Neither Minimum nor Maximum - Practice Coding Problems

Neither Minimum nor Maximum - Problem

Array Sorting Easy

Given an integer array nums containing distinct positive integers, you need to find and return any number from the array that is neither the minimum nor the maximum value in the array.

If no such number exists (meaning the array has fewer than 3 elements), return -1.

Goal: Find any "middle" value that isn't an extreme.

Key Points:

All integers are distinct and positive
You can return any valid middle value
Arrays with less than 3 elements should return -1

Input & Output

example_1.py — Basic Case

$ Input: [3, 2, 1, 4]

› Output: 2

💡 Note: The minimum is 1, maximum is 4. Both 2 and 3 are neither min nor max, so we can return either (2 is found first in most approaches).

example_2.py — Multiple Middle Values

$ Input: [1, 2, 3, 4, 5]

› Output: 2

💡 Note: Minimum is 1, maximum is 5. Elements 2, 3, and 4 are all valid answers. We return the first one encountered.

example_3.py — Edge Case

$ Input: [1, 2]

› Output: -1

💡 Note: Array has only 2 elements, so every element is either minimum or maximum. No middle element exists.

Constraints

1 ≤ nums.length ≤ 100
1 ≤ nums[i] ≤ 1000
All integers in nums are distinct
All integers are positive

Visualization

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Understanding the Visualization

Identify Extremes

First, determine who are the shortest and tallest students

Find Anyone In Between

Look for any student who isn't one of these extremes

Return First Found

As soon as you find someone in the middle, that's your answer

Key Takeaway

🎯 Key Insight: Since we only need ANY middle element (not a specific one), we can return immediately upon finding the first element that's neither the current minimum nor maximum!

Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 6

The optimal solution uses a single-pass approach with O(n) time complexity. Track minimum and maximum values while scanning, and return the first element that is neither an extreme. This approach can terminate early when a middle element is found, making it highly efficient for most practical cases.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Compare Each Element)	O(n²)	O(1)	For each element, check if it's neither min nor max by comparing with all others
One-Pass Optimal Solution	O(n)	O(1)	Track min/max while scanning, return first element that's neither extreme
Two-Pass Linear Scan	O(n)	O(1)	First pass finds min/max values, second pass finds any element that's neither

Brute Force (Compare Each Element) — Algorithm Steps

Iterate through each element in the array
For each element, check if it's the minimum by comparing with all others
For each element, check if it's the maximum by comparing with all others
If an element is neither min nor max, return it
If no such element exists, return -1

Visualization

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Step-by-Step Walkthrough

Check First Element

Compare first element with all others to see if it's min or max

Check Second Element

Compare second element with all others

Find Middle Element

Continue until we find an element that's neither min nor max

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

int findNonMinOrMax(int* nums, int numsSize) {
    if (numsSize < 3) {
        return -1;
    }
    
    for (int i = 0; i < numsSize; i++) {
        bool isMin = true;
        bool isMax = true;
        
        // Check if current element is minimum
        for (int j = 0; j < numsSize; j++) {
            if (nums[j] < nums[i]) {
                isMin = false;
                break;
            }
        }
        
        // Check if current element is maximum
        for (int j = 0; j < numsSize; j++) {
            if (nums[j] > nums[i]) {
                isMax = false;
                break;
            }
        }
        
        // If neither min nor max, return it
        if (!isMin && !isMax) {
            return nums[i];
        }
    }
    
    return -1;
}

int main() {
    int nums[1000];
    int size = 0;
    char c;
    int num = 0;
    bool inNumber = false;
    
    while ((c = getchar()) != '\n' && c != EOF) {
        if (c >= '0' && c <= '9') {
            num = num * 10 + (c - '0');
            inNumber = true;
        } else if (inNumber) {
            nums[size++] = num;
            num = 0;
            inNumber = false;
        }
    }
    if (inNumber) {
        nums[size++] = num;
    }
    
    printf("%d\n", findNonMinOrMax(nums, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n elements, we compare it with all n elements to determine if it's min/max

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables, no additional data structures

✓ Linear Space

28.4K Views

Medium Frequency

~8 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Compare Each Element) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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