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

• Difficulty Level : Medium
• Last Updated : 04 Jan, 2022

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":

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

CPP

 // C++ program to count all distinct palindromic // substrings of a string. #include using namespace std;    // Returns total number of palindrome substring of // length greater than equal to 2 int countPalindromes(string s) {     unordered_map 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; }

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

 // 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

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

Output:

4

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

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