Introduction

In this blog post, we will explore how to solve the LeetCode problem "2259" 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 number representing a positive integer and a character digit. Return the resulting string after removing exactly one occurrence of digit from number such that the value of the resulting string in decimal form is maximized. The test cases are generated such that digit occurs at least once in number. Example 1: Input: number = "123", digit = "3" Output: "12" Explanation: There is only one '3' in "123". After removing '3', the result is "12". Example 2: Input: number = "1231", digit = "1" Output: "231" Explanation: We can remove the first '1' to get "231" or remove the second '1' to get "123". Since 231 > 123, we return "231". Example 3: Input: number = "551", digit = "5" Output: "51" Explanation: We can remove either the first or second '5' from "551". Both result in the string "51". Constraints: 2 <= number.length <= 100 number consists of digits from '1' to '9'. digit is a digit from '1' to '9'. digit occurs at least once in number.

Explanation

Here's the solution:

• Greedy Approach: Iterate through the number string. If we find the digit to remove, we check if removing it results in a larger number. We remove the first such digit that satisfies this condition.

• Last Occurrence Optimization: If we reach the end of the string without finding a digit whose removal results in a larger number, it means the digit's removal leads to a smaller number irrespective of its position. In this scenario, we remove the last occurrence of the digit. This guarantees we maximize the result.

• Time and Space Complexity:

  • Runtime: O(n), where n is the length of the number string.
  • Storage: O(n), for storing the resulting string (at most).

Code

    def removeDigit(number: str, digit: str) -> str:
    """
    Removes exactly one occurrence of digit from number such that the value of
    the resulting string in decimal form is maximized.

    Args:
        number: The input string representing a positive integer.
        digit: The digit to remove.

    Returns:
        The resulting string after removing exactly one occurrence of digit
        such that the value is maximized.
    """
    n = len(number)
    last_occurrence = -1  # Store index of the last occurrence of digit
    result = ""

    for i in range(n):
        if number[i] == digit:
            last_occurrence = i
            if i < n - 1 and number[i] < number[i + 1]:
                result = number[:i] + number[i+1:]
                return result

    result = number[:last_occurrence] + number[last_occurrence + 1:]
    return result