LeetCode #2733 — EASY

Neither Minimum nor Maximum

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

Given an integer array nums containing distinct positive integers, find and return any number from the array that is neither the minimum nor the maximum value in the array, or -1 if there is no such number.

Return the selected integer.

Example 1:

Input: nums = [3,2,1,4]
Output: 2
Explanation: In this example, the minimum value is 1 and the maximum value is 4. Therefore, either 2 or 3 can be valid answers.

Example 2:

Input: nums = [1,2]
Output: -1
Explanation: Since there is no number in nums that is neither the maximum nor the minimum, we cannot select a number that satisfies the given condition. Therefore, there is no answer.

Example 3:

Input: nums = [2,1,3]
Output: 2
Explanation: Since 2 is neither the maximum nor the minimum value in nums, it is the only valid answer.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100
All values in nums are distinct

Step 01

Brute Force Baseline

Problem summary: Given an integer array nums containing distinct positive integers, find and return any number from the array that is neither the minimum nor the maximum value in the array, or -1 if there is no such number. Return the selected integer.

Baseline thinking

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

Pattern signal: Array

Example 1

[3,2,1,4]

Example 2

[1,2]

Example 3

[2,1,3]

Core Insight

What unlocks the optimal approach

Find any value in the array that is not the minimum or the maximum value.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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 #2733: Neither Minimum nor Maximum
class Solution {
    public int findNonMinOrMax(int[] nums) {
        int mi = 100, mx = 0;
        for (int x : nums) {
            mi = Math.min(mi, x);
            mx = Math.max(mx, x);
        }
        for (int x : nums) {
            if (x != mi && x != mx) {
                return x;
            }
        }
        return -1;
    }
}

// Accepted solution for LeetCode #2733: Neither Minimum nor Maximum
func findNonMinOrMax(nums []int) int {
	mi, mx := slices.Min(nums), slices.Max(nums)
	for _, x := range nums {
		if x != mi && x != mx {
			return x
		}
	}
	return -1
}

# Accepted solution for LeetCode #2733: Neither Minimum nor Maximum
class Solution:
    def findNonMinOrMax(self, nums: List[int]) -> int:
        mi, mx = min(nums), max(nums)
        return next((x for x in nums if x != mi and x != mx), -1)

// Accepted solution for LeetCode #2733: Neither Minimum nor Maximum
impl Solution {
    pub fn find_non_min_or_max(nums: Vec<i32>) -> i32 {
        let mut mi = 100;
        let mut mx = 0;

        for &ele in nums.iter() {
            if ele < mi {
                mi = ele;
            }
            if ele > mx {
                mx = ele;
            }
        }

        for &ele in nums.iter() {
            if ele != mi && ele != mx {
                return ele;
            }
        }

        -1
    }
}

// Accepted solution for LeetCode #2733: Neither Minimum nor Maximum
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2733: Neither Minimum nor Maximum
// class Solution {
//     public int findNonMinOrMax(int[] nums) {
//         int mi = 100, mx = 0;
//         for (int x : nums) {
//             mi = Math.min(mi, x);
//             mx = Math.max(mx, x);
//         }
//         for (int x : nums) {
//             if (x != mi && x != mx) {
//                 return x;
//             }
//         }
//         return -1;
//     }
// }

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.