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 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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Output:

4

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Output:

4

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.




My Personal Notes arrow_drop_up

Recommended Posts: