Count common subsequence in two strings
Last Updated :
08 Mar, 2024
Given two string S and T. 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 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
}
// 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
<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>
<?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
?>
# 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
Complexity Analysis:
- Time Complexity : O(n1 * n2)
- Auxiliary Space : O(n1 * n2)
Efficient approach : Space optimization
In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.
Implementation steps:
- Create a 1D vector prev of size n2+1 and initialize it with 0.
- Set a base case by initializing the values of prev.
- Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
- Now Create a temporary 1d vector curr used to store the current values from previous computations.
- After every iteration assign the value of curr to prev for further iteration.
- At last return and print the final answer stored in prev[n2]
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();
// to store previous computaions of subproblems
vector<int>prev(n2+1 , 0);
// for each character of S
for (int i = 1; i <= n1; i++) {
vector<int>curr(n2 +1 , 0);
// 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])
curr[j] = 1 + curr[j - 1] + prev[j];
else
curr[j] = curr[j - 1] + prev[j] - prev[j - 1];
}
// assigning values
// for iterate further
prev = curr;
}
// return final answer
return prev[n2];
}
// Driver Program
int main()
{
string s = "ajblqcpdz";
string t = "aefcnbtdi";
cout << CommonSubsequencesCount(s, t) << endl;
return 0;
}
import java.util.*;
public class Main {
// Function to count the number of common subsequences in two strings
public static int CommonSubsequencesCount(String s, String t) {
int n1 = s.length();
int n2 = t.length();
// To store previous computations of subproblems
int[] prev = new int[n2 + 1];
// For each character of S
for (int i = 1; i <= n1; i++) {
int[] curr = new int[n2 + 1];
// For each character in T
for (int j = 1; j <= n2; j++) {
// If characters are same in both the strings
if (s.charAt(i - 1) == t.charAt(j - 1)) {
curr[j] = 1 + curr[j - 1] + prev[j];
} else {
curr[j] = curr[j - 1] + prev[j] - prev[j - 1];
}
}
// Assigning values for further iterations
prev = curr;
}
// Return the final answer
return prev[n2];
}
// Driver program
public static void main(String[] args) {
String s = "ajblqcpdz";
String t = "aefcnbtdi";
System.out.println(CommonSubsequencesCount(s, t));
}
}
using System;
public class CommonSubsequenceCount {
// return the number of common subsequence in
// two strings
public static int Count(string s, string t)
{
int n1 = s.Length;
int n2 = t.Length;
// to store previous computations of subproblems
int[] prev = new int[n2 + 1];
// for each character of S
for (int i = 1; i <= n1; i++) {
int[] curr = new int[n2 + 1];
// for each character in T
for (int j = 1; j <= n2; j++) {
// if characters are the same in both
// strings
if (s[i - 1] == t[j - 1])
curr[j] = 1 + curr[j - 1] + prev[j];
else
curr[j] = curr[j - 1] + prev[j]
- prev[j - 1];
}
// assign values for further iteration
prev = curr;
}
// return final answer
return prev[n2];
}
public static void Main()
{
string s = "ajblqcpdz";
string t = "aefcnbtdi";
Console.WriteLine(Count(s, t));
}
}
// Javascript program to count common subsequence in two strings
// return the number of common subsequence in
// two strings
function CommonSubsequencesCount(s, t) {
let n1 = s.length;
let n2 = t.length;
// to store previous computaions of subproblems
let prev = new Array(n2+1).fill(0);
// for each character of s
for (let i = 1; i <= n1; i++) {
let curr = new Array(n2 + 1).fill(0);
// for each character in t
for (let j = 1; j <= n2; j++) {
// if character are same in both
// the string
if (s[i - 1] === t[j - 1])
curr[j] = 1 + curr[j - 1] + prev[j];
else
curr[j] = curr[j - 1] + prev[j] - prev[j - 1];
}
// assigning values
// for iterate further
prev = curr;
}
// return final answer
return prev[n2];
}
// Driver Program
let s = "ajblqcpdz";
let t = "aefcnbtdi";
console.log(CommonSubsequencesCount(s, t));
def CommonSubsequencesCount(s, t):
n1 = len(s)
n2 = len(t)
# to store previous computations of subproblems
prev = [0] * (n2 + 1)
# for each character of S
for i in range(1, n1 + 1):
curr = [0] * (n2 + 1)
# for each character in T
for j in range(1, n2 + 1):
# if characters are the same in both strings
if s[i - 1] == t[j - 1]:
curr[j] = 1 + curr[j - 1] + prev[j]
else:
curr[j] = curr[j - 1] + prev[j] - prev[j - 1]
# assigning values for iteration
prev = curr
# return the final answer
return prev[n2]
# Driver Program
if __name__ == "__main__":
s = "ajblqcpdz"
t = "aefcnbtdi"
print(CommonSubsequencesCount(s, t))
Time Complexity : O(n1 * n2)
Auxiliary Space : O(n2)
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