Solution Explanation for Palindrome Partitioning

This problem asks to find all possible ways to partition a given string s such that each substring in the partition is a palindrome. The solutions presented use a backtracking approach, optimized with dynamic programming to efficiently check for palindromes.

Approach:

1. Palindrome Check (Dynamic Programming): The core idea is to precompute a boolean table f where f[i][j] is true if the substring s[i:j+1] is a palindrome, and false otherwise. This is done efficiently using dynamic programming. The base cases are single characters (which are always palindromes), and the recursive relation is: f[i][j] = (s[i] == s[j]) and f[i+1][j-1]. This means a substring is a palindrome if its first and last characters are the same, and the inner substring is also a palindrome.

2. Backtracking: After precomputing the f table, a backtracking function dfs explores all possible partitions. It starts at index 0 of the string. At each index i, it iterates through all possible ending indices j (such that j >= i). If f[i][j] is true (meaning s[i:j+1] is a palindrome), it adds this substring to the current partition (t), recursively calls dfs to explore partitions starting from j+1, and then backtracks by removing the substring from t. This ensures that all possible combinations are explored.

Time and Space Complexity Analysis:

• Time Complexity: The dynamic programming part takes O(n²) time to fill the f table, where n is the length of the string. The backtracking part has a time complexity that's exponential in the worst case (e.g., string "aaaaaaaaa"), because it explores all possible partitions. However, it's bounded by the number of possible partitions, which is still exponential in the worst case. Overall, the dominant factor is the backtracking part, making the time complexity approximately O(2n) (exponential) in the worst case.

• Space Complexity: The dynamic programming part requires O(n²) space for the f table. The backtracking part uses the recursive call stack, whose depth is at most n. Additionally, the ans list to store the results can also grow to a size determined by the number of possible partitions (also exponential in the worst case). Thus the space complexity is O(n² + 2n), dominated by the space for storing all the partition results in the worst case.

Code Explanation (Python as example, other languages are similar):

class Solution:
    def partition(self, s: str) -> List[List[str]]:
        n = len(s)
        # DP table for palindrome checks
        f = [[True] * n for _ in range(n)]
        for i in range(n - 1, -1, -1):
            for j in range(i + 1, n):
                f[i][j] = s[i] == s[j] and f[i + 1][j - 1]
 
        ans = []  # List to store all partitions
        t = []     # Current partition
 
        def dfs(i):
            if i == n:
                ans.append(t[:])  # Add a copy of the current partition
                return
 
            for j in range(i, n):
                if f[i][j]:  # If substring is a palindrome
                    t.append(s[i : j + 1])  # Add to current partition
                    dfs(j + 1)              # Explore from next index
                    t.pop()                  # Backtrack: Remove the substring
 
        dfs(0)  # Start backtracking from index 0
        return ans
 

The code first performs dynamic programming to efficiently check for palindromes. Then the backtracking function dfs explores all combinations, adding palindromic substrings to the current partition and recursively calling itself until the end of the string is reached. Backtracking is used to explore other possibilities. The final list ans contains all possible palindrome partitions.