将给定字符串拆分为两个非空回文的方法数量

给定字符串S ，任务是找到将给定字符串S拆分为两个非空回文字符串。
例子：

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”.

天真的方法：天真的方法是在每个可能的索引处分割字符串，并检查两个子字符串是否都是回文。如果是，则增加该索引的计数。打印最终计数。
下面是上述方法的实现：

C++

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

 
    // 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));
 

输出：

4

时间复杂度： O(N ² )
高效的方法：可以使用哈希和Rabin-Karp算法来优化上述方法，以存储字符串的前缀和后缀哈希。请按照以下步骤解决问题：

• 计算给定字符串的前缀和后缀哈希。
• 对于[1，N – 1]范围内的每个索引i，检查两个子字符串[0，i – 1]和[i，N – 1]是否是回文。
• 要检查子字符串[l，r]是否是回文，只需检查：

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

• 对于发现有两个子字符串回文的每个索引i，增加计数。
• 打印count的最终值。

下面是上述方法的实现：

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;
}

输出：

4

时间复杂度： O(N * log(10 ⁹ ))
空间复杂度： O(N)