Merge pull request #270 from Jayesh-Waghmare/master

Added Longest Common Subsequence Code in C++

Author: Pradeepsingh61

CPP/dynamic_programming/longest_common_subsequence.cpp

@@ -0,0 +1,143 @@
+/*
+ * Longest Common Subsequence (LCS) Algorithm
+ * 
+ * Description:
+ * The Longest Common Subsequence algorithm finds the longest subsequence common to two sequences.
+ * A subsequence is a sequence that can be derived from another sequence by deleting some or no elements
+ * without changing the order of the remaining elements.
+ * 
+ * This implementation uses dynamic programming with a 2D table to solve the problem:
+ * 1. Create a table to store lengths of LCS for all subproblems
+ * 2. Fill the table in bottom-up manner
+ * 3. The value in dp[m][n] contains the length of LCS
+ * 4. Reconstruct the LCS using the filled table
+ * 
+ * Time Complexity: O(m*n) where m and n are lengths of the two sequences
+ * Space Complexity: O(m*n) for the DP table
+ */
+
+#include <iostream>
+#include <vector>
+#include <string>
+#include <algorithm>
+
+using namespace std;
+
+// Function to find the length of LCS and the actual LCS
+pair<int, string> longestCommonSubsequence(const string& X, const string& Y) {
+    int m = X.length();
+    int n = Y.length();
+    
+    // Create a table to store lengths of LCS for all subproblems
+    vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
+    
+    // Fill dp[][] 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;
+            } else {
+                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
+            }
+        }
+    }
+    
+    // Length of LCS is dp[m][n]
+    int lcsLength = dp[m][n];
+    
+    // Reconstruct the LCS
+    string lcs = "";
+    int i = m, j = n;
+    
+    while (i > 0 && j > 0) {
+        // If current characters match, include in LCS
+        if (X[i - 1] == Y[j - 1]) {
+            lcs = X[i - 1] + lcs;
+            i--;
+            j--;
+        }
+        // Move in the direction of larger value
+        else if (dp[i - 1][j] > dp[i][j - 1]) {
+            i--;
+        } else {
+            j--;
+        }
+    }
+    
+    return {lcsLength, lcs};
+}
+
+// Function to print the DP table (for educational purposes)
+void printDPTable(const string& X, const string& Y, const vector<vector<int>>& dp) {
+    int m = X.length();
+    int n = Y.length();
+    
+    cout << "DP Table:" << endl;
+    cout << "    ";
+    for (int j = 0; j < n; j++) {
+        cout << Y[j] << " ";
+    }
+    cout << endl;
+    
+    for (int i = 0; i <= m; i++) {
+        if (i == 0) {
+            cout << "  ";
+        } else {
+            cout << X[i - 1] << " ";
+        }
+        
+        for (int j = 0; j <= n; j++) {
+            cout << dp[i][j] << " ";
+        }
+        cout << endl;
+    }
+}
+
+// Main function with test cases
+int main() {
+    // Test Case 1: Basic example
+    cout << "Test Case 1:" << endl;
+    string X1 = "ABCDGH";
+    string Y1 = "AEDFHR";
+    cout << "String 1: " << X1 << endl;
+    cout << "String 2: " << Y1 << endl;
+    
+    auto result1 = longestCommonSubsequence(X1, Y1);
+    cout << "Length of LCS: " << result1.first << endl;
+    cout << "LCS: " << result1.second << endl << endl;
+    
+    // Test Case 2: No common characters
+    cout << "Test Case 2:" << endl;
+    string X2 = "ABC";
+    string Y2 = "DEF";
+    cout << "String 1: " << X2 << endl;
+    cout << "String 2: " << Y2 << endl;
+    
+    auto result2 = longestCommonSubsequence(X2, Y2);
+    cout << "Length of LCS: " << result2.first << endl;
+    cout << "LCS: " << result2.second << endl << endl;
+    
+    // Test Case 3: One string is a subsequence of the other
+    cout << "Test Case 3:" << endl;
+    string X3 = "ABCDEFG";
+    string Y3 = "ACG";
+    cout << "String 1: " << X3 << endl;
+    cout << "String 2: " << Y3 << endl;
+    
+    auto result3 = longestCommonSubsequence(X3, Y3);
+    cout << "Length of LCS: " << result3.first << endl;
+    cout << "LCS: " << result3.second << endl << endl;
+    
+    // Test Case 4: Both strings are identical
+    cout << "Test Case 4:" << endl;
+    string X4 = "PROGRAMMING";
+    string Y4 = "PROGRAMMING";
+    cout << "String 1: " << X4 << endl;
+    cout << "String 2: " << Y4 << endl;
+    
+    auto result4 = longestCommonSubsequence(X4, Y4);
+    cout << "Length of LCS: " << result4.first << endl;
+    cout << "LCS: " << result4.second << endl;
+    
+    return 0;
+}