Neither Minimum nor Maximum

Neither Minimum nor Maximum - Problem

Array Sorting Easy

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.

Input & Output

Example 1 — Basic Case

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

› Output: 2

💡 Note: Array has min=1, max=4. Elements 2 and 3 are neither min nor max, so we can return either one. Return 2.

Example 2 — Small Array

$ Input: nums = [1,2]

› Output: -1

💡 Note: Array has only 2 elements: min=1 and max=2. No element exists that is neither min nor max.

Example 3 — Larger Array

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

› Output: 2

💡 Note: Array has min=1, max=3. Element 2 is neither the minimum nor maximum, so return 2.

Constraints

1 ≤ nums.length ≤ 100
1 ≤ nums[i] ≤ 100
All the numbers of nums are unique.

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that we need to find any element that's not an extreme value. Best approach is the two-pass solution: first find min/max values, then find any element that's neither. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Find Min/Max by Comparison

⏱️ Time: O(n³) Space: O(1)

For each element, check if it's the minimum by comparing with all others, then check if it's the maximum. Return the first element that is neither.

Sort and Pick Middle

⏱️ Time: O(n log n) Space: O(1)

Sort the array in ascending order, then return any element that's not the first (minimum) or last (maximum). The simplest choice is the second element.

Two-Pass Solution

⏱️ Time: O(n) Space: O(1)

Make one pass through the array to find the minimum and maximum values, then make another pass to find any element that is neither the min nor max.

Brute Force - Find Min/Max by Comparison — Algorithm Steps

Iterate through each element
For each element, check if it's minimum by comparing with all others
Check if it's maximum by comparing with all others
Return first element that is neither min nor max

Visualization

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Step-by-Step Walkthrough

Check First Element

Compare nums[0] with all elements to see if it's min or max

Check Second Element

Compare nums[1] with all elements to see if it's min or max

Found Answer

Return first element that is neither minimum nor maximum

Code -

solution.c — C

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

int solution(int* nums, int n) {
    if (n <= 2) return -1;
    
    for (int i = 0; i < n; i++) {
        bool isMin = true;
        bool isMax = true;
        
        // Check if nums[i] is minimum
        for (int j = 0; j < n; j++) {
            if (nums[j] < nums[i]) {
                isMin = false;
                break;
            }
        }
        
        // Check if nums[i] is maximum
        for (int j = 0; j < n; 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() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse JSON array manually
    int nums[100];
    int n = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

For each element, we compare with all n elements twice (min check + max check)

⚡ Linearithmic

Space Complexity

O(1)

Only using a few variables, no extra data structures

✓ Linear Space

23.4K 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 - Find Min/Max by Comparison — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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