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

Difficulty Level : Medium
Last Updated : 19 Mar, 2021

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:

Time Complexity: O(N²)
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;
}
 
// Precompte 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;
}

Output:

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes