Average Salary Excluding the Minimum and Maximum Salary

Average Salary Excluding the Minimum and Maximum Salary - Problem

Array Sorting Easy

You are given an array of unique integers salary where salary[i] is the salary of the i-th employee.

Return the average salary of employees excluding the minimum and maximum salary. Answers within 10^-5 of the actual answer will be accepted.

Input & Output

Example 1 — Basic Case

$ Input: salary = [4000,3000,1000,2000]

› Output: 2500.0

💡 Note: Minimum salary is 1000, maximum salary is 4000. Average of remaining salaries: (3000 + 2000) / 2 = 2500.0

Example 2 — Minimum Size

$ Input: salary = [1000,2000,3000]

› Output: 2000.0

💡 Note: Minimum salary is 1000, maximum salary is 3000. Average of remaining salaries: 2000 / 1 = 2000.0

Example 3 — Larger Array

$ Input: salary = [8000,9000,2000,3000,6000,1000]

› Output: 4750.0

💡 Note: Minimum salary is 1000, maximum salary is 9000. Average of remaining: (8000+2000+3000+6000) / 4 = 4750.0

Constraints

3 ≤ salary.length ≤ 100
1000 ≤ salary[i] ≤ 10⁶
All the integers of salary are unique

Visualization

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Asked in

🍎 Apple 15 🏢 Microsoft 12

The key insight is that we only need to find the minimum and maximum salaries to exclude them from the average calculation. The optimal approach uses a single pass to track min, max, and total sum simultaneously. Time: O(n), Space: O(1)

Common Approaches

✓ Single Pass with Min/Max Tracking

⏱️ Time: O(n) Space: O(1)

Iterate through the array once to find minimum, maximum, and total sum. Then subtract min and max from total and divide by (n-2) to get the average.

Sort and Remove Extremes

⏱️ Time: O(n log n) Space: O(1)

Sort the entire salary array, remove the minimum (first element) and maximum (last element), then calculate the average of remaining elements.

Single Pass with Min/Max Tracking — Algorithm Steps

Iterate through array once to find min, max, and total sum
Subtract min and max from total sum
Divide result by (n-2) to get average

Visualization

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Step-by-Step Walkthrough

Single Pass

Track min, max, and sum simultaneously

Calculate

Subtract min and max from total

Average

Divide by count of remaining elements

Code -

solution.c — C

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

double solution(int* salary, int n) {
    int minSalary = INT_MAX;
    int maxSalary = INT_MIN;
    int total = 0;
    
    for (int i = 0; i < n; i++) {
        if (salary[i] < minSalary) {
            minSalary = salary[i];
        }
        if (salary[i] > maxSalary) {
            maxSalary = salary[i];
        }
        total += salary[i];
    }
    
    return (double) (total - minSalary - maxSalary) / (n - 2);
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    int salary[1000];
    int n = 0;
    char *token = strtok(line + 1, ",]");
    while (token != NULL) {
        salary[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    double result = solution(salary, n);
    printf("%f\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array to find min, max, and sum

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra space for min, max, sum variables

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Input & Output

Constraints

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Related Problems

Common Approaches

Single Pass with Min/Max Tracking — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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