Neither Minimum nor Maximum — Easy Problem & Solution

Problem statement

nums holds distinct positive integers. Return the smallest number in it that is neither the minimum nor the maximum of the array.

If no such number exists, return -1.

Example 1

Input: nums = [3,2,1,4]
Output: 2
Explanation: The minimum is 1 and the maximum 4; the smallest of what is left is 2.

Example 2

Input: nums = [1,2]
Output: -1
Explanation: Every element is either the minimum or the maximum.

Example 3

Input: nums = [2,1,3]
Output: 2

Constraints

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100
• All values of nums are distinct.
• Among the qualifying numbers, the smallest is returned.

How to solve Neither Minimum nor Maximum

Find the extremes, then scan again for the smallest value that is neither.

Approach

1. Return -1 immediately when the array is shorter than three.
2. Find the minimum and maximum.
3. Scan for the smallest value different from both.

Why it works

Distinctness is what makes the two-pass approach exact: with repeats, a value equal to the minimum might still be a different element, and the != test would wrongly exclude it. Sorting and reading index 1 is the classic one-liner, but the two passes are linear.

Complexity

• Time — O(n)
• Space — O(1)

Pitfalls

• Arrays of length 1 or 2 always answer -1.
• Returning the second element of the unsorted array is not the same as the second smallest.
• The problem asks for a value, not an index.

Reference solution

Python

from typing import List

def findNonMinOrMax(nums: List[int]) -> int:
    if len(nums) < 3:
        return -1
    mn, mx = min(nums), max(nums)
    rest = [v for v in nums if v != mn and v != mx]
    return min(rest) if rest else -1

JavaScript

var findNonMinOrMax = function(nums) {
    if (nums.length < 3) return -1;
    var mn = nums[0], mx = nums[0], i;
    for (i = 1; i < nums.length; i++) {
        if (nums[i] < mn) mn = nums[i];
        if (nums[i] > mx) mx = nums[i];
    }
    var best = -1;
    for (i = 0; i < nums.length; i++) {
        if (nums[i] === mn || nums[i] === mx) continue;
        if (best < 0 || nums[i] < best) best = nums[i];
    }
    return best;
};

Also on the editorial tab: C, C#, C++, Go, Java, Kotlin, PHP, Ruby, Rust, Swift, TypeScript.

More arrays problems

• Special Array I Easy
• Type of Triangle Easy
• Find the Losers of the Circular Game Easy
• Form Smallest Number From Two Digit Arrays Easy
• Delete Greatest Value in Each Row Easy
• The Employee That Worked on the Longest Task Easy