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Count common subsequence in two strings

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  • Difficulty Level : Hard
  • Last Updated : 02 Sep, 2022
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Given two string S and Q. The task is to count the number of the common subsequence in S and T.

Examples:

Input : S = “ajblqcpdz”, T = “aefcnbtdi” 
Output : 11 
Common subsequences are : { “a”, “b”, “c”, “d”, “ab”, “bd”, “ad”, “ac”, “cd”, “abd”, “acd” }

Input : S = “a”, T = “ab” 
Output : 1

To find the number of common subsequences in two string, say S and T, we use Dynamic Programming by defining a 2D array dp[][], where dp[i][j] is the number of common subsequences in the string S[0…i-1] and T[0….j-1]. 

Now, we can define dp[i][j] as = dp[i][j-1] + dp[i-1][j] + 1, when S[i-1] is equal to T[j-1] 
This is because when S[i-1] == S[j-1], using the above fact all the previous common sub-sequences are doubled as they get appended by one more character. Both dp[i][j-1] and dp[i-1][j] contain dp[i-1][j-1] and hence it gets added two times in our recurrence which takes care of doubling of count of all previous common sub-sequences. Addition of 1 in recurrence is for the latest character match : common sub-sequence made up of s1[i-1] and s2[j-1]  = dp[i-1][j] + dp[i][j-1] – dp[i-1][j-1], when S[i-1] is not equal to T[j-1] 
Here we subtract dp[i-1][j-1] once because it is present in both dp[i][j – 1] and dp[i – 1][j] and gets added twice.

Implementation:

C++




// C++ program to count common subsequence in two strings
#include <bits/stdc++.h>
using namespace std;
 
// return the number of common subsequence in
// two strings
int CommonSubsequencesCount(string s, string t)
{
    int n1 = s.length();
    int n2 = t.length();
    int dp[n1+1][n2+1];
 
    for (int i = 0; i <= n1; i++) {
        for (int j = 0; j <= n2; j++) {
            dp[i][j] = 0;
        }
    }
 
    // for each character of S
    for (int i = 1; i <= n1; i++) {
 
        // for each character in T
        for (int j = 1; j <= n2; j++) {
 
            // if character are same in both
            // the string
            if (s[i - 1] == t[j - 1])
                dp[i][j] = 1 + dp[i][j - 1] + dp[i - 1][j];           
            else
                dp[i][j] = dp[i][j - 1] + dp[i - 1][j] -
                                        dp[i - 1][j - 1];           
        }
    }
 
    return dp[n1][n2];
}
 
// Driver Program
int main()
{
    string s = "ajblqcpdz";
    string t = "aefcnbtdi";
 
    cout << CommonSubsequencesCount(s, t) << endl;
    return 0;
}

Java




// Java program to count common subsequence in two strings
public class GFG {
     
    // return the number of common subsequence in
    // two strings
    static int CommonSubsequencesCount(String s, String t)
    {
        int n1 = s.length();
        int n2 = t.length();
        int dp[][] = new int [n1+1][n2+1];
        char ch1,ch2 ;
       
        for (int i = 0; i <= n1; i++) {
            for (int j = 0; j <= n2; j++) {
                dp[i][j] = 0;
            }
        }
       
        // for each character of S
        for (int i = 1; i <= n1; i++) {
       
            // for each character in T
            for (int j = 1; j <= n2; j++) {
                 
                ch1 = s.charAt(i - 1);
                ch2 = t.charAt(j - 1);
                 
                // if character are same in both 
                // the string                
                if (ch1 == ch2) 
                    dp[i][j] = 1 + dp[i][j - 1] + dp[i - 1][j];            
                else
                    dp[i][j] = dp[i][j - 1] + dp[i - 1][j] - 
                                            dp[i - 1][j - 1];            
            }
        }
       
        return dp[n1][n2];
    }
    // Driver code
    public static void main (String args[]){
         
          String s = "ajblqcpdz";
          String t = "aefcnbtdi";
           
        System.out.println(CommonSubsequencesCount(s, t));
           
    }
 
// This code is contributed by ANKITRAI1
}

Python3




# Python3 program to count common
# subsequence in two strings
 
# return the number of common subsequence
# in two strings
def CommonSubsequencesCount(s, t):
 
    n1 = len(s)
    n2 = len(t)
    dp = [[0 for i in range(n2 + 1)]
             for i in range(n1 + 1)]
 
    # for each character of S
    for i in range(1, n1 + 1):
 
        # for each character in T
        for j in range(1, n2 + 1):
 
            # if character are same in both
            # the string
            if (s[i - 1] == t[j - 1]):
                dp[i][j] = (1 + dp[i][j - 1] +
                                dp[i - 1][j])        
            else:
                dp[i][j] = (dp[i][j - 1] + dp[i - 1][j] -
                            dp[i - 1][j - 1])        
         
    return dp[n1][n2]
 
# Driver Code
s = "ajblqcpdz"
t = "aefcnbtdi"
 
print(CommonSubsequencesCount(s, t))
 
# This code is contributed by Mohit Kumar

C#




// C# program to count common
// subsequence in two strings
using System;
 
class GFG
{
 
// return the number of common
// subsequence in two strings
static int CommonSubsequencesCount(string s,
                                   string t)
{
    int n1 = s.Length;
    int n2 = t.Length;
    int[,] dp = new int [n1 + 1, n2 + 1];
     
    for (int i = 0; i <= n1; i++)
    {
        for (int j = 0; j <= n2; j++)
        {
            dp[i, j] = 0;
        }
    }
     
    // for each character of S
    for (int i = 1; i <= n1; i++)
    {
     
        // for each character in T
        for (int j = 1; j <= n2; j++)
        {
             
            // if character are same in
            // both the string                
            if (s[i - 1] == t[j - 1])
                dp[i, j] = 1 + dp[i, j - 1] +
                               dp[i - 1, j];            
            else
                dp[i, j] = dp[i, j - 1] +
                           dp[i - 1, j] -
                           dp[i - 1, j - 1];            
        }
    }
     
    return dp[n1, n2];
}
 
// Driver code
public static void Main ()
{
    string s = "ajblqcpdz";
    string t = "aefcnbtdi";
         
    Console.Write(CommonSubsequencesCount(s, t));
}
}
 
// This code is contributed
// by ChitraNayal

PHP




<?php
// PHP program to count common subsequence
// in two strings
 
// return the number of common subsequence
// in two strings
function CommonSubsequencesCount($s, $t)
{
    $n1 = strlen($s);
    $n2 = strlen($t);
    $dp = array();
 
    for ($i = 0; $i <= $n1; $i++)
    {
        for ($j = 0; $j <= $n2; $j++)
        {
            $dp[$i][$j] = 0;
        }
    }
 
    // for each character of S
    for ($i = 1; $i <= $n1; $i++)
    {
 
        // for each character in T
        for ($j = 1; $j <= $n2; $j++)
        {
 
            // if character are same in both
            // the string
            if ($s[$i - 1] == $t[$j - 1])
                $dp[$i][$j] = 1 + $dp[$i][$j - 1] +
                                  $dp[$i - 1][$j];    
            else
                $dp[$i][$j] = $dp[$i][$j - 1] +
                              $dp[$i - 1][$j] -
                              $dp[$i - 1][$j - 1];    
        }
    }
 
    return $dp[$n1][$n2];
}
 
// Driver Code
$s = "ajblqcpdz";
$t = "aefcnbtdi";
 
echo CommonSubsequencesCount($s, $t) ."\n";
 
// This code is contributed
// by Akanksha Rai
?>

Javascript




<script>
 
// Javascript program to count common subsequence in two strings
 
// return the number of common subsequence in
// two strings
function CommonSubsequencesCount(s, t)
{
    var n1 = s.length;
    var n2 = t.length;
    var dp = Array.from(Array(n1+1), ()=> Array(n2+1));
 
    for (var i = 0; i <= n1; i++) {
        for (var j = 0; j <= n2; j++) {
            dp[i][j] = 0;
        }
    }
 
    // for each character of S
    for (var i = 1; i <= n1; i++) {
 
        // for each character in T
        for (var j = 1; j <= n2; j++) {
 
            // if character are same in both
            // the string
            if (s[i - 1] == t[j - 1])
                dp[i][j] = 1 + dp[i][j - 1] + dp[i - 1][j];           
            else
                dp[i][j] = dp[i][j - 1] + dp[i - 1][j] -
                                        dp[i - 1][j - 1];           
        }
    }
 
    return dp[n1][n2];
}
 
// Driver Program
var s = "ajblqcpdz";
var t = "aefcnbtdi";
document.write( CommonSubsequencesCount(s, t));
 
 
</script>

Output

11

Complexity Analysis:

  • Time Complexity : O(n1 * n2) 
  • Auxiliary Space : O(n1 * n2)

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