# Solving Leetcode Interviews in Seconds with AI: Apply Operations to Make String Empty


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3039" 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. Consider performing the following operation until s becomes empty:  For every alphabet character from 'a' to 'z', remove the first occurrence of that character in s (if it exists).  For example, let initially s = "aabcbbca". We do the following operations:  Remove the underlined characters s = "aabcbbca". The resulting string is s = "abbca". Remove the underlined characters s = "abbca". The resulting string is s = "ba". Remove the underlined characters s = "ba". The resulting string is s = "".  Return the value of the string s right before applying the last operation. In the example above, answer is "ba".   Example 1:  Input: s = "aabcbbca" Output: "ba" Explanation: Explained in the statement.  Example 2:  Input: s = "abcd" Output: "abcd" Explanation: We do the following operation: - Remove the underlined characters s = "abcd". The resulting string is s = "". The string just before the last operation is "abcd".    Constraints:  1 <= s.length <= 5 * 105 s consists only of lowercase English letters.  

	# Explanation
	Here's an efficient solution to the problem:

*   **Iterate and Mark:**  Iterate through the input string `s`. For each character, record its first occurrence index in a separate array (or dictionary).
*   **Simulate Rounds:** Simulate the removal rounds. In each round, iterate through the alphabet ('a' to 'z'). If a character's first occurrence index exists and is still valid (not already removed in a prior round), mark it for removal.
*   **Build Result:**  After each round, construct a new string by excluding characters marked for removal. The loop continues until the string becomes empty. The string from the previous round is the result.

*   **Runtime Complexity:** O(N), where N is the length of the input string. The string is iterated a limited number of times that depends on the number of alphabet characters, i.e., 26 times in the worst case.
*   **Storage Complexity:** O(N) to store the first occurrence indices and temporarily store strings.

	
	# Code
	```python
	def solve():
    s = input()
    n = len(s)
    first_occurrence = {}
    for i in range(n):
        if s[i] not in first_occurrence:
            first_occurrence[s[i]] = i
    
    current_string = s
    previous_string = ""
    
    while current_string:
        previous_string = current_string
        removed_indices = set()
        
        for char_code in range(ord('a'), ord('z') + 1):
            char = chr(char_code)
            if char in first_occurrence and first_occurrence[char] < len(current_string):
                
                first_index = current_string.find(char)
                if first_index != -1:
                  removed_indices.add(first_index)

        new_string = ""
        for i in range(len(current_string)):
            if i not in removed_indices:
                new_string += current_string[i]

        current_string = new_string
        
    return previous_string
print(solve())
	```