Introduction

In this blog post, we will explore how to solve the LeetCode problem "1745" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given a string s, return true if it is possible to split the string s into three non-empty palindromic substrings. Otherwise, return false.​​​​​ A string is said to be palindrome if it the same string when reversed. Example 1: Input: s = "abcbdd" Output: true Explanation: "abcbdd" = "a" + "bcb" + "dd", and all three substrings are palindromes. Example 2: Input: s = "bcbddxy" Output: false Explanation: s cannot be split into 3 palindromes. Constraints: 3 <= s.length <= 2000 s​​​​​​ consists only of lowercase English letters.

Explanation

Here's the solution to the problem, along with explanations:

• Approach: The core idea is to iterate through all possible split points in the string s. For each pair of split points, we check if the three resulting substrings are palindromes. We use a helper function is_palindrome to efficiently determine if a substring is a palindrome.
• Optimization: The solution is optimized by directly checking palindromes within the loop without storing intermediate results. It terminates early as soon as a valid split is found.
• Complexity:
  • Runtime Complexity: O(n^3), where n is the length of the string s (due to the nested loops and palindrome checks)
  • Storage Complexity: O(1) (constant extra space)

Code

    def is_palindrome(s):
    """Checks if a string is a palindrome."""
    return s == s[::-1]

def split_into_three_palindromes(s):
    """
    Given a string s, return true if it is possible to split the string s into three non-empty palindromic substrings.
    Otherwise, return false.
    """
    n = len(s)
    for i in range(1, n - 1):
        for j in range(i + 1, n):
            s1 = s[:i]
            s2 = s[i:j]
            s3 = s[j:]

            if is_palindrome(s1) and is_palindrome(s2) and is_palindrome(s3):
                return True

    return False