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

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

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++

// 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; 
} 

Java

// 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 

Python3

# 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 

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 

PHP

<?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 
?> 

Output:

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My Personal Notes

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

C++

Java

Python3

C#

PHP

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