Introduction

In this blog post, we will explore how to solve the LeetCode problem "2430" 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

You are given a string s consisting of only lowercase English letters. In one operation, you can: Delete the entire string s, or Delete the first i letters of s if the first i letters of s are equal to the following i letters in s, for any i in the range 1 <= i <= s.length / 2. For example, if s = "ababc", then in one operation, you could delete the first two letters of s to get "abc", since the first two letters of s and the following two letters of s are both equal to "ab". Return the maximum number of operations needed to delete all of s. Example 1: Input: s = "abcabcdabc" Output: 2 Explanation: - Delete the first 3 letters ("abc") since the next 3 letters are equal. Now, s = "abcdabc". - Delete all the letters. We used 2 operations so return 2. It can be proven that 2 is the maximum number of operations needed. Note that in the second operation we cannot delete "abc" again because the next occurrence of "abc" does not happen in the next 3 letters. Example 2: Input: s = "aaabaab" Output: 4 Explanation: - Delete the first letter ("a") since the next letter is equal. Now, s = "aabaab". - Delete the first 3 letters ("aab") since the next 3 letters are equal. Now, s = "aab". - Delete the first letter ("a") since the next letter is equal. Now, s = "ab". - Delete all the letters. We used 4 operations so return 4. It can be proven that 4 is the maximum number of operations needed. Example 3: Input: s = "aaaaa" Output: 5 Explanation: In each operation, we can delete the first letter of s. Constraints: 1 <= s.length <= 4000 s consists only of lowercase English letters.

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: The core idea is to use dynamic programming to store the maximum number of operations needed to delete substrings of s. dp[i] represents the maximum operations to delete s[i:].

• Base Case & Transitions: The base case is dp[n] = 1 (deleting an empty string takes 1 operation). The transition involves considering deleting the entire substring or deleting a prefix that matches the following prefix.

• Iterative Calculation: The dp array is filled iteratively from right to left, exploring all possible prefix deletions at each position.

• Runtime & Storage Complexity: O(n^2) runtime, O(n) storage.

Code

    def solve():
    s = input()
    n = len(s)
    dp = [0] * (n + 1)
    dp[n] = 1

    for i in range(n - 1, -1, -1):
        dp[i] = dp[i + 1] + 1  # Option 1: Delete the remaining string

        for j in range(1, (n - i) // 2 + 1):
            if s[i:i+j] == s[i+j:i+2*j]:
                dp[i] = max(dp[i], dp[i + j] + 1)

    print(dp[0])

solve()