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”.

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

Naive Approach: The naive approach is to split the string at each possible index and check if both the subtrings 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 

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 <bits/stdc++.h> 
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<int> prefixHash(MAXN); 
  
// Stores suffix hash 
vector<int> suffixHash(MAXN); 
  
// Stores inverse modulo 
// of P for prefix 
vector<int> inversePrefix(MAXN); 
  
// Stores inverse modulo 
// of P  for suffix 
vector<int> 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 < n; i++) { 
  
        // Calculate Prefix Hash 
        prefixHash[i] = (1LL 
                         * int(s[i] 
                               - 'a' + 1) 
                         * x) 
                        % MOD; 
  
        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 = (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 < n - 1) 
            suffixHash[i] 
                = (suffixHash[i] 
                   + suffixHash[i + 1]) 
                  % MOD; 
  
        // computingCompute inverse modulo 
        // of P ^ i for division 
        // using Fermat Little theorem 
        inverseSuffix[i] = power(x, MOD - 2, 
                                 MOD); 
  
        x = (1LL * x * P) % MOD; 
    } 
} 
  
// Function to return Prefix 
// Hash of substring 
int getPrefixHash(int l, int r) 
{ 
    // Calculate Prefix Hash 
    // from l to r 
    int h = prefixHash[r] 
            - (l > 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)

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My Personal Notes

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

C++

Java

C++

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