LeetCode #1491 — EASY

Average Salary Excluding the Minimum and Maximum Salary

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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.

Example 1:

Input: salary = [4000,3000,1000,2000]
Output: 2500.00000
Explanation: Minimum salary and maximum salary are 1000 and 4000 respectively.
Average salary excluding minimum and maximum salary is (2000+3000) / 2 = 2500

Example 2:

Input: salary = [1000,2000,3000]
Output: 2000.00000
Explanation: Minimum salary and maximum salary are 1000 and 3000 respectively.
Average salary excluding minimum and maximum salary is (2000) / 1 = 2000

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: You are given an array of unique integers salary where salary[i] is the salary of the ith employee. Return the average salary of employees excluding the minimum and maximum salary. Answers within 10-5 of the actual answer will be accepted.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[4000,3000,1000,2000]

Example 2

[1000,2000,3000]

Step 02

Core Insight

What unlocks the optimal approach

Get the total sum and subtract the minimum and maximum value in the array. Finally divide the result by n - 2.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1491: Average Salary Excluding the Minimum and Maximum Salary
class Solution {
    public double average(int[] salary) {
        int s = 0;
        int mi = 10000000, mx = 0;
        for (int v : salary) {
            mi = Math.min(mi, v);
            mx = Math.max(mx, v);
            s += v;
        }
        s -= (mi + mx);
        return s * 1.0 / (salary.length - 2);
    }
}

// Accepted solution for LeetCode #1491: Average Salary Excluding the Minimum and Maximum Salary
func average(salary []int) float64 {
	s := 0
	mi, mx := 10000000, 0
	for _, v := range salary {
		s += v
		mi = min(mi, v)
		mx = max(mx, v)
	}
	s -= (mi + mx)
	return float64(s) / float64(len(salary)-2)
}

# Accepted solution for LeetCode #1491: Average Salary Excluding the Minimum and Maximum Salary
class Solution:
    def average(self, salary: List[int]) -> float:
        s = sum(salary) - min(salary) - max(salary)
        return s / (len(salary) - 2)

// Accepted solution for LeetCode #1491: Average Salary Excluding the Minimum and Maximum Salary
impl Solution {
    pub fn average(salary: Vec<i32>) -> f64 {
        let n = salary.len() as i32;
        let mut min = i32::MAX;
        let mut max = i32::MIN;
        let mut sum = 0;
        for &num in salary.iter() {
            min = min.min(num);
            max = max.max(num);
            sum += num;
        }
        f64::from(sum - min - max) / f64::from(n - 2)
    }
}

// Accepted solution for LeetCode #1491: Average Salary Excluding the Minimum and Maximum Salary
function average(salary: number[]): number {
    let max = -Infinity;
    let min = Infinity;
    let sum = 0;
    for (const v of salary) {
        sum += v;
        max = Math.max(max, v);
        min = Math.min(min, v);
    }
    return (sum - max - min) / (salary.length - 2);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.