Unique Numbers ||

Last Updated : 29 Mar, 2026

Given an array arr[] containing 2*n + 2 positive numbers, out of which 2*n numbers exist in pairs whereas the other two number occur exactly once and are distinct. The task is to find the other two numbers.

Note: Return the numbers in increasing order.

Example:

Input: arr[] = [1, 2, 3, 2, 1, 4]
Output: 3 4
Explanation: 3 and 4 occur exactly once.
Input: arr[] = [2, 1, 3, 2]
Output: 1 3
Explanation: 1 and 3 occur exactly once.

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Table of Content

[Naive Approach] Nested Loop Frequency Counting - O(n^2) Time and O(1) Space

This approach iterates through the array and counts the frequency of each element using a nested loop. For each element, the inner loop counts how many times it appears in the array. If an element appears exactly once, it is returned as the result.

Output

3 4

Note : We can also use sorting and hashing. With sorting in O(n Log n) time, we get all same elements together and can count frequency easily. With hashing we can count frequency in O(n) time.

[Expected Approach] Using Bit Manipulation - O(n) time and O(1) space

The key idea is that by XOR-ing all elements in the array, we eliminate the effect of all numbers that appear twice (since a ^ a = 0), leaving us with x ^ y, the XOR of the two unique elements. We then find any set bit (typically the rightmost set bit) in this result to determine a position where these two numbers differ. Using this bit, we split the numbers into two groups, XOR each group separately, and get the two unique numbers.

To find the rightmost set bit in xorVal, we use the expression xorVal & (-xorVal). This is a common bit manipulation trick that isolates the lowest (rightmost) bit that is set to 1. It works because in two's complement, -xorVal is the bitwise inverse of xorVal plus one, which flips all bits after the rightmost 1, thus leaving only that 1 bit in the AND operation. This bit tells us a position where the two numbers (say, x and y) differ , one has a 1 at this position, and the other has a 0.

This trick is particularly efficient and often used in problems involving two unique elements among duplicates. However, if we don’t specifically need the rightmost set bit, we can alternatively find any set bit by scanning bit positions, or using other logical operations

Illustration:

Given the array [2, 4, 7, 9, 2, 4], XOR all elements to get 14. The rightmost set bit of 14 is at position 1. Divide the array into two groups based on this bit:

Group 1 (bit set at position 1): [2, 7, 2]
Group 2 (bit not set): [4, 9, 4]

XOR the elements in each group: