Solution Explanation: Palindrome Partitioning IV

This problem asks whether a given string s can be partitioned into three non-empty palindromic substrings. The solution uses dynamic programming to efficiently determine if substrings are palindromes.

Approach:

1. Palindrome Check: The core idea is to efficiently determine if any substring of s is a palindrome. We use a boolean matrix g where g[i][j] is true if the substring s[i...j] (inclusive) is a palindrome, and false otherwise. This is filled using dynamic programming:

   • The base case is that single characters (i == j) are always palindromes (g[i][i] = true).
   • For substrings of length 2 or greater, g[i][j] is true if and only if:
     • s[i] == s[j] (the first and last characters are equal), and
     • g[i+1][j-1] is true (the inner substring is also a palindrome).

2. Three-Partition Check: Once the g matrix is built, we iterate through all possible partitions of s into three substrings. For each partition at indices i and j, we check if all three substrings s[0...i], s[i+1...j], and s[j+1...n-1] are palindromes using the g matrix. If we find such a partition, we immediately return true.

3. Result: If no such three-palindrome partition is found after checking all possibilities, we return false.

Time Complexity Analysis:

• Building the g matrix takes O(n^2) time, where n is the length of the string. This is because we iterate over all possible substring starting and ending indices.
• Checking the three-partition possibilities takes O(n^2) time in the worst case because we have a nested loop iterating over possible i and j values.

Therefore, the overall time complexity is O(n^2).

Space Complexity Analysis:

The space complexity is dominated by the g matrix, which has dimensions n x n. Hence, the space complexity is O(n^2).

Code Explanation (Python):

class Solution:
    def checkPartitioning(self, s: str) -> bool:
        n = len(s)
        g = [[True] * n for _ in range(n)]  # Initialize g matrix
 
        # Build the g matrix using dynamic programming
        for i in range(n - 1, -1, -1):
            for j in range(i + 1, n):
                g[i][j] = s[i] == s[j] and (i + 1 == j or g[i + 1][j - 1])
 
        # Check for three-palindrome partitions
        for i in range(n - 2):
            for j in range(i + 1, n - 1):
                if g[0][i] and g[i + 1][j] and g[j + 1][-1]:  # Check if all three substrings are palindromes
                    return True
 
        return False  # No three-palindrome partition found
 

The code in other languages (Java, C++, Go) follows the same algorithm and has the same time and space complexities. The only difference is the syntax and data structure usage specific to each language.