# Solving Leetcode Interviews in Seconds with AI: Maximum Deletions on a String


	# 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](https://www.chatmagic.app), 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
	```python
	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()
	```