Count all Prime Length Palindromic Substrings

Given a string str, the task is to count all the sub-strings of str which are palindromes and their length is prime.

Examples:

Input: str = “geeksforgeeks”
Output: 2
“ee” and “ee” are the only valid sub-strings

Input: str = “abccc”
Output: 3

Approach: Using Sieve of Eratosthenes, find all the primes till the length of str because that is the maximum length a sub-string of str can have. Now starting from the smallest prime i.e. j = 2 till j ≤ len(str). If j is prime then count all the palindromic sub-strings of str whose length = j. Print the total count in the end.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns true if sub-string
// starting at i and ending at j in str is a palindrome
bool isPalindrome(string str, int i, int j)
{
    while (i < j) {
        if (str[i] != str[j])
            return false;
        i++;
        j--;
    }
  
    return true;
}
  
// Function to count all palindromic substring
// whose lwngth is a prime number
int countPrimePalindrome(string str, int len)
{
  
    bool prime[len + 1];
    memset(prime, true, sizeof(prime));
  
    // 0 and 1 are non-primes
    prime[0] = prime[1] = false;
    for (int p = 2; p * p <= len; p++) {
  
        // If prime[p] is not changed, then it is a prime
        if (prime[p]) {
  
            // Update all multiples of p greater than or
            // equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked.
            for (int i = p * p; i <= len; i += p)
                prime[i] = false;
        }
    }
  
    // To store the required number of sub-strings
    int count = 0;
  
    // Starting from the smallest prime till 
    // the largest length of the sub-string possible
    for (int j = 2; j <= len; j++) {
  
        // If j is prime
        if (prime[j]) {
  
            // Check all the sub-strings of length j
            for (int i = 0; i + j - 1 < len; i++) {
  
                // If current sub-string is a palindrome
                if (isPalindrome(str, i, i + j - 1))
                    count++;
            }
        }
    }
  
    return count;
}
  
// Driver Code
int main()
{
    string s = "geeksforgeeks";
    int len = s.length();
  
    cout << countPrimePalindrome(s, len);
  
    return 0;
}

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Java

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// Java implementation of the approach 
import java.util.Arrays;
  
class GfG
{
  
    // Function that returns true if 
    // sub-string starting at i and 
    // ending at j in str is a palindrome 
    static boolean isPalindrome(String str, int i, int j) 
    
        while (i < j) 
        
            if (str.charAt(i) != str.charAt(j)) 
                return false
            i++; 
            j--; 
        
      
        return true
    
      
    // Function to count all palindromic substring 
    // whose lwngth is a prime number 
    static int countPrimePalindrome(String str, int len) 
    
      
        boolean[] prime = new boolean[len + 1]; 
        Arrays.fill(prime, true); 
      
        // 0 and 1 are non-primes 
        prime[0] = prime[1] = false
        for (int p = 2; p * p <= len; p++) 
        
      
            // If prime[p] is not changed,
            // then it is a prime 
            if (prime[p])
            
      
                // Update all multiples of p greater than or 
                // equal to the square of it 
                // numbers which are multiple of p and are 
                // less than p^2 are already been marked. 
                for (int i = p * p; i <= len; i += p) 
                    prime[i] = false
            
        
      
        // To store the required number of sub-strings 
        int count = 0
      
        // Starting from the smallest prime till 
        // the largest length of the sub-string possible 
        for (int j = 2; j <= len; j++) 
        
      
            // If j is prime 
            if (prime[j]) 
            
      
                // Check all the sub-strings of length j 
                for (int i = 0; i + j - 1 < len; i++) 
                
      
                    // If current sub-string is a palindrome 
                    if (isPalindrome(str, i, i + j - 1)) 
                        count++; 
                
            
        
        return count; 
    
      
    // Driver code
    public static void main(String []args)
    {
        String s = "geeksforgeeks"
        int len = s.length(); 
  
        System.out.println(countPrimePalindrome(s, len));
    }
}
  
// This code is contributed by Rituraj Jain

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Python3

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# Python3 implementation of the approach
import math as mt
  
# Function that returns True if sub-str1ing
# starting at i and ending at j in str1 
# is a palindrome
def isPalindrome(str1, i, j):
  
    while (i < j):
        if (str1[i] != str1[j]):
            return False
        i += 1
        j -= 1
      
    return True
      
# Function to count all palindromic substr1ing
# whose lwngth is a prime number
def countPrimePalindrome(str1, Len):
  
    prime = [True for i in range(Len + 1)]
  
    # 0 and 1 are non-primes
    prime[0], prime[1] = False, False
    for p in range(2, mt.ceil(mt.sqrt(Len + 1))):
  
        # If prime[p] is not changed, 
        # then it is a prime
        if (prime[p]):
  
            # Update all multiples of p greater 
            # than or equal to the square of it
            # numbers which are multiple of p 
            # and are less than p^2 are already 
            # been marked.
            for i in range(2 * p, Len + 1, p):
                prime[i] = False
          
    # To store the required number 
    # of sub-str1ings
    count = 0
  
    # Starting from the smallest prime 
    # till the largest Length of the 
    # sub-str1ing possible
    for j in range(2, Len + 1):
  
        # If j is prime
        if (prime[j]):
  
            # Check all the sub-str1ings of 
            # Length j
            for i in range(Len + 1 - j):
  
                # If current sub-str1ing is a palindrome
                if (isPalindrome(str1, i, i + j - 1)):
                    count += 1
              
    return count
  
# Driver Code
s = "geeksforgeeks"
Len = len(s)
  
print( countPrimePalindrome(s, Len))
  
# This code is contributed by 
# Mohit kumar 29

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C#

// C# implementation of the approach
using System;

class GfG
{

// Function that returns true if
// sub-string starting at i and
// ending at j in str is a palindrome
static bool isPalindrome(string str,
int i, int j)
{
while (i < j) { if (str[i] != str[j]) return false; i++; j--; } return true; } // Function to count all palindromic // substring whose lwngth is a prime number static int countPrimePalindrome(string str, int len) { bool[] prime = new bool[len + 1]; Array.Fill(prime, true); // 0 and 1 are non-primes prime[0] = prime[1] = false; for (int p = 2; p * p <= len; p++) { // If prime[p] is not changed, // then it is a prime if (prime[p]) { // Update all multiples of p greater // than or equal to the square of it // numbers which are multiple of p // and are less than p^2 are already // been marked. for (int i = p * p; i <= len; i += p) prime[i] = false; } } // To store the required number // of sub-strings int count = 0; // Starting from the smallest prime // till the largest length of the // sub-string possible for (int j = 2; j <= len; j++) { // If j is prime if (prime[j]) { // Check all the sub-strings of // length j for (int i = 0; i + j - 1 < len; i++) { // If current sub-string is a // palindrome if (isPalindrome(str, i, i + j - 1)) count++; } } } return count; } // Driver code public static void Main() { string s = "geeksforgeeks"; int len = s.Length; Console.WriteLine(countPrimePalindrome(s, len)); } } // This code is contributed by Code_Mech [tabby title = "PHP"]

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<?php
// PHP implementation of the approach 
  
// Function that returns true if sub-string 
// starting at i and ending at j in str is 
// a palindrome 
function isPalindrome($str, $i, $j
    while ($i < $j
    
        if ($str[$i] != $str[$j]) 
            return false; 
        $i++; 
        $j--; 
    
  
    return true; 
  
// Function to count all palindromic substring 
// whose lwngth is a prime number 
function countPrimePalindrome($str, $len
    $prime = array_fill(0, $len + 1, true);
  
    // 0 and 1 are non-primes 
    $prime[0] = false ;
    $prime[1] = false; 
    for ($p = 2; $p * $p <= $len; $p++)
    
  
        // If prime[p] is not changed, 
        // then it is a prime 
        if ($prime[$p]) 
        
  
            // Update all multiples of p greater 
            // than or equal to the square of it 
            // numbers which are multiple of p and are 
            // less than p^2 are already been marked. 
            for ($i = $p * $p; $i <= $len; $i += $p
                $prime[$i] = false; 
        
    
  
    // To store the required number 
    // of sub-strings 
    $count = 0; 
  
    // Starting from the smallest prime till 
    // the largest length of the sub-string possible 
    for ($j = 2; $j <= $len; $j++) 
    
  
        // If j is prime 
        if ($prime[$j])
        
  
            // Check all the sub-strings of length j 
            for ($i = 0; $i + $j - 1 < $len; $i++) 
            
  
                // If current sub-string is a palindrome 
                if (isPalindrome($str, $i, $i + $j - 1)) 
                    $count++; 
            
        
    
  
    return $count
  
// Driver Code 
$s = "geeksforgeeks"
$len = strlen($s); 
  
echo countPrimePalindrome($s, $len); 
  
// This code is contributed by Ryuga
?>

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Output:

2


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