Leetcode #2091: Removing Minimum and Maximum From Array
In this guide, we solve Leetcode #2091 Removing Minimum and Maximum From Array in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Problem Statement
You are given a 0-indexed array of distinct integers nums. There is an element in nums that has the lowest value and an element that has the highest value.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Array
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: nums = [2,10,7,5,4,1,8,6]
Output: 5
Explanation:
The minimum element in the array is nums[5], which is 1.
The maximum element in the array is nums[1], which is 10.
We can remove both the minimum and maximum by removing 2 elements from the front and 3 elements from the back.
This results in 2 + 3 = 5 deletions, which is the minimum number possible.
Python Solution
class Solution:
def minimumDeletions(self, nums: List[int]) -> int:
mi = mx = 0
for i, num in enumerate(nums):
if num < nums[mi]:
mi = i
if num > nums[mx]:
mx = i
if mi > mx:
mi, mx = mx, mi
return min(mx + 1, len(nums) - mi, mi + 1 + len(nums) - mx)
Complexity
The time complexity is O(n log n). The space complexity is O(1) to O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.