Problem

Given an integer array arr, return every element that appears exactly once, in the order of its first appearance in the array.

Input / Output

• Input: integer array arr.
• Output: list of integers (may be empty).

Examples

Example 1 Input: arr = [4, 3, 2, 4, 1, 3, 5] Output: [2, 1, 5]

Example 2 Input: arr = [1, 1, 2, 2] Output: []

Example 3 Input: arr = [9] Output: [9]

Constraints

• 1 <= n <= 10^5

Approach

Two passes with a hash map:

1. Count occurrences of each element.
2. Walk the array again and emit elements whose count is 1. Walking the original array (rather than the map) preserves input order.

Why not XOR? XOR-ing all elements cancels pairs and finds a single unique element when every other value appears exactly twice; it cannot list several unique elements or handle values that appear three times. Mention it as a related trick, not as the solution.

Order-preserving with one structure: an insertion-ordered map (Python dict, Java LinkedHashMap) lets you iterate the map itself in first-appearance order.

Complexity

• Time: O(n)
• Space: O(k)

Reference solution (Python)

from collections import Counter

def non_repeating(arr):
    freq = Counter(arr)
    return [v for v in arr if freq[v] == 1]

Test cases

Input	Expected
[4,3,2,4,1,3,5]	[2,1,5]
[1,1,2,2]	[]
[9]	[9]
[2,2,2,3]	[3]