To solve the problem of finding the number of ways to partition a string such that every substring of the partition is a palindrome, we can use dynamic programming (DP) with two key tables: one to check if a substring is a palindrome and another to count the number of valid partitions.

Approach

1. Palindrome Check Table:

   • Create a 2D boolean table is_pal where is_pal[i][j] is True if the substring s[i..j] (inclusive) is a palindrome.
   • Fill this table for substrings of length 1 (all are palindromes), length 2 (check if both characters are the same), and lengths greater than 2 (check if the first and last characters are the same and the substring between them is a palindrome).

2. Partition Count Table:

   • Create a DP array dp where dp[i] is the number of ways to partition the substring s[0..i-1].
   • dp[0] = 1 (empty string has one way to partition).
   • For each i (end of substring), sum dp[k] for all k < i where s[k..i-1] is a palindrome (using the is_pal table).

Solution Code

def count_palindromic_partitions(s):
    n = len(s)
    if n == 0:
        return 0

    # Step 1: Precompute palindrome table
    is_pal = [[False] * n for _ in range(n)]

    # Substrings of length 1
    for i in range(n):
        is_pal[i][i] = True

    # Substrings of length 2
    for i in range(n - 1):
        is_pal[i][i + 1] = (s[i] == s[i + 1])

    # Substrings of length >=3
    for length in range(3, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            is_pal[i][j] = (s[i] == s[j]) and is_pal[i + 1][j - 1]

    # Step 2: Compute number of partitions using DP
    dp = [0] * (n + 1)
    dp[0] = 1  # Base case: empty string

    for i in range(1, n + 1):
        for k in range(i):
            if is_pal[k][i - 1]:
                dp[i] += dp[k]

    return dp[n]

Explanation

• Palindrome Table: This table helps quickly check if any substring is a palindrome, which is essential for the DP step.
• DP Array: The DP array builds up the solution by considering all possible partitions ending at each position. For each position i, we look back to all previous positions k where the substring from k to i-1 is a palindrome, and add the number of partitions up to k to the count for i.

This approach efficiently computes the result in O(n²) time and space complexity, which is feasible for most practical string lengths.

Example: For s = "aab", the result is 2 (partitions: ["a", "a", "b"] and ["aa", "b"]). For s = "aaa", the result is 4 (partitions: ["a", "a", "a"], ["a", "aa"], ["aa", "a"], ["aaa"]).

This solution should handle all valid input cases correctly. If the original problem had different constraints or requirements, the approach might need adjustments, but this is the standard solution for the palindromic partition count problem.

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