), then:
+   $$
+   dp[i][j] = dp[i-1][j-1] + 1
+   $$
+2. If the characters do not match, we take the maximum value from either ignoring the current character from X or Y:
+   $$
+   dp[i][j] = \max(dp[i-1][j], dp[i][j-1])
+   $$
+
+### Base Case
+
+- If either string is empty, then `dp[i][0] = 0` and `dp[0][j] = 0` since there can be no common subsequence.
+
+## Code Implementation in C++
+
+Here’s a C++ implementation of the LCS problem:
+
+```cpp
+#include <iostream>
+#include <vector>
+#include <string>
+
+using namespace std;
+
+int lcs(string X, string Y) {
+    int m = X.length();
+    int n = Y.length();
+    
+    // Create a 2D dp table to store lengths of LCS
+    vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
+
+    // Fill dp table in bottom-up manner
+    for (int i = 1; i <= m; i++) {
+        for (int j = 1; j <= n; j++) {
+            if (X[i - 1] == Y[j - 1]) {
+                dp[i][j] = dp[i - 1][j - 1] + 1; // Characters match
+            } else {
+                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]); // Characters do not match
+            }
+        }
+    }
+
+    // The length of LCS is found at dp[m][n]
+    return dp[m][n];
+}
+
+int main() {
+    string X = "ABCBDAB";
+    string Y = "BDCAB";
+    cout << "Length of LCS: " << lcs(X, Y) << endl;  // Output: 4
+
+    // Additional test cases
+    string X1 = "AGGTAB";
+    string Y1 = "GXTXAYB";
+    cout << "Length of LCS (AGGTAB, GXTXAYB): " << lcs(X1, Y1) << endl;  // Output: 4
+
+    string X2 = "AA";
+    string Y2 = "AB";
+    cout << "Length of LCS (AA, AB): " << lcs(X2, Y2) << endl;  // Output: 1
+
+    string X3 = "ABC";
+    string Y3 = "DEF";
+    cout << "Length of LCS (ABC, DEF): " << lcs(X3, Y3) << endl;  // Output: 0
+
+    string X4 = "AAB";
+    string Y4 = "AAAB";
+    cout << "Length of LCS (AAB, AAAB): " << lcs(X4, Y4) << endl;  // Output: 2
+
+    return 0;
+}
+```
+## Explanation of the Code
+
+1.  **Initial Setup:**
+    
+    -   We first determine the lengths of the input strings X and Y using `length()`.
+    -   We create a 2D vector `dp` of size `(m + 1) x (n + 1)`, initializing all elements to 0.
+2.  **Filling the Table:**
+    
+    -   We iterate through each character of both strings X and Y using nested loops.
+    -   If `X[i - 1]` is equal to `Y[j - 1]`, we set `dp[i][j]` to `dp[i - 1][j - 1] + 1`, indicating that we've found a matching character.
+    -   If the characters do not match, we set `dp[i][j]` to the maximum of the values from the cell above (`dp[i - 1][j]`) and the cell to the left (`dp[i][j - 1]`).
+3.  **Returning the Result:**
+    
+    -   Once the table is completely filled, the value at `dp[m][n]` contains the length of the longest common subsequence.
+
+
+
+## Time and Space Complexity
+
+-   **Time Complexity:** The time complexity is $O(m \times n)$ because we fill up a table of size `(m + 1) x (n + 1)` in nested loops.
+-   **Space Complexity:** The space complexity is also $O(m \times n)$ due to the 2D table used for storing intermediate results.
+
+## Conclusion
+
+The Longest Common Subsequence (LCS) is a key problem in dynamic programming with various applications, including file comparison and DNA sequence analysis. The approach described here utilizes a 2D table to efficiently solve the problem, allowing for easy retrieval of the LCS length.
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