Count All Palindrome Sub-Strings in a String | Set 2

Given a string, the task is to count all palindrome substring in a given string. Length of palindrome substring is greater then or equal to 2.

Examples:
Input : str = "abaab"
Output: 3
Explanation : 
All palindrome substring are :
 "aba", "aa", "baab" 

Input : str = "abbaeae"
Output: 4
Explanation : 
All palindrome substring are : 
"bb", "abba", "aea", "eae":

We have discussed a dynamic programming based solution in below post.



Count All Palindrome Sub-Strings in a String | Set 1

The solution discussed here is extension of Longest Palindromic Substring problem. The idea is for each character in the given input string, we consider it as midpoint of a palindrome and expand it in both directions to find all palindromes of even and odd lengths. We use hashmap to keep track of all the distinct palindromes of length greater than 1 and return map size which have count of all possible palindrome substrings.

The C++ implementation is given below.

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// C++ program to count all distinct palindromic
// substrings of a string.
#include <bits/stdc++.h>
using namespace std;
  
// Returns total number of palindrome substring of
// length greater than equal to 2
int countPalindromes(string s)
{
    unordered_map<string, int> m;
    for (int i = 0; i < s.length(); i++) {
  
        // check for odd length palindromes
        for (int j = 0; j <= i; j++) {
  
            if (!s[i + j])
                break;
  
            if (s[i - j] == s[i + j]) {
  
                // check for palindromes of length
                // greater than 1
                if ((i + j + 1) - (i - j) > 1)
                    m[s.substr(i - j, 
                        (i + j + 1) - (i - j))]++;
  
            } else
                break;
        }
  
        // check for even length palindromes
        for (int j = 0; j <= i; j++) {
            if (!s[i + j + 1])
                break;
            if (s[i - j] == s[i + j + 1]) {
  
                // check for palindromes of length 
                // greater than 1
                if ((i + j + 2) - (i - j) > 1)
                    m[s.substr(i - j, 
                         (i + j + 2) - (i - j))]++;
  
            } else
                break;
        }
    }
    return m.size();
}
  
// Driver code
int main()
{
    string s = "abbaeae";
    cout << countPalindromes(s) << endl;
    return 0;
}

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Output:

4

Time complexity O(n2)
Auxiliary Space O(n)

We can easily extend this solution to print all palindromic substrings also. We need to simply traverse the map m and print its content.

Another Approach is to use Java String Class, to do so,

  1. Iterate the loop twice for substring, get the substring of a string using substring() method.
  2. Reverse the substring using StringBuffer Class method reverse()
  3. Check for palindrome with substring and reverse substring
  4. If it is palindrome increment the count and return the count at last

Java

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// Java Program to count palindrome substring
// in a string
public class PalindromeSubstring {
      
    // Method which return count palindrome substring
    static int countPS(String str){
        String temp = "";
        StringBuffer stf;
        int count = 0;
        // Iterate the loop twice
        for (int i = 0; i < str.length(); i++) {
            for (int j = i + 1; j <= str.length(); j++) {
                // Get each substring
                temp = str.substring(i, j);
                  
                // If length is greater than or equal to two
                // Check for palindrome    
                if (temp.length() >= 2) {
                    // Use StringBuffer class to reverse the string
                    stf = new StringBuffer(temp);
                    stf.reverse();
                    // Compare substring wih reverse of substring
                    if (stf.toString().compareTo(temp) == 0)
                        count++;
                }
            }
        }
        // return the count
        return count;
    }
      
    // Driver Code
    public static void main(String args[]) throws Exception {
        // Declare and initialize the string
        String str = "abbaeae";
        // Call the method
        System.out.println(countPS(str));
    }
} // This code is contributes by hungundji

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C#

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// C# Program to count palindrome substring
// in a string 
using System;
  
class GFG
{
      
    // Method which return count palindrome substring
    static int countPS(String str)
    {
        String temp = "";
        String stf;
        int count = 0;
          
        // Iterate the loop twice
        for (int i = 0; i < str.Length; i++) 
        {
            for (int j = i + 1; 
                     j <= str.Length; j++)
            {
                // Get each substring
                temp = str.Substring(i, j-i);
                  
                // If length is greater than or equal to two
                // Check for palindrome 
                if (temp.Length >= 2)
                {
                    // Use StringBuffer class to reverse 
                    // the string
                    stf = temp;
                    stf = reverse(temp);
                      
                    // Compare substring wih reverse of substring
                    if (stf.ToString().CompareTo(temp) == 0)
                        count++;
                }
            }
        }
          
        // return the count
        return count;
    }
      
    static String reverse(String input) 
    {
        char[] a = input.ToCharArray();
        int l, r = 0;
        r = a.Length - 1;
  
        for (l = 0; l < r; l++, r--)
        {
            // Swap values of l and r 
            char temp = a[l];
            a[l] = a[r];
            a[r] = temp;
        }
        return String.Join("",a);
    }
      
    // Driver Code
    public static void Main(String []args)
    {
        // Declare and initialize the string
        String str = "abbaeae";
          
        // Call the method
        Console.WriteLine(countPS(str));
    }
  
// This code is contributed by Rajput-Ji

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Output:

4

Time complexity O(n2)
Auxiliary Space O(n)

This article is contributed by Ankur Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : bilal-hungund, Rajput-Ji