Introduction

In this blog post, we will explore how to solve the LeetCode problem "1910" 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 two strings s and part, perform the following operation on s until all occurrences of the substring part are removed: Find the leftmost occurrence of the substring part and remove it from s. Return s after removing all occurrences of part. A substring is a contiguous sequence of characters in a string. Example 1: Input: s = "daabcbaabcbc", part = "abc" Output: "dab" Explanation: The following operations are done: - s = "daabcbaabcbc", remove "abc" starting at index 2, so s = "dabaabcbc". - s = "dabaabcbc", remove "abc" starting at index 4, so s = "dababc". - s = "dababc", remove "abc" starting at index 3, so s = "dab". Now s has no occurrences of "abc". Example 2: Input: s = "axxxxyyyyb", part = "xy" Output: "ab" Explanation: The following operations are done: - s = "axxxxyyyyb", remove "xy" starting at index 4 so s = "axxxyyyb". - s = "axxxyyyb", remove "xy" starting at index 3 so s = "axxyyb". - s = "axxyyb", remove "xy" starting at index 2 so s = "axyb". - s = "axyb", remove "xy" starting at index 1 so s = "ab". Now s has no occurrences of "xy". Constraints: 1 <= s.length <= 1000 1 <= part.length <= 1000 s​​​​​​ and part consists of lowercase English letters.

Explanation

Here's the breakdown of the solution:

• Iterative Removal: Repeatedly search for the part substring within s and remove the leftmost occurrence until no more occurrences are found.
• String Manipulation: Use string slicing for efficient removal of the substring.

• While Loop: Use a while loop to continue the search and removal process until part is no longer found in s.

• Runtime Complexity: O(n*m), where n is the length of s and m is the length of part.

• Storage Complexity: O(1) (excluding the space for the final string, which is unavoidable)

Code

    def removeOccurrences(s: str, part: str) -> str:
    while part in s:
        index = s.find(part)
        if index != -1:
            s = s[:index] + s[index + len(part):]
        else:
            break
    return s