Count Elements With Strictly Smaller and Greater Elements

Count Elements With Strictly Smaller and Greater Elements - Problem

Given an integer array nums, return the number of elements that have both a strictly smaller and a strictly greater element appear in nums.

In other words, for each element nums[i], count it if there exists some element nums[j] where nums[j] < nums[i] and some element nums[k] where nums[k] > nums[i].

Input & Output

Example 1 — Basic Case

$ Input: nums = [11,7,2,15]

› Output: 2

💡 Note: Element 11: has smaller (7,2) and greater (15) ✓. Element 7: has smaller (2) and greater (11,15) ✓. Element 2: has greater (7,11,15) but no smaller ✗. Element 15: has smaller (11,7,2) but no greater ✗. Count = 2

Example 2 — No Valid Elements

$ Input: nums = [6,2,6,5,10,5]

› Output: 3

💡 Note: Min=2, Max=10. Elements between: first 6 (2<6<10), 5 (2<5<10), second 5 (2<5<10). Count = 3

Example 3 — Minimum Size

$ Input: nums = [1,2]

› Output: 0

💡 Note: With only 2 elements, neither can have both smaller and greater elements. Count = 0

Constraints

1 ≤ nums.length ≤ 100
-10⁵ ≤ nums[i] ≤ 10⁵

Visualization

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Understanding the Visualization

Input Array

Given array [11,7,2,15] with various values

Check Conditions

For each element, verify both smaller and greater exist

Count Valid

Count elements satisfying both conditions = 2

Key Takeaway

🎯 Key Insight: Only elements that are neither minimum nor maximum can have both smaller and greater neighbors

Asked in

a Amazon 15 M Microsoft 8

The key insight is that only elements that are neither the minimum nor maximum value can have both smaller and greater elements. The optimal approach finds min/max in one pass, then counts elements strictly between them. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Check Each Element	O(n²)	O(1)	For each element, scan the entire array to find if smaller and greater elements exist
Sort First - Find Min/Max	O(n)	O(1)	Find minimum and maximum values, then count elements that are neither

Brute Force - Check Each Element — Algorithm Steps

Iterate through each element
For each element, scan all others to find smaller and greater
Count elements that have both conditions satisfied

Visualization

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Step-by-Step Walkthrough

Select Element

Pick current element to check

Scan Array

Compare with all elements to find smaller/greater

Count Valid

Increment count if both conditions met

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int count = 0;
    for (int i = 0; i < numsSize; i++) {
        int hasSmaller = 0;
        int hasGreater = 0;
        
        for (int j = 0; j < numsSize; j++) {
            if (nums[j] < nums[i]) {
                hasSmaller = 1;
            }
            if (nums[j] > nums[i]) {
                hasGreater = 1;
            }
        }
        
        if (hasSmaller && hasGreater) {
            count++;
        }
    }
    
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, numsSize);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n elements, we scan through all n elements to find smaller/greater

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables for counting, no extra data structures

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check Each Element — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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