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

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  • Difficulty Level : Hard
  • Last Updated : 05 Oct, 2022
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Given a string, the task is to count all palindrome substring in a given string. Length of palindrome substring is greater than 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. 

Implementation:

CPP




// 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;
}

Java




// 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 with 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

Python3




# Python program to count all distinct palindromic
# substrings of a string.
 
# Returns total number of palindrome substring of
# length greater than equal to 2
def countPalindromes(s):
 
    m = {}
    for i in range(len(s)):
 
        # check for odd length palindromes
        for j in range(i+1):
 
            if (i + j >= len(s)):
                break
 
            if (s[i - j] == s[i + j]):
 
                # check for palindromes of length
                # greater than 1
                if ((i + j + 1) - (i - j) > 1):
                    if(s[i - j:(i + j + 1)] in m):
                        m[s[i - j:(i + j + 1)]] += 1
                    else:
                        m[s[i - j:(i + j + 1)]] = 1
 
            else:
                break
 
        # check for even length palindromes
        for j in range(i+1):
            if (i + j + 1 >= len(s)):
                break
            if (s[i - j] == s[i + j + 1]):
 
                # check for palindromes of length
                # greater than 1
                if ((i + j + 2) - (i - j) > 1):
                    if(s[i - j:i + j + 2] in m):
                        m[s[i - j:(i + j + 2)]] += 1
                    else:
                        m[s[i - j:(i + j + 2)]] = 1
 
            else:
                break
         
    return len(m)
 
# Driver code
s = "abbaeae"
print(countPalindromes(s))
 
# This code is contributed by shinjanpatra

C#




// 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 with 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

Javascript




<script>
 
// JavaScript program to count all distinct palindromic
// substrings of a string.
 
// Returns total number of palindrome substring of
// length greater than equal to 2
function countPalindromes(s)
{
    let m = new Map();
    for (let i = 0; i < s.length; i++) {
 
        // check for odd length palindromes
        for (let 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){
                    if(m.has(s.substr(i - j,(i + j + 1) - (i - j)))){
                        m.set(s.substr(i - j,(i + j + 1) - (i - j)),m.get(s.substr(i - j,(i + j + 1) - (i - j))) + 1)
                    }
                    else m.set(s.substr(i - j,(i + j + 1) - (i - j)),1)
                }
 
            } else
                break;
        }
 
        // check for even length palindromes
        for (let 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){
                    if(m.has(s.substr(i - j,(i + j + 2) - (i - j)))){
                        m.set(s.substr(i - j,(i + j + 2) - (i - j)),m.get(s.substr(i - j,(i + j + 2) - (i - j))) + 1)
                    }
                    else m.set(s.substr(i - j,(i + j + 2) - (i - j)),1)
                }
 
            } else
                break;
        }
    }
    return m.size;
}
 
// Driver code
 
let s = "abbaeae"
document.write(countPalindromes(s),"</br>")
 
// This code is contributed by shinjanpatra
 
</script>

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.

In the Approach discussed above we are using a unordered map which is taking O(n) space but we can do the same thing without using any extra space.Instead of making a hashmap we can directly count the palindromic substring.

Implementation:

C++




#include <bits/stdc++.h>
using namespace std;
int countPalindromes(string s)
{
    int n = s.size();
    int count = 0;
    for (int i = 0; i < s.size(); i++) {
        int left = i - 1;
        int right = i + 1;
        while (left >= 0 and right < n) {
            if (s[left] == s[right])
                count++;
            else
                break;
            left--;
            right++;
        }
    }
 
    for (int i = 0; i < s.size(); i++) {
        int left = i;
        int right = i + 1;
        while (left >= 0 and right < n) {
            if (s[left] == s[right])
                count++;
            else
                break;
            left--;
            right++;
        }
    }
    return count;
}
int main()
{
 
    string s = "abbaeae";
    cout << countPalindromes(s) << endl;
    return 0;
}
// This  code is contributed by Arpit Jain

Java




class GFG {
   
      public static int countPalindromes(String s){
          int count = 0;
        for(int i=0; i<s.length(); i++){
          //Taking every node as a center and expanding it from there to check if it is a palindrome or not.
          //This will make sure that the length of the string to be checked > 2.
            count += expandFromCenter(s, i-1, i+1) +  expandFromCenter(s, i , i+1);
        }
         
        return count;  
    }
   
      public static int expandFromCenter(String s, int left, int right){
        int count = 0;
          //As we expand from the center and find if the character matches, we increase the count.
        while(left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)){
              count++;
              left--;
              right++;
        }
        return count;
    }
   
    public static void main (String[] args) {
        String s = "abbaeae";
           System.out.println(countPalindromes(s));
    }
}

Python




# Python program to count all distinct palindromic
# substrings of a string.
 
# Returns total number of palindrome substring of
# length greater than equal to 2
 
 
def countPalindromes(s):
    n = len(s)
    count = 0
    for i in range(0, n):
        left = i-1
        right = i+1
        while left >= 0 and right < n:
            if s[left] == s[right]:
                count += 1
            else:
                break
            left -= 1
            right += 1
    for i in range(0, n):
        left = i
        right = i+1
        while left >= 0 and right < n:
            if s[left] == s[right]:
                count += 1
            else:
                break
            left -= 1
            right += 1
    return count
 
   # Driver code
s = "abbaeae"
print(countPalindromes(s))
 
# This code is contributed by Arpit Jain

C#




using System;
 
public class GFG{
 
  public static int countPalindromes(string s){
    int count = 0;
    for(int i = 0; i < s.Length; i++)
    {
      // Taking every node as a center and expanding
      // it from there to check if it is a palindrome or not.
      // This will make sure that the length of the string to be checked > 2.
      count += expandFromCenter(s, i - 1, i + 1) +  expandFromCenter(s, i , i + 1);
    }
 
    return count;  
  }
 
  public static int expandFromCenter(string s, int left, int right){
    int count = 0;
 
    // As we expand from the center and find
    // if the character matches, we increase the count.
    while(left >= 0 && right < s.Length && s[left] == s[right]){
      count++;
      left--;
      right++;
    }
    return count;
  }
 
  public static void Main (string[] args) {
    string s = "abbaeae";
    Console.WriteLine(countPalindromes(s));
  }
}
 
// This code is contributed by Shubham singh

Javascript




<script>
function countPalindromes(s)
{
    n = s.length;
    count = 0;
    for (i = 0; i < s.length; i++) {
        left = i - 1;
        right = i + 1;
        while (left >= 0 && right < n) {
            if (s[left] == s[right])
                count++;
            else
                break;
            left--;
            right++;
        }
    }
 
    for (i = 0; i < s.length; i++) {
        left = i;
        right = i + 1;
        while (left >= 0 && right < n) {
            if (s[left] == s[right])
                count++;
            else
                break;
            left--;
            right++;
        }
    }
    return count;
}
 
s = "abbaeae";
document.write(countPalindromes(s));
 
// This  code is contributed by Shubham Singh
</script>

Output

4

Time complexity is O(n^2) in worst case because we are running one while loop inside a for loop because in while loop we are expanding to both directions so that’s why it’s  O(n/2) in worst case when all characters are same eg : “aaaaaaaa” .

But we are not using any extra space , Space Complexity is O(1) . 


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