LeetCode #2091 — MEDIUM

Removing Minimum and Maximum From Array

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
The Problem

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. We call them the minimum and maximum respectively. Your goal is to remove both these elements from the array.

A deletion is defined as either removing an element from the front of the array or removing an element from the back of the array.

Return the minimum number of deletions it would take to remove both the minimum and maximum element from the array.

Example 1:

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.

Example 2:

Input: nums = [0,-4,19,1,8,-2,-3,5]
Output: 3
Explanation: 
The minimum element in the array is nums[1], which is -4.
The maximum element in the array is nums[2], which is 19.
We can remove both the minimum and maximum by removing 3 elements from the front.
This results in only 3 deletions, which is the minimum number possible.

Example 3:

Input: nums = [101]
Output: 1
Explanation:  
There is only one element in the array, which makes it both the minimum and maximum element.
We can remove it with 1 deletion.

Constraints:

  • 1 <= nums.length <= 105
  • -105 <= nums[i] <= 105
  • The integers in nums are distinct.
Patterns Used

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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. We call them the minimum and maximum respectively. Your goal is to remove both these elements from the array. A deletion is defined as either removing an element from the front of the array or removing an element from the back of the array. Return the minimum number of deletions it would take to remove both the minimum and maximum element from the array.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[2,10,7,5,4,1,8,6]

Example 2

[0,-4,19,1,8,-2,-3,5]

Example 3

[101]

Related Problems

  • Maximum Points You Can Obtain from Cards (maximum-points-you-can-obtain-from-cards)
  • Minimum Deletions to Make Character Frequencies Unique (minimum-deletions-to-make-character-frequencies-unique)
Step 02

Core Insight

What unlocks the optimal approach

  • There can only be three scenarios for deletions such that both minimum and maximum elements are removed:
  • Scenario 1: Both elements are removed by only deleting from the front.
  • Scenario 2: Both elements are removed by only deleting from the back.
  • Scenario 3: Delete from the front to remove one of the elements, and delete from the back to remove the other element.
  • Compare which of the three scenarios results in the minimum number of moves.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2091: Removing Minimum and Maximum From Array
class Solution {
    public int minimumDeletions(int[] nums) {
        int mi = 0, mx = 0, n = nums.length;
        for (int i = 0; i < n; ++i) {
            if (nums[i] < nums[mi]) {
                mi = i;
            }
            if (nums[i] > nums[mx]) {
                mx = i;
            }
        }
        if (mi > mx) {
            int t = mx;
            mx = mi;
            mi = t;
        }
        return Math.min(Math.min(mx + 1, n - mi), mi + 1 + n - mx);
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n log n)
Space
O(1)

Approach Breakdown

EXHAUSTIVE
O(2ⁿ) time
O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY
O(n log n) time
O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.