Find how many palindromic subsequences (need not necessarily be distinct) can be formed in a given string. Note that the empty string is not considered a palindrome.
Examples:
Input : str = "abcd"
Output : 4
Explanation :- palindromic subsequence are : "a" ,"b", "c" ,"d"
Input : str = "aab"
Output : 4
Explanation :- palindromic subsequence are :"a", "a", "b", "aa"
Input : str = "aaaa"
Output : 15
The above problem can be recursively defined.
Initial Values : i= 0, j= n-1;
CountPS(i,j)
// Every single character of a string is a palindrome
// subsequence
if i == j
return 1 // palindrome of length 1
// If first and last characters are same, then we
// consider it as palindrome subsequence and check
// for the rest subsequence (i+1, j), (i, j-1)
Else if (str[i] == str[j])
return countPS(i+1, j) + countPS(i, j-1) + 1;
else
// check for rest sub-sequence and remove common
// palindromic subsequences as they are counted
// twice when we do countPS(i+1, j) + countPS(i,j-1)
return countPS(i+1, j) + countPS(i, j-1) - countPS(i+1, j-1)
Recursive Approach
Here is the recursive code to the above approach
C++
#include <iostream>#include <string>using namespace std;
// Function to return the total palindromic subsequenceint Solve(string str, int i, int j) {
if (i == j) // Base case: when the indices are the same, there is one palindrome
return 1;
if (i > j) // Base case: when the first index is greater than the second, there are no palindromes
return 0;
if (str[i] == str[j]) {
// If the characters at indices i and j are the same, we can form palindromic subsequences
// by including or excluding both characters, so we add 1 and make recursive calls.
return 1 + Solve(str, i + 1, j) + Solve(str, i, j - 1);
} else {
// If the characters at indices i and j are different, we exclude one of them and make recursive calls.
return Solve(str, i + 1, j) + Solve(str, i, j - 1) - Solve(str, i + 1, j - 1);
}
}// Driver programint main() {
string str = "abcb";
cout << "Total palindromic subsequence are: " << Solve(str, 0, str.length() - 1) << endl;
return 0;
} |
Java
// Java code to Count Palindromic Subsequence// in a given Stringpublic class GFG {
// Function return the total palindromic
// subsequence
static int solve(String str, int i, int j)
{
if (i == j) // base case when index is same
return 1;
if (i > j)
return 0;
if (str.charAt(i) == str.charAt(j)) {
return 1 + solve(str, i + 1, j)
+ solve(str, i, j - 1);
}
else
return solve(str, i + 1, j)
+ solve(str, i, j - 1)
- solve(str, i + 1, j - 1);
}
// Driver program
public static void main(String args[])
{
String str = "abcb";
System.out.println(
"Total palindromic "
+ "subsequence are : "
+ solve(str, 0, str.length() - 1));
}
}// This code is contributed by Ankit Jha |
Python3
# Function to return the total palindromic subsequencedef solve(s, i, j):
if i == j:
# Base case: when the indices are the same, there is one palindrome
return 1
if i > j:
# Base case: when the first index is greater than the second, there are no palindromes
return 0
if s[i] == s[j]:
# If the characters at indices i and j are the same, we can form palindromic subsequences
# by including or excluding both characters, so we add 1 and make recursive calls.
return 1 + solve(s, i + 1, j) + solve(s, i, j - 1)
else:
# If the characters at indices i and j are different, we exclude one of them and make recursive calls.
return solve(s, i + 1, j) + solve(s, i, j - 1) - solve(s, i + 1, j - 1)
# Driver programif __name__ == "__main__":
s = "abcb"
print("Total palindromic subsequences are:", solve(s, 0, len(s) - 1))
# This code is contributed by shivamgupta310570 |
C#
using System;
public class GFG
{ // Function to return the total palindromic subsequence
static int Solve(string str, int i, int j)
{
if (i == j) // Base case: when the indices are the same, there is one palindrome
return 1;
if (i > j) // Base case: when the first index is greater
// than the second, there are no palindromes
return 0;
if (str[i] == str[j])
{
// If the characters at indices i and j are the same,
// we can form palindromic subsequences
// by including or excluding both characters, so we add 1 and make recursive calls.
return 1 + Solve(str, i + 1, j) + Solve(str, i, j - 1);
}
else
{
// If the characters at indices i and j are different,
// we exclude one of them and make recursive calls.
return Solve(str, i + 1, j) + Solve(str, i, j - 1) - Solve(str, i + 1, j - 1);
}
}
// Driver program
public static void Main(string[] args)
{
string str = "abcb";
Console.WriteLine("Total palindromic subsequence are: " + Solve(str, 0, str.Length - 1));
}
} |
Javascript
// Function to return the total palindromic subsequencefunction solve(str, i, j) {
if (i === j) // Base case: when the indices are the same, there is one palindrome
return 1;
if (i > j) // Base case: when the first index is greater than the second, there are no palindromes
return 0;
if (str[i] === str[j]) {
// If the characters at indices i and j are the same, we can form palindromic subsequences
// by including or excluding both characters, so we add 1 and make recursive calls.
return 1 + solve(str, i + 1, j) + solve(str, i, j - 1);
} else {
// If the characters at indices i and j are different, we exclude one of them and make recursive calls.
return solve(str, i + 1, j) + solve(str, i, j - 1) - solve(str, i + 1, j - 1);
}
}// Driver programconst str = "abcb";
console.log("Total palindromic subsequences are: " + solve(str, 0, str.length - 1));
|
Output
Total palindromic subsequence are : 6
If we draw recursion tree of the above recursive solution, we can observe overlapping Subproblems. Since the problem has overlapping subproblems, we can solve it efficiently using Dynamic Programming. Below is Dynamic Programming based solution.
C++
// Counts Palindromic Subsequence in a given String#include <cstring>#include <iostream>using namespace std;
// Function return the total palindromic subsequenceint countPS(string str)
{ int N = str.length();
// create a 2D array to store the count of palindromic
// subsequence
int cps[N + 1][N + 1];
memset(cps, 0, sizeof(cps));
// palindromic subsequence of length 1
for (int i = 0; i < N; i++)
cps[i][i] = 1;
// check subsequence of length L is palindrome or not
for (int L = 2; L <= N; L++) {
for (int i = 0; i <= N-L; i++) {
int k = L + i - 1;
if (str[i] == str[k])
cps[i][k]
= cps[i][k - 1] + cps[i + 1][k] + 1;
else
cps[i][k] = cps[i][k - 1] + cps[i + 1][k]
- cps[i + 1][k - 1];
}
}
// return total palindromic subsequence
return cps[0][N - 1];
}// Driver programint main()
{ string str = "abcb";
cout << "Total palindromic subsequence are : "
<< countPS(str) << endl;
return 0;
} |
Java
// Java code to Count Palindromic Subsequence// in a given Stringpublic class GFG {
// Function return the total palindromic
// subsequence
static int countPS(String str)
{
int N = str.length();
// create a 2D array to store the count
// of palindromic subsequence
int[][] cps = new int[N][N];
// palindromic subsequence of length 1
for (int i = 0; i < N; i++)
cps[i][i] = 1;
// check subsequence of length L is
// palindrome or not
for (int L = 2; L <= N; L++) {
for (int i = 0; i <= N-L; i++) {
int k = L + i - 1;
if (str.charAt(i) == str.charAt(k)) {
cps[i][k] = cps[i][k - 1]
+ cps[i + 1][k] + 1;
}else{
cps[i][k] = cps[i][k - 1]
+ cps[i + 1][k]
- cps[i + 1][k - 1];
}
}
}
// return total palindromic subsequence
return cps[0][N - 1];
}
// Driver program
public static void main(String args[])
{
String str = "abcb";
System.out.println("Total palindromic "
+ "subsequence are : "
+ countPS(str));
}
}// This code is contributed by Sumit Ghosh |
Python3
# Python3 code to Count Palindromic# Subsequence in a given String# Function return the total# palindromic subsequencedef countPS(str):
N = len(str)
# Create a 2D array to store the count
# of palindromic subsequence
cps = [[0 for i in range(N + 2)]for j in range(N + 2)]
# palindromic subsequence of length 1
for i in range(N):
cps[i][i] = 1
# check subsequence of length L
# is palindrome or not
for L in range(2, N + 1):
for i in range(N):
k = L + i - 1
if (k < N):
if (str[i] == str[k]):
cps[i][k] = (cps[i][k - 1] +
cps[i + 1][k] + 1)
else:
cps[i][k] = (cps[i][k - 1] +
cps[i + 1][k] -
cps[i + 1][k - 1])
# return total palindromic subsequence
return cps[0][N - 1]
# Driver programstr = "abcb"
print("Total palindromic subsequence are : ", countPS(str))
# This code is contributed by Anant Agarwal. |
C#
// C# code to Count Palindromic Subsequence// Subsequence in a given Stringusing System;
class GFG {
// Function return the total
// palindromic subsequence
static int countPS(string str)
{
int N = str.Length;
// create a 2D array to store the
// count of palindromic subsequence
int[, ] cps = new int[N + 1, N + 1];
// palindromic subsequence
// of length 1
for (int i = 0; i < N; i++)
cps[i, i] = 1;
// check subsequence of length
// L is palindrome or not
for (int L = 2; L <= N; L++) {
for (int i = 0; i <= N-L; i++) {
int k = L + i - 1;
if (k < N) {
if (str[i] == str[k])
cps[i, k] = cps[i, k - 1]
+ cps[i + 1, k] + 1;
else
cps[i, k] = cps[i, k - 1]
+ cps[i + 1, k]
- cps[i + 1, k - 1];
}
}
}
// return total palindromic
// subsequence
return cps[0, N - 1];
}
// Driver Code
public static void Main()
{
string str = "abcb";
Console.Write("Total palindromic "
+ "subsequence are : "
+ countPS(str));
}
}// This code is contributed by nitin mittal. |
Javascript
<script>// Javascript code to Count Palindromic Subsequence// in a given String // Function return the total palindromic
// subsequence
function countPS(str)
{
let N = str.length;
// create a 2D array to store the count
// of palindromic subsequence
let cps = new Array(N);
for(let i=0;i<N;i++)
{
cps[i]=new Array(N);
for(let j=0;j<N;j++)
{
cps[i][j]=0;
}
}
// palindromic subsequence of length 1
for (let i = 0; i < N; i++)
cps[i][i] = 1;
// check subsequence of length L is
// palindrome or not
for (let L = 2; L <= N; L++) {
for (let i = 0; i <= N-L; i++) {
let k = L + i - 1;
if (str[i] == str[k]) {
cps[i][k] = cps[i][k - 1]
+ cps[i + 1][k] + 1;
}else{
cps[i][k] = cps[i][k - 1]
+ cps[i + 1][k]
- cps[i + 1][k - 1];
}
}
}
// return total palindromic subsequence
return cps[0][N - 1];
}
// Driver program
let str = "abcb";
document.write("Total palindromic "
+ "subsequence are : "
+ countPS(str));
// This code is contributed by avanitrachhadiya2155
</script> |
PHP
<?php // Counts Palindromic Subsequence in // a given String// Function return the total // palindromic subsequencefunction countPS($str)
{ $N = strlen($str);
// create a 2D array to store the
// count of palindromic subsequence
$cps = array_fill(0, $N + 1,
array_fill(0, $N + 1, NULL));
// palindromic subsequence of length 1
for ($i = 0; $i < $N; $i++)
$cps[$i][$i] = 1;
// check subsequence of length L
// is palindrome or not
for ($L = 2; $L <= $N; $L++)
{
for ($i = 0; $i <= $N-$L; $i++)
{
$k = $L + $i - 1;
if ($str[$i] == $str[$k])
$cps[$i][$k] = $cps[$i][$k - 1] +
$cps[$i + 1][$k] + 1;
else
$cps[$i][$k] = $cps[$i][$k - 1] +
$cps[$i + 1][$k] -
$cps[$i + 1][$k - 1];
}
}
// return total palindromic subsequence
return $cps[0][$N - 1];
}// Driver Code$str = "abcb";
echo "Total palindromic subsequence are : " .
countPS($str) . "\n";
// This code is contributed by ita_c?> |
Output
Total palindromic subsequence are : 6
Time Complexity: O(N2)
Auxiliary Space: O(N2)
Another approach: (Using memoization)
C++
// C++ program to counts Palindromic Subsequence// in a given String using recursion#include <bits/stdc++.h>using namespace std;
int n, dp[1000][1000];
string str = "abcb";
// Function return the total// palindromic subsequenceint countPS(int i, int j)
{ if (i > j)
return 0;
if (dp[i][j] != -1)
return dp[i][j];
if (i == j)
return dp[i][j] = 1;
else if (str[i] == str[j])
return dp[i][j]
= countPS(i + 1, j) +
countPS(i, j - 1) + 1;
else
return dp[i][j] = countPS(i + 1, j) +
countPS(i, j - 1) -
countPS(i + 1, j - 1);
}// Driver codeint main()
{ memset(dp, -1, sizeof(dp));
n = str.size();
cout << "Total palindromic subsequence are : "
<< countPS(0, n - 1) << endl;
return 0;
}// this code is contributed by Kushdeep Mittal |
Java
// Java program to counts Palindromic Subsequence// in a given String using recursionclass GFG {
static int n;
static int[][] dp = new int[1000][1000];
static String str = "abcb";
// Function return the total
// palindromic subsequence
static int countPS(int i, int j)
{
if (i > j)
return 0;
if (dp[i][j] != -1)
return dp[i][j];
if (i == j)
return dp[i][j] = 1;
else if (str.charAt(i) == str.charAt(j))
return dp[i][j]
= countPS(i + 1, j) +
countPS(i, j - 1) + 1;
else
return dp[i][j] = countPS(i + 1, j) +
countPS(i, j - 1) -
countPS(i + 1, j - 1);
}
// Driver code
public static void main(String[] args)
{
for (int i = 0; i < 1000; i++)
for (int j = 0; j < 1000; j++)
dp[i][j] = -1;
n = str.length();
System.out.println("Total palindromic subsequence"
+ "are : " + countPS(0, n - 1));
}
}// This code is contributed by Ryuga |
Python 3
# Python 3 program to counts Palindromic# Subsequence in a given String using recursionstr = "abcb"
# Function return the total# palindromic subsequencedef countPS(i, j):
if(i > j):
return 0
if(dp[i][j] != -1):
return dp[i][j]
if(i == j):
dp[i][j] = 1
return dp[i][j]
else if (str[i] == str[j]):
dp[i][j] = (countPS(i + 1, j) +
countPS(i, j - 1) + 1)
return dp[i][j]
else:
dp[i][j] = (countPS(i + 1, j) +
countPS(i, j - 1) -
countPS(i + 1, j - 1))
return dp[i][j]
# Driver codeif __name__ == "__main__":
dp = [[-1 for x in range(1000)]
for y in range(1000)]
n = len(str)
print("Total palindromic subsequence are :",
countPS(0, n - 1))
# This code is contributed by ita_c |
C#
// C# program to counts Palindromic Subsequence// in a given String using recursionusing System;
class GFG {
static int n;
static int[, ] dp = new int[1000, 1000];
static string str = "abcb";
// Function return the total
// palindromic subsequence
static int countPS(int i, int j)
{
if (i > j)
return 0;
if (dp[i, j] != -1)
return dp[i, j];
if (i == j)
return dp[i, j] = 1;
else if (str[i] == str[j])
return dp[i, j]
= countPS(i + 1, j) +
countPS(i, j - 1) + 1;
else
return dp[i, j] = countPS(i + 1, j)
+ countPS(i, j - 1)
- countPS(i + 1, j - 1);
}
// Driver code
static void Main()
{
for (int i = 0; i < 1000; i++)
for (int j = 0; j < 1000; j++)
dp[i, j] = -1;
n = str.Length;
Console.Write("Total palindromic subsequence"
+ "are : " + countPS(0, n - 1));
}
}// This code is contributed by DrRoot_ |
Javascript
<script>// Javascript program to counts Palindromic Subsequence// in a given String using recursion let n;
let dp=new Array(1000);
for(let i=0;i<1000;i++)
{
dp[i]=new Array(1000);
for(let j=0;j<1000;j++)
{
dp[i][j]=-1;
}
}
let str = "abcb";
// Function return the total
// palindromic subsequence
function countPS(i,j)
{
if (i > j)
return 0;
if (dp[i][j] != -1)
return dp[i][j];
if (i == j)
return dp[i][j] = 1;
else if (str[i] == str[j])
return dp[i][j]
= countPS(i + 1, j) +
countPS(i, j - 1) + 1;
else
return dp[i][j] = countPS(i + 1, j) +
countPS(i, j - 1) -
countPS(i + 1, j - 1);
}
// Driver code
n = str.length;
document.write("Total palindromic subsequence"
+ "are : " + countPS(0, n - 1));
// This code is contributed by rag2127
</script> |
PHP
<?php// PHP program to counts Palindromic Subsequence // in a given String using recursion$dp = array_fill(0, 100,
array_fill(0, 1000, -1));
$str = "abcb";
$n = strlen($str);
// Function return the total // palindromic subsequencefunction countPS($i, $j)
{ global $str, $dp, $n;
if($i > $j)
return 0;
if($dp[$i][$j] != -1)
return $dp[$i][$j];
if($i == $j)
return $dp[$i][$j] = 1;
else if ($str[$i] == $str[$j])
return $dp[$i][$j] = countPS($i + 1, $j) +
countPS($i, $j - 1) + 1;
else
return $dp[$i][$j] = countPS($i + 1, $j) +
countPS($i, $j - 1) -
countPS($i + 1, $j - 1);
}// Driver codeecho "Total palindromic subsequence are : " .
countPS(0, $n - 1);
// This code is contributed by mits?> |
Output
Total palindromic subsequence are : 6
Time Complexity : O(N2),
Auxiliary Space: O(N2)
This article is contributed by Aarti_Rathi and Nishant_sing.