Count all Prime Length Palindromic Substrings
Last Updated :
30 Mar, 2023
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++
#include <bits/stdc++.h>
using namespace std;
bool isPalindrome(string str, int i, int j)
{
while (i < j) {
if (str[i] != str[j])
return false;
i++;
j--;
}
return true;
}
int countPrimePalindrome(string str, int len)
{
bool prime[len + 1];
memset(prime, true, sizeof(prime));
prime[0] = prime[1] = false;
for (int p = 2; p * p <= len; p++) {
if (prime[p]) {
for (int i = p * p; i <= len; i += p)
prime[i] = false;
}
}
int count = 0;
for (int j = 2; j <= len; j++) {
if (prime[j]) {
for (int i = 0; i + j - 1 < len; i++) {
if (isPalindrome(str, i, i + j - 1))
count++;
}
}
}
return count;
}
int main()
{
string s = "geeksforgeeks";
int len = s.length();
cout << countPrimePalindrome(s, len);
return 0;
}
|
Java
import java.util.Arrays;
class GfG
{
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;
}
static int countPrimePalindrome(String str, int len)
{
boolean[] prime = new boolean[len + 1];
Arrays.fill(prime, true);
prime[0] = prime[1] = false;
for (int p = 2; p * p <= len; p++)
{
if (prime[p])
{
for (int i = p * p; i <= len; i += p)
prime[i] = false;
}
}
int count = 0;
for (int j = 2; j <= len; j++)
{
if (prime[j])
{
for (int i = 0; i + j - 1 < len; i++)
{
if (isPalindrome(str, i, i + j - 1))
count++;
}
}
}
return count;
}
public static void main(String []args)
{
String s = "geeksforgeeks";
int len = s.length();
System.out.println(countPrimePalindrome(s, len));
}
}
|
Python3
import math as mt
def isPalindrome(str1, i, j):
while (i < j):
if (str1[i] != str1[j]):
return False
i += 1
j -= 1
return True
def countPrimePalindrome(str1, Len):
prime = [True for i in range(Len + 1)]
prime[0], prime[1] = False, False
for p in range(2, mt.ceil(mt.sqrt(Len + 1))):
if (prime[p]):
for i in range(2 * p, Len + 1, p):
prime[i] = False
count = 0
for j in range(2, Len + 1):
if (prime[j]):
for i in range(Len + 1 - j):
if (isPalindrome(str1, i, i + j - 1)):
count += 1
return count
s = "geeksforgeeks"
Len = len(s)
print( countPrimePalindrome(s, Len))
|
C#
using System;
class GfG
{
static bool isPalindrome(string str,
int i, int j)
{
while (i < j)
{
if (str[i] != str[j])
return false;
i++;
j--;
}
return true;
}
static int countPrimePalindrome(string str,
int len)
{
bool[] prime = new bool[len + 1];
Array.Fill(prime, true);
prime[0] = prime[1] = false;
for (int p = 2; p * p <= len; p++)
{
if (prime[p])
{
for (int i = p * p; i <= len; i += p)
prime[i] = false;
}
}
int count = 0;
for (int j = 2; j <= len; j++)
{
if (prime[j])
{
for (int i = 0;
i + j - 1 < len; i++)
{
if (isPalindrome(str, i, i + j - 1))
count++;
}
}
}
return count;
}
public static void Main()
{
string s = "geeksforgeeks";
int len = s.Length;
Console.WriteLine(countPrimePalindrome(s, len));
}
}
|
Javascript
<script>
function isPalindrome(str, i, j)
{
while (i < j) {
if (str[i] != str[j])
return false;
i++;
j--;
}
return true;
}
function countPrimePalindrome(str, len)
{
var prime = Array(len + 1).fill(true);
prime[0] = prime[1] = false;
for (var p = 2; p * p <= len; p++) {
if (prime[p]) {
for (var i = p * p; i <= len; i += p)
prime[i] = false;
}
}
var count = 0;
for (var j = 2; j <= len; j++) {
if (prime[j]) {
for (var i = 0; i + j - 1 < len; i++) {
if (isPalindrome(str, i, i + j - 1))
count++;
}
}
}
return count;
}
var s = "geeksforgeeks";
var len = s.length;
document.write( countPrimePalindrome(s, len));
</script>
|
PHP
<?php
function isPalindrome($str, $i, $j)
{
while ($i < $j)
{
if ($str[$i] != $str[$j])
return false;
$i++;
$j--;
}
return true;
}
function countPrimePalindrome($str, $len)
{
$prime = array_fill(0, $len + 1, true);
$prime[0] = false ;
$prime[1] = false;
for ($p = 2; $p * $p <= $len; $p++)
{
if ($prime[$p])
{
for ($i = $p * $p; $i <= $len; $i += $p)
$prime[$i] = false;
}
}
$count = 0;
for ($j = 2; $j <= $len; $j++)
{
if ($prime[$j])
{
for ($i = 0; $i + $j - 1 < $len; $i++)
{
if (isPalindrome($str, $i, $i + $j - 1))
$count++;
}
}
}
return $count;
}
$s = "geeksforgeeks";
$len = strlen($s);
echo countPrimePalindrome($s, $len);
?>
|
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;
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();
for (int p = 2; p <= n; p++) {
bool is_prime = true;
for (int i = 2; i <= sqrt(p); i++) {
if (p % i == 0) {
is_prime = false;
break;
}
}
if (is_prime) {
for (int i = 0; i <= n - p; i++) {
if (is_palindrome(s.substr(i, p))) {
count++;
}
}
}
}
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 {
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();
for (int p = 2; p <= n; p++) {
boolean isPrime = true;
for (int i = 2; i <= Math.sqrt(p); i++) {
if (p % i == 0) {
isPrime = false;
break;
}
}
if (isPrime) {
for (int i = 0; i <= n - p; i++) {
if (isPalindrome(s.substring(i, i + p))) {
count++;
}
}
}
}
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
def count_palindromic_substrings(s):
def is_palindrome(s):
return s == s[::-1]
count = 0
n = len(s)
for p in range(2, n+1):
is_prime = True
for i in range(2, int(p**0.5)+1):
if p % i == 0:
is_prime = False
break
if is_prime:
for i in range(n-p+1):
if is_palindrome(s[i:i+p]):
count += 1
return count
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) {
function is_palindrome(s) {
return s === s.split("").reverse().join("");
}
let count = 0;
const n = s.length;
for (let p = 2; p <= n; p++) {
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) {
for (let i = 0; i <= n-p; i++) {
if (is_palindrome(s.substring(i, i+p))) {
count += 1;
}
}
}
}
return count;
}
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 {
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;
for (int p = 2; p <= n; p++) {
bool isPrime = true;
for (int i = 2; i <= Math.Sqrt(p); i++) {
if (p % i == 0) {
isPrime = false;
break;
}
}
if (isPrime) {
for (int i = 0; i <= n - p; i++) {
if (isPalindrome(s.Substring(i, p))) {
count++;
}
}
}
}
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).
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...