Count of ways to split given string into two non-empty palindromes

Given a string S, the task is to find the number of ways to split the given string S into two non-empty palindromic strings.
Examples:

Input: S = “aaaaa” 
Output: 4 
Explanation:
Possible Splits: {“a”, “aaaa”}, {“aa”, “aaa”}, {“aaa”, “aa”}, {“aaaa”, “a”}
Input: S = “abacc” 
Output: 1 
Explanation:
Only possible split is “aba”, “cc”.

Naive Approach: The naive approach is to split the string at each possible index and check if both the substrings are palindromic or not. If yes then increment the count for that index. Print the final count.
Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include<bits/stdc++.h>
using namespace std;

// Function to check whether the
// substring from l to r is
// palindrome or not
bool isPalindrome(int l, int r,
                  string& s)
{

    while (l <= r) {

        // If characters at l and
        // r differ
        if (s[l] != s[r])

            // Not a palindrome
            return false;

        l++;
        r--;
    }

    // If the string is
    // a palindrome
    return true;
}

// Function to count and return
// the number of possible splits
int numWays(string& s)
{
    int n = s.length();

    // Stores the count
    // of splits
    int ans = 0;
    for (int i = 0;
         i < n - 1; i++) {

        // Check if the two substrings
        // after the split are
        // palindromic or not
        if (isPalindrome(0, i, s)
            && isPalindrome(i + 1,
                            n - 1, s)) {

            // If both are palindromes
            ans++;
        }
    }

    // Print the final count
    return ans;
}

// Driver Code
int main()
{
    string S = "aaaaa";

    cout << numWays(S);
    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{

// Function to check whether the
// substring from l to r is
// palindrome or not
public static boolean isPalindrome(int l, int r,
                                   String s)
{
    while (l <= r)
    {

        // If characters at l and
        // r differ
        if (s.charAt(l) != s.charAt(r))

            // Not a palindrome
            return false;

        l++;
        r--;
    }

    // If the string is
    // a palindrome
    return true;
}

// Function to count and return
// the number of possible splits
public static int numWays(String s)
{
    int n = s.length();

    // Stores the count
    // of splits
    int ans = 0;
    for(int i = 0; i < n - 1; i++)
    {

       // Check if the two substrings
       // after the split are
       // palindromic or not
       if (isPalindrome(0, i, s) &&
           isPalindrome(i + 1, n - 1, s))
       {

           // If both are palindromes
           ans++;
       }
    }

    // Print the final count
    return ans;
}

// Driver Code
public static void main(String args[])
{
    String S = "aaaaa";

    System.out.println(numWays(S));
}
}

// This code is contributed by SoumikMondal

Python3

# Python3 program to implement
# the above approach

# Function to check whether the
# substring from l to r is
# palindrome or not
def isPalindrome(l, r, s):

    while (l <= r):

        # If characters at l and
        # r differ
        if (s[l] != s[r]):

            # Not a palindrome
            return bool(False)

        l += 1
        r -= 1

    # If the string is
    # a palindrome
    return bool(True)

# Function to count and return
# the number of possible splits
def numWays(s):

    n = len(s)

    # Stores the count
    # of splits
    ans = 0
    for i in range(n - 1):

        # Check if the two substrings
        # after the split are
        # palindromic or not
        if (isPalindrome(0, i, s) and
            isPalindrome(i + 1, n - 1, s)):

            # If both are palindromes
            ans += 1

    # Print the final count
    return ans

# Driver Code
S = "aaaaa"

print(numWays(S))

# This code is contributed by divyeshrabadiya07

C#

// C# program to implement
// the above approach
using System;
class GFG{

// Function to check whether the
// substring from l to r is
// palindrome or not
public static bool isPalindrome(int l, int r,
                                string s)
{
    while (l <= r)
    {

        // If characters at l and
        // r differ
        if (s[l] != s[r])

            // Not a palindrome
            return false;

        l++;
        r--;
    }

    // If the string is
    // a palindrome
    return true;
}

// Function to count and return
// the number of possible splits
public static int numWays(string s)
{
    int n = s.Length;

    // Stores the count
    // of splits
    int ans = 0;
    for(int i = 0; i < n - 1; i++)
    {

       // Check if the two substrings
       // after the split are
       // palindromic or not
       if (isPalindrome(0, i, s) &&
           isPalindrome(i + 1, n - 1, s))
       {

           // If both are palindromes
           ans++;
       }
    }

    // Print the final count
    return ans;
}

// Driver Code
public static void Main(string []args)
{
    string S = "aaaaa";

    Console.Write(numWays(S));
}
}

// This code is contributed by Rutvik

Javascript

<script>

    // Javascript program to implement
    // the above approach

    // Function to check whether the
    // substring from l to r is
    // palindrome or not
    function isPalindrome(l, r, s)
    {
        while (l <= r)
        {

            // If characters at l and
            // r differ
            if (s[l] != s[r])

                // Not a palindrome
                return false;

            l++;
            r--;
        }

        // If the string is
        // a palindrome
        return true;
    }

    // Function to count and return
    // the number of possible splits
    function numWays(s)
    {
        let n = s.length;

        // Stores the count
        // of splits
        let ans = 0;
        for(let i = 0; i < n - 1; i++)
        {

           // Check if the two substrings
           // after the split are
           // palindromic or not
           if (isPalindrome(0, i, s) &&
               isPalindrome(i + 1, n - 1, s))
           {

               // If both are palindromes
               ans++;
           }
        }

        // Print the final count
        return ans;
    }

    let S = "aaaaa";

    document.write(numWays(S));

</script>

Output

4

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using the Hashing and Rabin-Karp Algorithm to store Prefix and Suffix Hashes of the string. Follow the steps below to solve the problem:

• Compute prefix and suffix hash of the given string.
• For every index i in the range [1, N – 1], check if the two substrings [0, i – 1] and [i, N – 1] are palindrome or not.
• To check if a substring [l, r] is a palindrome or not, simply check:

PrefixHash[l - r] = SuffixHash[l - r]

• For every index i for which two substrings are found to be palindromic, increase the count.
• Print the final value of count.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach

#include
using namespace std;

// Modulo for rolling hash
const int MOD = 1e9 + 9;

// Small prime for rolling hash
const int P = 37;

// Maximum length of string
const int MAXN = 1e5 + 5;

// Stores prefix hash
vector prefixHash(MAXN);

// Stores suffix hash
vector suffixHash(MAXN);

// Stores inverse modulo
// of P for prefix
vector inversePrefix(MAXN);

// Stores inverse modulo
// of P for suffix
vector inverseSuffix(MAXN);

int n;
int power(int x, int y, int mod)
{
    // Function to compute
    // power under modulo
    if (x == 0)
        return 0;

    int ans = 1;
    while (y > 0) {
        if (y & 1)
            ans = (1LL * ans * x)
                % MOD;

        x = (1LL * x * x) % MOD;
        y >>= 1;
    }
    return ans;
}

// Precompute hashes for the
// given string
void preCompute(string& s)
{

    int x = 1;
    for (int i = 0; i 0)
            prefixHash[i]
                = (prefixHash[i]
                + prefixHash[i - 1])
                % MOD;

        // Compute inverse modulo
        // of P ^ i for division
        // using Fermat Little theorem
        inversePrefix[i] = power(x, MOD - 2,
                                MOD);

        x = (1LL * x * P) % MOD;
    }

    x = 1;

    // Calculate suffix hash
    for (int i = n - 1; i >= 0; i--) {

        // Calculate and store hash
        suffixHash[i]
            = (1LL * int(s[i]
                        - 'a' + 1)
            * x)
            % MOD;

        if (i 0
                ? prefixHash[l - 1]
                : 0);
    h = (h + MOD) % MOD;
    h = (1LL * h * inversePrefix[l])
        % MOD;

    return h;
}

// Function to return Suffix
// Hash of substring
int getSuffixHash(int l, int r)
{
    // Calculate suffix hash
    // from l to r
    int h = suffixHash[l]
            - (r < n - 1
                ? suffixHash[r + 1]
                : 0);

    h = (h + MOD) % MOD;
    h = (1LL * h * inverseSuffix[r])
        % MOD;

    return h;
}

int numWays(string& s)
{
    n = s.length();

    // Compute prefix and
    // suffix hashes
    preCompute(s);

    // Stores the number of
    // possible splits
    int ans = 0;
    for (int i = 0;
        i < n - 1; i++) {

        int preHash = getPrefixHash(0, i);
        int sufHash = getSuffixHash(0, i);

        // If the substring s[0]...s[i]
        // is not palindromic
        if (preHash != sufHash)
            continue;

        preHash = getPrefixHash(i + 1,
                                n - 1);
        sufHash = getSuffixHash(i + 1,
                                n - 1);

        // If the substring (i + 1, n - 1)
        // is not palindromic
        if (preHash != sufHash)
            continue;

        // If both are palindromic
        ans++;
    }

    return ans;
}

// Driver Code
int main()
{
    string s = "aaaaa";

    int ans = numWays(s);

    cout << ans << endl;

    return 0;
}

Java

// Java Program to implement
// the above approach

import java.util.*;

public class Main {

    // Modulo for rolling hash
    static final int MOD = 1_000_000_007;

    // Small prime for rolling hash
    static final int P = 37;

    // Maximum length of string
    static final int MAXN = 100_005;

    // Stores prefix hash
    static int[] prefixHash = new int[MAXN];

    // Stores suffix hash
    static int[] suffixHash = new int[MAXN];

    // Stores inverse modulo
    // of P for prefix
    static int[] inversePrefix = new int[MAXN];


    // Stores inverse modulo
    // of P for suffix
    static int[] inverseSuffix = new int[MAXN];

    static int n;


    static int power(int x, int y, int mod) {
        // Function to compute
        // power under modulo
        if (x == 0)
            return 0;

        int ans = 1;
        while (y > 0) {
            if ((y & 1) == 1)
                ans = (int)(((long)ans * x) % mod);

            x = (int)(((long)x * x) % mod);
            y >>= 1;
        }
        return ans;
    }

    // Precompute hashes for the
    // given string
    static void preCompute(String s) {
        int x = 1;
        for (int i = 0; i < n; i++) {
            prefixHash[i] = (int)(((long)(s.charAt(i) - 'a' + 1) * x) % MOD);

            // Compute inverse modulo
            // of P ^ i for division
            // using Fermat Little theorem
            if (i > 0)
                prefixHash[i] = (prefixHash[i] + prefixHash[i - 1]) % MOD;

            inversePrefix[i] = power(x, MOD - 2, MOD);

            x = (int)(((long)x * P) % MOD);
        }

        x = 1;

        // Calculate suffix hash
        for (int i = n - 1; i >= 0; i--) {


            // Calculate and store hash
            suffixHash[i] = (int)(((long)(s.charAt(i) - 'a' + 1) * x) % MOD);
            if (i < n - 1)
                suffixHash[i] = (suffixHash[i] + suffixHash[i + 1]) % MOD;

            inverseSuffix[i] = power(x, MOD - 2, MOD);

            x = (int)(((long)x * P) % MOD);
        }
    }

    static int getPrefixHash(int l, int r) {
        int h = prefixHash[r];
        if (l > 0)
            h = (h - prefixHash[l - 1] + MOD) % MOD;

        h = (int)(((long)h * inversePrefix[l]) % MOD);

        return h;
    }

    // Function to return Suffix
    // Hash of substring
    static int getSuffixHash(int l, int r) {

        int h = suffixHash[l];
        if (r < n - 1)
            h = (h - suffixHash[r + 1] + MOD) % MOD;

        h = (int)(((long)h * inverseSuffix[r]) % MOD);

        return h;
    }

    static int numWays(String s) {
        n = s.length();

        // Calculate suffix hash
        // from l to r
        preCompute(s);


        // Stores the number of
        // possible splits
        int ans = 0;
        for (int i = 0; i < n - 1; i++) {
            int preHash = getPrefixHash(0, i);
            int sufHash = getSuffixHash(0, i);

            // If the substring s[0]...s[i]
            // is not palindromic
            if (preHash != sufHash)
                continue;

            preHash = getPrefixHash(i + 1, n - 1);
            sufHash = getSuffixHash(i + 1, n - 1);


            // If the substring (i + 1, n - 1)
            // is not palindromic
            if (preHash != sufHash)
                continue;

            // If both are palindromic
            ans++;
        }
        return ans;
    }

    // Driver Code
    public static void main(String[] args) {
        String s = "aaaaa";

        int ans = numWays(s);

        System.out.println(ans);
    }
}

// Contributed by adityasha4x71

Python3

# Python Program to implement
# the above approach

# Modulo for rolling hash
MOD = 10**9 + 9

# Small prime for rolling hash
P = 37

# Maximum length of string
MAXN = 10**5 + 5

# Stores prefix hash
prefixHash = [0] * MAXN

# Stores suffix hash
suffixHash = [0] * MAXN

# Stores inverse modulo
# of P for prefix
inversePrefix = [0] * MAXN

# Stores inverse modulo
# of P for suffix
inverseSuffix = [0] * MAXN


def power(x, y, mod):
    # Function to compute
    # power under modulo
    if x == 0:
        return 0

    ans = 1
    while y > 0:
        if y & 1:
            ans = (ans * x) % mod

        x = (x * x) % mod
        y >>= 1
    return ans

# Precompute hashes for the
# given string
def preCompute(s):

    global prefixHash, suffixHash, inversePrefix, inverseSuffix, P, MOD

    x = 1
    for i in range(len(s)):

        # Calculate and store hash
        prefixHash[i] = ((ord(s[i]) - ord('a') + 1) * x) % MOD

        # Calculate prefix sum
        if i > 0:
            prefixHash[i] = (prefixHash[i] + prefixHash[i - 1]) % MOD

        # Compute inverse modulo
        # of P ^ i for division
        # using Fermat Little theorem
        inversePrefix[i] = power(x, MOD - 2, MOD)

        x = (x * P) % MOD

    x = 1

    # Calculate suffix hash
    for i in range(len(s) - 1, -1, -1):

        # Calculate and store hash
        suffixHash[i] = ((ord(s[i]) - ord('a') + 1) * x) % MOD

        if i < len(s) - 1:
            suffixHash[i] = (suffixHash[i] + suffixHash[i + 1]) % MOD

        # Compute inverse modulo
        # of P ^ i for division
        # using Fermat Little theorem
        inverseSuffix[i] = power(x, MOD - 2, MOD)

        x = (x * P) % MOD

# Function to return Prefix
# Hash of substring
def getPrefixHash(l, r):

    global prefixHash, inversePrefix, P, MOD

    # Calculate prefix hash
    # from l to r
    h = prefixHash[r] - (prefixHash[l - 1] if l > 0 else 0)
    h = (h + MOD) % MOD
    h = (h * inversePrefix[l]) % MOD

    return h

# Function to return Suffix
# Hash of substring
def getSuffixHash(l, r):

    global suffixHash, inverseSuffix, P, MOD

    # Calculate suffix hash
    # from l to r
    h = suffixHash[l] - (suffixHash[r + 1] if r < len(suffixHash) - 1 else 0)
    h = (h + MOD) % MOD
    h = (h * inverseSuffix[r]) % MOD

    return h

def numWays(s):

    global n, preHash, sufHash
    n = len(s)

    # Compute prefix and
    # suffix hashes
    preCompute(s)

    # Stores the number of
    # possible splits
    ans = 0
    for i in range(n - 1):

        preHash = getPrefixHash(0, i)
        sufHash = getSuffixHash(0, i)

        # If the substring s[0]...s[i]
        # is not palindromic
        if (preHash != sufHash):
            continue

        preHash = getPrefixHash(i + 1,
                                n - 1)
        sufHash = getSuffixHash(i + 1,
                                n - 1)

        # If the substring (i + 1, n - 1)
        # is not palindromic
        if (preHash != sufHash):
            continue

        # If both are palindromic
        ans += 1
    return ans

# Driver Code
s = "aaaaa"
ans = numWays(s)
print(ans)

C#

// C# program for the above approach
using System;

public class GFG
{
  // Modulo for rolling hash
  static readonly int MOD = 1_000_000_007;

  // Small prime for rolling hash
  static readonly int P = 37;

  // Maximum length of string
  static readonly int MAXN = 100_005;

  // Stores prefix hash
  static int[] prefixHash = new int[MAXN];

  // Stores suffix hash
  static int[] suffixHash = new int[MAXN];

  // Stores inverse modulo
  // of P for prefix
  static int[] inversePrefix = new int[MAXN];


  // Stores inverse modulo
  // of P for suffix
  static int[] inverseSuffix = new int[MAXN];

  static int n;

  // Function to compute
  // power under modulo
  static int power(int x, int y, int mod)
  {
    if (x == 0)
      return 0;

    int ans = 1;
    while (y > 0)
    {
      if ((y & 1) == 1)
        ans = (int)(((long)ans * x) % mod);

      x = (int)(((long)x * x) % mod);
      y >>= 1;
    }
    return ans;
  }

  // Precompute hashes for the
  // given string
  static void preCompute(string s)
  {
    int x = 1;
    for (int i = 0; i < n; i++)
    {
      prefixHash[i] = (int)(((long)(s[i] - 'a' + 1) * x) % MOD);

      // Compute inverse modulo
      // of P ^ i for division
      // using Fermat Little theorem
      if (i > 0)
        prefixHash[i] = (prefixHash[i] + prefixHash[i - 1]) % MOD;

      inversePrefix[i] = power(x, MOD - 2, MOD);

      x = (int)(((long)x * P) % MOD);
    }

    x = 1;

    // Calculate suffix hash
    for (int i = n - 1; i >= 0; i--)
    {
      // Calculate and store hash
      suffixHash[i] = (int)(((long)(s[i] - 'a' + 1) * x) % MOD);
      if (i < n - 1)
        suffixHash[i] = (suffixHash[i] + suffixHash[i + 1]) % MOD;

      inverseSuffix[i] = power(x, MOD - 2, MOD);

      x = (int)(((long)x * P) % MOD);
    }
  }


  // Function to return Prefix
  // Hash of substring
  static int getPrefixHash(int l, int r)
  {
    int h = prefixHash[r];
    if (l > 0)
      h = (h - prefixHash[l - 1] + MOD) % MOD;

    h = (int)(((long)h * inversePrefix[l]) % MOD);

    return h;
  }

  // Function to return Suffix
  // Hash of the substring
  static int getSuffixHash(int l, int r)
  {
    int h = suffixHash[l];
    if (r < n - 1)
      h = (h - suffixHash[r + 1] + MOD) % MOD;

    h = (int)(((long)h * inverseSuffix[r]) % MOD);

    return h;
  }

  static int numWays(string s)
  {
    n = s.Length;

    // Calculate suffix hash
    // from l to r
    preCompute(s);

    // Stores the number of
    // possible splits
    int ans = 0;
    for (int i = 0; i < n - 1; i++)
    {
      int preHash = getPrefixHash(0, i);
      int sufHash = getSuffixHash(0, i);

      // If the substring s[0]...s[i]
      // is not palindromic
      if (preHash != sufHash)
        continue;

      preHash = getPrefixHash(i + 1, n - 1);
      sufHash = getSuffixHash(i + 1, n - 1);


      // If the substring (i + 1, n - 1)
      // is not palindromic
      if (preHash != sufHash)
        continue;


      // If both are palindromic
      ans++;
    }
    return ans;
  }

  // Driver Code
  public static void Main(string[] args)
  {
    string s = "aaaaa";

    int ans = numWays(s);

    Console.WriteLine(ans);
  }
}


// This code is contributed by princekumaras

Javascript

// Modulo for rolling hash
const MOD = BigInt(10 ** 9 + 9);

// Small prime for rolling hash
const P = BigInt(37);

// Maximum length of string
const MAXN = 10 ** 5 + 5;

// Stores prefix hash
const prefixHash = new Array(MAXN).fill(0n);

// Stores suffix hash
const suffixHash = new Array(MAXN).fill(0n);

// Stores inverse modulo
// of P for prefix
const inversePrefix = new Array(MAXN).fill(0n);

// Stores inverse modulo
// of P for suffix
const inverseSuffix = new Array(MAXN).fill(0n);

function power(x, y, mod) {
  // Function to compute
  // power under modulo
  if (x == 0n) {
    return 0n;
  }

  let ans = 1n;
  while (y > 0) {
    if (y & 1n) {
      ans = (ans * x) % mod;
    }

    x = (x * x) % mod;
    y >>= 1n;
  }
  return ans;
}

// Precompute hashes for the
// given string
function preCompute(s) {
  let x = 1n;
  for (let i = 0; i < s.length; i++) {
    // Calculate and store hash
    prefixHash[i] = ((BigInt(s.charCodeAt(i) - "a".charCodeAt(0) + 1) * x) % MOD);

    // Calculate prefix sum
    if (i > 0) {
      prefixHash[i] = (prefixHash[i] + prefixHash[i - 1]) % MOD;
    }

    // Compute inverse modulo
    // of P ^ i for division
    // using Fermat Little theorem
    inversePrefix[i] = power(x, MOD - 2n, MOD);

    x = (x * P) % MOD;
  }

  x = 1n;

  // Calculate suffix hash
  for (let i = s.length - 1; i >= 0; i--) {
    // Calculate and store hash
    suffixHash[i] = ((BigInt(s.charCodeAt(i) - "a".charCodeAt(0) + 1) * x) % MOD);

    if (i < s.length - 1) {
      suffixHash[i] = (suffixHash[i] + suffixHash[i + 1]) % MOD;
    }

    // Compute inverse modulo
    // of P ^ i for division
    // using Fermat Little theorem
    inverseSuffix[i] = power(x, MOD - 2n, MOD);

    x = (x * P) % MOD;
  }
}

// Function to return Prefix
// Hash of substring
function getPrefixHash(l, r) {
  // Calculate prefix hash
  // from l to r
  let h = prefixHash[r] - (prefixHash[l - 1n] || 0n);
  h = (h + MOD) % MOD;
  h = (h * inversePrefix[l]) % MOD;

  return h;
}

// Function to return Suffix
// Hash of substring
function getSuffixHash(l, r) {
  // Calculate suffix hash
  // from l to r
  let h = suffixHash[l] - (suffixHash[r + 1] || 0n);
  h = (h + MOD) % MOD;
  h = (h * inverseSuffix[r]) % MOD;

  return h;
}

function numW

Time Complexity: O(N * log(10⁹)) 
Auxiliary Space: O(N)

Approach Name: Split String into Palindromes

Steps:

1. Define a function named count_palindrome_splits that takes a string S as input.
2. Initialize a variable count to 0.
3. Loop through each possible index i to split the string from 1 to len(S)-1.
4. Check if both the substrings formed by the split are palindromes.
5. If yes, increment count.
6. Return the count.

C++

#include <iostream>
#include <string>

using namespace std;

bool is_palindrome(string s)
{
    return s == string(s.rbegin(), s.rend());
}

int count_palindrome_splits(string S)
{
    int count = 0;
    for (int i = 1; i < S.length(); i++) {
        string left_substring = S.substr(0, i);
        string right_substring = S.substr(i);
        if (is_palindrome(left_substring)
            && is_palindrome(right_substring)) {
            count++;
        }
    }
    return count;
}

int main()
{
    string S = "aaaaa";
    cout << count_palindrome_splits(S) << endl; // Output: 4
    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 countPalindromeSplits(String S)
    {
        int count = 0;
        for (int i = 1; i < S.length(); i++) {
            String leftSubstring = S.substring(0, i);
            String rightSubstring = S.substring(i);
            if (isPalindrome(leftSubstring)
                && isPalindrome(rightSubstring)) {
                count++;
            }
        }
        return count;
    }

    public static void main(String[] args)
    {
        String S = "aaaaa";
        System.out.println(
            countPalindromeSplits(S)); // Output: 4
    }
}

Python3

def count_palindrome_splits(S):
    count = 0
    for i in range(1, len(S)):
        left_substring = S[:i]
        right_substring = S[i:]
        if is_palindrome(left_substring) and is_palindrome(right_substring):
            count += 1
    return count


def is_palindrome(s):
    return s == s[::-1]


# Example usage
S = "aaaaa"
print(count_palindrome_splits(S))  # Output: 4

C#

using System;

public class Program {
    public static int CountPalindromeSplits(string s)
    {
        int count = 0;
        for (int i = 1; i < s.Length; i++) {
            string leftSubstring = s.Substring(0, i);
            string rightSubstring = s.Substring(i);
            if (IsPalindrome(leftSubstring)
                && IsPalindrome(rightSubstring)) {
                count++;
            }
        }
        return count;
    }

    public static bool IsPalindrome(string s)
    {
        char[] charArray = s.ToCharArray();
        Array.Reverse(charArray);
        string reversedString = new string(charArray);
        return s == reversedString;
    }

    public static void Main()
    {
        string s = "aaaaa";
        Console.WriteLine(
            CountPalindromeSplits(s)); // Output: 4
    }
}

Javascript

//Funtion to check palindrome
function isPalindrome(s) {
  return s === s.split('').reverse().join('');
}

function countPalindromeSplits(S) {
  let count = 0;
  for (let i = 1; i < S.length; i++) {
  // finding the left and right substring
    let leftSubstring = S.substring(0, i);
    let rightSubstring = S.substring(i);
    if (isPalindrome(leftSubstring) && isPalindrome(rightSubstring)) {
      count++;
    }
  }
  return count;
}

// Main function
function main() {
  let S = "aaaaa";

  // Function call
  console.log(countPalindromeSplits(S)); // Output: 4
}

main();

Output

4

Time Complexity: O(n^2) where n is the length of the input string S.

Auxiliary Space: O(n) where n is the length of the input string S.