Naive Solution

The most straightforward approach is to iterate through the string number, identify all occurrences of the digit digit, and for each occurrence, create a new string by removing that digit. Then, compare all the resulting strings and return the lexicographically largest one.

Code:

def remove_digit_naive(number: str, digit: str) -> str:
    results = []
    for i in range(len(number)):
        if number[i] == digit:
            results.append(number[:i] + number[i+1:])
    return max(results)

Time Complexity: O(n^2)

• Iterating through the string takes O(n) time.
• Creating a new string by slicing takes O(n) time.
• Finding the maximum among the generated strings takes O(n*n) in the worst case.

Space Complexity: O(n)

• Storing the resulting strings takes O(n) space in the worst case where the digit appears multiple times.

Optimal Solution

A more efficient approach is to iterate through the string number and remove the digit such that the remaining string is maximized. This can be achieved by removing the first occurrence of digit where the next digit is greater than the current digit. If no such digit is found, remove the last occurrence of digit.

Code:

def remove_digit_optimal(number: str, digit: str) -> str:
    last_occurrence = -1
    for i in range(len(number)):
        if number[i] == digit:
            last_occurrence = i
            if i < len(number) - 1 and number[i] < number[i + 1]:
                return number[:i] + number[i + 1:]

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

Time Complexity: O(n)

• Iterating through the string takes O(n) time.
• String slicing takes O(n) time.

Space Complexity: O(n)

• O(n) due to string slicing to create new strings.

Edge Cases

1. Empty String: The problem statement says that the number is always a positive integer so we can ignore this case.
2. Single Digit: The problem statement says that the length is between 2 and 100, so it's not a concern.
3. The digit occurs at the end of the string: The logic handles it correctly by the last_occurrence variable.
4. The digit only occurs once: The logic handles it correctly. The last_occurrence variable ensures this works.
5. All digits are the same: The logic handles it by the last_occurrence variable.

Summary

The optimal solution provides a more efficient way to solve the problem by iterating through the string once and making a decision on the fly. The naive solution, while simpler to understand, has a higher time complexity because it involves creating multiple strings and comparing them.