Create longestCommonSubsequence.cpp

Finding the longest common subsequence using the dynamic programming concept.

Author: NirdeshGothania

longestCommonSubsequence.cpp

@@ -0,0 +1,83 @@
+#include <iostream>
+#include <math.h>
+using namespace std;
+
+void print(string s1, int **dp, int **s, int m, int n)
+{
+    if (n >= 1 && m >= 1)
+    {
+        if (s[m][n] == 1)
+        {
+            print(s1, dp, s, m - 1, n - 1);
+            cout << s1[m - 1];
+        }
+        else if (s[m][n] == 2)
+        {
+            print(s1, dp, s, m, n - 1);
+        }
+        else
+        {
+            print(s1, dp, s, m - 1, n);
+        }
+        return;
+    }
+}
+
+int main()
+{
+    string s1, s2;
+    cin >> s1;
+    cin >> s2;
+    int m = s1.length();
+    int n = s2.length();
+
+    // int dp[m + 1][n + 1];
+    int **dp = (int **)malloc((m + 1) * sizeof(int *));
+    for (int i = 0; i < m + 1; i++)
+    {
+        dp[i] = (int *)malloc((n + 1) * sizeof(int));
+    }
+
+    for (int i = 0; i < m + 1; i++)
+    {
+        dp[i][0] = 0;
+    }
+    for (int j = 0; j < n + 1; j++)
+    {
+        dp[0][j] = 0;
+    }
+    // int s[m + 1][n + 1];
+    int **s = (int **)malloc((m + 1) * sizeof(int *));
+    for (int i = 0; i < m + 1; i++)
+    {
+        s[i] = (int *)malloc((n + 1) * sizeof(int));
+    }
+    for (int i = 1; i < m + 1; i++)
+    {
+        for (int j = 1; j < n + 1; j++)
+        {
+
+            if (s1[i - 1] == s2[j - 1])
+            {
+                dp[i][j] = dp[i - 1][j - 1] + 1;
+                s[i][j] = 1;
+            }
+            else
+            {
+                if (dp[i - 1][j] < dp[i][j - 1])
+                {
+                    dp[i][j] = dp[i][j - 1];
+                    s[i][j] = 2;
+                }
+                else
+                {
+                    dp[i][j] = dp[i - 1][j];
+                    s[i][j] = 3;
+                }
+            }
+        }
+    }
+    cout << dp[m][n] << "\n";
+    print(s1, dp, s, m, n);
+    return 0;
+}