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
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 |
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. |
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?> |
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> |
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_ |
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?> |
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> |
Output
Total palindromic subsequence are : 6
Time Complexity : O(N2),
Auxiliary Space: O(N2)
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