longest-common-subsequence-implementation

Author: adarsh1mehra

LongestCommonSubsequence/C++/Longest_common_subsequence.cpp

@@ -0,0 +1,48 @@
+//Begin
+//Take the array of strings as input.
+//function matchedPrefixtill(): find the matched prefix between string s1 and s2 :
+  // n1 = store length of string s1.
+   //n2 = store length of string s2.
+   //for i = 0, j = 0 to i <= n1 – 1 && j <= n2 - 1
+     // if s1[i] != s2[j]
+       //  break
+    //  result.push_back(s1[i])
+  // return result
+//End
+//Begin
+//function matchedPrefix(): returns the longest matched prefix from the array of strings:
+  // for int i = 1 to n - 1
+    //     pre = matchedPrefixtill(pre, a[i])
+   //return pre.
+//End
+
+#include<bits/stdc++.h>
+using namespace std;
+
+string matchedPrefixtill(string s1, string s2) {
+   string res;
+   int n1 = s1.length(); //store length of string s1.
+   int n2 = s2.length(); //store length of string s2.
+   for (int i = 0, j = 0; i <= n1 - 1 && j <= n2 - 1; i++, j++) {
+      if (s1[i] != s2[j])
+         break;
+      res.push_back(s1[i]);
+   }
+   return (res);
+}
+string matchedPrefix (string a[], int n) {
+   string pre = a[0];
+   for (int i = 1; i <= n - 1; i++)
+   pre = matchedPrefixtill(pre, a[i]);
+   return (pre);
+}
+int main() {
+   string a[] = {"Tutorialspoint", "Tutor", "Tutorials"}; //taking inputs
+   int n = sizeof(a) / sizeof(a[0]);
+   string res = matchedPrefix(a, n);
+   if (res.length())
+      cout<<"Longest common subsequence is matched - "<<res.c_str();
+   else
+      cout<<"No matched prefix";
+   return (0);
+}

LongestCommonSubsequence/longest-common-subsequence.java

@@ -0,0 +1,28 @@
+public class LongestCommonSubsequence { 
+  
+    /* Returns length of LCS for X[0..m-1], Y[0..n-1] */
+    int lcs(char[] X, char[] Y, int m, int n) 
+    { 
+        int L[][] = new int[m + 1][n + 1]; 
+  
+        /* Following steps build L[m+1][n+1] in bottom up fashion. Note 
+         that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */
+        for (int i = 0; i <= m; i++) { 
+            for (int j = 0; j <= n; j++) { 
+                if (i == 0 || j == 0) 
+                    L[i][j] = 0; 
+                else if (X[i - 1] == Y[j - 1]) 
+                    L[i][j] = L[i - 1][j - 1] + 1; 
+                else
+                    L[i][j] = max(L[i - 1][j], L[i][j - 1]); 
+            } 
+        } 
+        return L[m][n]; 
+    } 
+  
+    /* Utility function to get max of 2 integers */
+    int max(int a, int b) 
+    { 
+        return (a > b) ? a : b; 
+    } 
+