How do you solve Remove Digit From Number to Maximize Result on LeetCode?

Remove Digit From Number to Maximize Result (LeetCode #2259) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Removing a digit can lead to different results based on its position, especially if there are multiple occurrences.

What is the brute force solution for Remove Digit From Number to Maximize Result?

The brute force approach for Remove Digit From Number to Maximize Result is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach involves generating all possible strings by removing each occurrence of the specified digit and then comparing them to find the maximum. It's straightforward but inefficient as it checks every possibility. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Remove Digit From Number to Maximize Result?

Remove Digit From Number to Maximize Result uses the following concepts: String, Greedy, Enumeration. The recommended patterns to study are: Greedy, Enumeration.

What are the interview tips for Remove Digit From Number to Maximize Result?

Think about edge cases, such as when the digit appears multiple times. Clarify whether the input will always contain the digit at least once. Discuss your thought process and the trade-offs of different approaches during the interview.

What are common mistakes when solving Remove Digit From Number to Maximize Result?

Not considering all occurrences of the digit properly. Failing to compare results correctly to find the maximum.

#2259

Remove Digit From Number to Maximize Result

Easy

StringGreedyEnumerationGreedyEnumeration

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution iterates through the string once, checking each occurrence of the digit and deciding whether removing it yields a larger number. This avoids unnecessary string creations and comparisons.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable to store the best result as an empty string.
2Step 2: Iterate through each character in the string number.
3Step 3: When the character matches the digit, create a new string without that character.
4Step 4: Compare this new string with the current best result and update if it's larger.
5Step 5: Return the best result after the loop.

solution.py8 lines

1def removeDigit(number: str, digit: str) -> str:
2    best_result = ''
3    for i in range(len(number)):
4        if number[i] == digit:
5            new_number = number[:i] + number[i+1:]
6            if new_number > best_result:
7                best_result = new_number
8    return best_result

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once. The space complexity is O(1) since we only store a few variables, not dependent on the input size.

1Removing a digit can lead to different results based on its position, especially if there are multiple occurrences.
2The best result is determined by comparing all possible outcomes, but an optimal approach minimizes unnecessary comparisons.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.