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.
Build confidence with an intuition-first walkthrough focused on array fundamentals.
You have an array of floating point numbers averages which is initially empty. You are given an array nums of n integers where n is even.
You repeat the following procedure n / 2 times:
minElement, and the largest element maxElement, from nums.(minElement + maxElement) / 2 to averages.Return the minimum element in averages.
Example 1:
Input: nums = [7,8,3,4,15,13,4,1]
Output: 5.5
Explanation:
| step | nums | averages |
|---|---|---|
| 0 | [7,8,3,4,15,13,4,1] | [] |
| 1 | [7,8,3,4,13,4] | [8] |
| 2 | [7,8,4,4] | [8,8] |
| 3 | [7,4] | [8,8,6] |
| 4 | [] | [8,8,6,5.5] |
Example 2:
Input: nums = [1,9,8,3,10,5]
Output: 5.5
Explanation:
| step | nums | averages |
|---|---|---|
| 0 | [1,9,8,3,10,5] | [] |
| 1 | [9,8,3,5] | [5.5] |
| 2 | [8,5] | [5.5,6] |
| 3 | [] | [5.5,6,6.5] |
Example 3:
Input: nums = [1,2,3,7,8,9]
Output: 5.0
Explanation:
| step | nums | averages |
|---|---|---|
| 0 | [1,2,3,7,8,9] | [] |
| 1 | [2,3,7,8] | [5] |
| 2 | [3,7] | [5,5] |
| 3 | [] | [5,5,5] |
Constraints:
2 <= n == nums.length <= 50n is even.1 <= nums[i] <= 50Problem summary: You have an array of floating point numbers averages which is initially empty. You are given an array nums of n integers where n is even. You repeat the following procedure n / 2 times: Remove the smallest element, minElement, and the largest element maxElement, from nums. Add (minElement + maxElement) / 2 to averages. Return the minimum element in averages.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Two Pointers
[7,8,3,4,15,13,4,1]
[1,9,8,3,10,5]
[1,2,3,7,8,9]
number-of-distinct-averages)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3194: Minimum Average of Smallest and Largest Elements
class Solution {
public double minimumAverage(int[] nums) {
Arrays.sort(nums);
int n = nums.length;
int ans = 1 << 30;
for (int i = 0; i < n / 2; ++i) {
ans = Math.min(ans, nums[i] + nums[n - i - 1]);
}
return ans / 2.0;
}
}
// Accepted solution for LeetCode #3194: Minimum Average of Smallest and Largest Elements
func minimumAverage(nums []int) float64 {
sort.Ints(nums)
n := len(nums)
ans := 1 << 30
for i, x := range nums[:n/2] {
ans = min(ans, x+nums[n-i-1])
}
return float64(ans) / 2
}
# Accepted solution for LeetCode #3194: Minimum Average of Smallest and Largest Elements
class Solution:
def minimumAverage(self, nums: List[int]) -> float:
nums.sort()
n = len(nums)
return min(nums[i] + nums[-i - 1] for i in range(n // 2)) / 2
// Accepted solution for LeetCode #3194: Minimum Average of Smallest and Largest Elements
impl Solution {
pub fn minimum_average(mut nums: Vec<i32>) -> f64 {
nums.sort();
let n = nums.len();
let ans = (0..n / 2).map(|i| nums[i] + nums[n - i - 1]).min().unwrap();
ans as f64 / 2.0
}
}
// Accepted solution for LeetCode #3194: Minimum Average of Smallest and Largest Elements
function minimumAverage(nums: number[]): number {
nums.sort((a, b) => a - b);
const n = nums.length;
let ans = Infinity;
for (let i = 0; i * 2 < n; ++i) {
ans = Math.min(ans, nums[i] + nums[n - 1 - i]);
}
return ans / 2;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
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.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.