How do you solve Removing Minimum and Maximum From Array on LeetCode?

Removing Minimum and Maximum From Array (LeetCode #2091) 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: Understanding the positions of the minimum and maximum elements is crucial for determining the number of deletions.

What is the time complexity of Removing Minimum and Maximum From Array?

The optimal solution for Removing Minimum and Maximum From Array runs in O(n) time and uses O(1) space. This complexity is linear because we only traverse the array once to find the indices of the minimum and maximum elements.

What is the brute force solution for Removing Minimum and Maximum From Array?

The brute force approach for Removing Minimum and Maximum From Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible way to remove the minimum and maximum elements from the array. This means calculating the number of deletions required for each possible combination of removing elements from the front and back. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Removing Minimum and Maximum From Array?

Removing Minimum and Maximum From Array uses the following concepts: Array, Greedy. The recommended patterns to study are: Array, Greedy.

What are the interview tips for Removing Minimum and Maximum From Array?

Always clarify the problem statement and constraints before jumping into coding. Discuss your thought process and approach with the interviewer to ensure you are on the right track. Consider edge cases, such as arrays with only one element, and explain how your solution handles them.

What are common mistakes when solving Removing Minimum and Maximum From Array?

Not considering all three scenarios for deletions. Confusing the indices of minimum and maximum elements, leading to incorrect calculations.

#2091

Removing Minimum and Maximum From Array

Medium

ArrayGreedyArrayGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the fact that we only need to consider the positions of the minimum and maximum elements and calculate the deletions required in a single pass, leading to a more efficient solution.

⚙️

Algorithm

3 steps

1Step 1: Find the indices of the minimum and maximum elements in the array.
2Step 2: Calculate the number of deletions required for three scenarios: removing both from the front, both from the back, and one from each end.
3Step 3: Return the minimum deletions from the calculated scenarios.

solution.py12 lines

1# Full working Python code
2
3def min_deletions(nums):
4    min_index = nums.index(min(nums))
5    max_index = nums.index(max(nums))
6    front_only = max(min_index, max_index) + 1
7    back_only = len(nums) - min(min_index, max_index)
8    both_ends = min_index + 1 + (len(nums) - max_index)
9    return min(front_only, back_only, both_ends)
10
11# Example usage
12print(min_deletions([2, 10, 7, 5, 4, 1, 8, 6]))  # Output: 5

ℹ

Complexity note: This complexity is linear because we only traverse the array once to find the indices of the minimum and maximum elements.

1Understanding the positions of the minimum and maximum elements is crucial for determining the number of deletions.
2Only three scenarios need to be considered for deletions: from the front, from the back, or a combination of both.

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