Count all Prime Length Palindromic Substrings

Given 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 at 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 length 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 length 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-string
# 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 substring
# whose length 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-strings
    count = 0

    # Starting from the smallest prime
    # till the largest Length of the
    # sub-string possible
    for j in range(2, Len + 1):

        # If j is prime
        if (prime[j]):

            # Check all the sub-strings of
            # Length j
            for i in range(Len + 1 - j):

                # If current sub-string 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 length 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

Javascript

<script>

// Javascript 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 length is a prime number
function countPrimePalindrome(str, len)
{

    var prime = Array(len + 1).fill(true);

    // 0 and 1 are non-primes
    prime[0] = prime[1] = false;
    for (var 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 (var i = p * p; i <= len; i += p)
                prime[i] = false;
        }
    }

    // To store the required number of sub-strings
    var count = 0;

    // Starting from the smallest prime till
    // the largest length of the sub-string possible
    for (var j = 2; j <= len; j++) {

        // If j is prime
        if (prime[j]) {

            // Check all the sub-strings of length j
            for (var 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
var s = "geeksforgeeks";
var len = s.length;
document.write( countPrimePalindrome(s, len));

</script>

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 length 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:

2

 Python program to count all prime length palindromic substrings :

Approach steps:

1.Define a function count_palindromic_substrings that takes a string s as input. The function will return the count of all prime length palindromic substrings in s.
2.Define a helper function is_palindrome that takes a string s as input and returns True if s is a palindrome, and False otherwise. This function checks if the string is equal to its reverse.
3.Initialize a variable count to 0 to keep track of the total count of prime length palindromic substrings.
4.Iterate over all possible substring lengths p from 2 to n (length of s), and check if p is a prime number using a loop that runs from 2 to the square root of p. If p is a prime number, iterate over all substrings of length p and check if each substring is a palindrome using the is_palindrome helper function. If the substring is a palindrome, increment the count variable.
5.Return the total count of prime length palindromic substrings.
6.In the example usage, create a string s and call the count_palindromic_substrings function with this argument. Finally, print the total count of prime length palindromic substrings.

C++

#include <iostream>
#include <string>
#include <cmath>

using namespace std;

// helper function to check if a string is a palindrome
bool is_palindrome(string s) {
    return s == string(s.rbegin(), s.rend());
}

int count_palindromic_substrings(string s) {
    int count = 0;
    int n = s.length();

    // iterate over all possible substrings of length `p`
    for (int p = 2; p <= n; p++) {
        // check if `p` is a prime number
        bool is_prime = true;
        for (int i = 2; i <= sqrt(p); i++) {
            if (p % i == 0) {
                is_prime = false;
                break;
            }
        }
        if (is_prime) {
            // iterate over all substrings of length `p`
            for (int i = 0; i <= n - p; i++) {
                if (is_palindrome(s.substr(i, p))) {
                    count++;
                }
            }
        }
    }

    // return the total count of prime length palindromic substrings
    return count;
}

int main() {
    string s = "abccba";
    int count = count_palindromic_substrings(s);
    cout << "Total number of prime length palindromic substrings in " << s << " is " << count << endl;

    return 0;
}

Java

import java.util.*;

public class Main {

    // helper function to check if a string is a palindrome
    public static boolean isPalindrome(String s) {
        return s.equals(new StringBuilder(s).reverse().toString());
    }

    public static int countPalindromicSubstrings(String s) {
        int count = 0;
        int n = s.length();

        // iterate over all possible substrings of length `p`
        for (int p = 2; p <= n; p++) {
            // check if `p` is a prime number
            boolean isPrime = true;
            for (int i = 2; i <= Math.sqrt(p); i++) {
                if (p % i == 0) {
                    isPrime = false;
                    break;
                }
            }
            if (isPrime) {
                // iterate over all substrings of length `p`
                for (int i = 0; i <= n - p; i++) {
                    if (isPalindrome(s.substring(i, i + p))) {
                        count++;
                    }
                }
            }
        }

        // return the total count of prime length palindromic substrings
        return count;
    }

    public static void main(String[] args) {
        String s = "abccba";
        int count = countPalindromicSubstrings(s);
        System.out.println("Total number of prime length palindromic substrings in " + s + " is " + count);
    }
}

Python3

# program to count all prime length palindromic substrings in a given string

def count_palindromic_substrings(s):
    # helper function to check if a string is a palindrome
    def is_palindrome(s):
        return s == s[::-1]

    count = 0
    n = len(s)

    # iterate over all possible substrings of length `p`
    for p in range(2, n+1):
        # check if `p` is a prime number
        is_prime = True
        for i in range(2, int(p**0.5)+1):
            if p % i == 0:
                is_prime = False
                break
        if is_prime:
            # iterate over all substrings of length `p`
            for i in range(n-p+1):
                if is_palindrome(s[i:i+p]):
                    count += 1

    # return the total count of prime length palindromic substrings
    return count

# example usage
s = "abccba"
count = count_palindromic_substrings(s)
print("Total number of prime length palindromic substrings in", s, "is", count)

Javascript

function count_palindromic_substrings(s) {
    // helper function to check if a string is a palindrome
    function is_palindrome(s) {
        return s === s.split("").reverse().join("");
    }

    let count = 0;
    const n = s.length;

    // iterate over all possible substrings of length `p`
    for (let p = 2; p <= n; p++) {
        // check if `p` is a prime number
        let is_prime = true;
        for (let i = 2; i <= Math.floor(Math.sqrt(p)); i++) {
            if (p % i === 0) {
                is_prime = false;
                break;
            }
        }
        if (is_prime) {
            // iterate over all substrings of length `p`
            for (let i = 0; i <= n-p; i++) {
                if (is_palindrome(s.substring(i, i+p))) {
                    count += 1;
                }
            }
        }
    }

    // return the total count of prime length palindromic substrings
    return count;
}

// example usage
const s = "abccba";
const count = count_palindromic_substrings(s);
console.log(`Total number of prime length palindromic substrings in ${s} is ${count}`);

C#

using System;

class MainClass {
    // helper function to check if a string is a palindrome
    public static bool isPalindrome(string s) {
        char[] charArray = s.ToCharArray();
        Array.Reverse(charArray);
        return s == new string(charArray);
    }

    public static int countPalindromicSubstrings(string s) {
        int count = 0;
        int n = s.Length;

        // iterate over all possible substrings of length `p`
        for (int p = 2; p <= n; p++) {
            // check if `p` is a prime number
            bool isPrime = true;
            for (int i = 2; i <= Math.Sqrt(p); i++) {
                if (p % i == 0) {
                    isPrime = false;
                    break;
                }
            }
            if (isPrime) {
                // iterate over all substrings of length `p`
                for (int i = 0; i <= n - p; i++) {
                    if (isPalindrome(s.Substring(i, p))) {
                        count++;
                    }
                }
            }
        }

        // return the total count of prime length palindromic substrings
        return count;
    }

    public static void Main (string[] args) {
        string s = "abccba";
        int count = countPalindromicSubstrings(s);
        Console.WriteLine("Total number of prime length palindromic substrings in " + s + " is " + count);
    }
}

Output

Total number of prime length palindromic substrings in abccba is 1

Time complexity: O(n^2 log n)

Auxiliary Space: O(1).