The “Single Number III” problem asks you to find two numbers in an array that appear exactly once, while every other number appears exactly twice. The array can be in any order, and you must identify those two unique values. Unlike simpler versions of the problem where only one number appears once, here the presence of two unique numbers adds an extra layer of complexity.
At first glance, this problem looks like it might require counting frequencies or using additional memory to track occurrences. However, the problem is designed to be solved efficiently using bit manipulation, without extra data structures that scale with input size.
Why frequency counting is not the intended solution
A straightforward approach would be to use a map or dictionary to count how many times each number appears, then return the two numbers with a count of one. While this works logically, it uses additional space proportional to the number of unique elements.
The challenge of this problem is not correctness but efficiency. The goal is to leverage the special structure of the input, where most numbers appear exactly twice, to eliminate unnecessary memory usage and achieve an optimal solution.
Revisiting the XOR property
The key to solving this problem lies in understanding the properties of the XOR operation. XOR has two crucial characteristics that make it ideal here. First, a number XORed with itself results in zero. Second, a number XORed with zero remains unchanged.
If you XOR all numbers in the array together, every number that appears twice cancels itself out. What remains is the XOR of the two unique numbers. At this point, you do not yet know the two numbers individually, but you know that their combined XOR value is nonzero, because the two numbers are different.
Want to explore more coding problem solutions? Check out the Diagonal Traverse and Remove Duplicate Letters.
Why the combined XOR is not enough
After XORing all elements, you are left with a single value that represents the XOR of the two unique numbers. However, this value alone does not tell you what the two numbers are. You need a way to separate them.
The insight is that if two numbers are different, they must differ in at least one bit position. That differing bit is the key to splitting the array into two groups in a meaningful way.
Using a distinguishing bit to split the array
From the combined XOR value, you can isolate a bit that is set to one. This bit indicates a position where the two unique numbers differ. One of the unique numbers has this bit set, and the other does not.
You can now use this bit as a partition rule. Divide the original array into two groups: numbers that have this bit set, and numbers that do not. Because duplicate numbers are identical, both occurrences of any duplicated number will fall into the same group and cancel out when XORed within that group.
Recovering the two unique numbers
Once the array is split into two groups, you XOR the numbers within each group separately. In each group, all duplicated numbers cancel out, leaving exactly one unique number per group.
At the end of this process, you have recovered the two numbers that appear only once in the array. No extra memory proportional to input size was needed, and the array was processed in linear time.
Why this approach is guaranteed to work
The correctness of this solution relies on two facts. First, XOR reliably cancels out pairs of identical numbers. Second, the two unique numbers must differ in at least one bit, which guarantees that they can be separated into different groups.
Because the partition is based on a bit where the two unique numbers differ, they cannot end up in the same group. This ensures that each group contains exactly one unique number after all cancellations.
Time and space complexity considerations
The algorithm runs in linear time because it scans the array a constant number of times. Space usage is constant, as it only uses a few integer variables to track XOR values and bit positions.
This efficiency is what makes the solution elegant and optimal for the problem’s constraints.
Why this problem is popular in interviews
Single Number III is a favorite interview problem because it tests whether candidates truly understand bitwise operations rather than treating them as obscure tricks. Many people know that XOR can find a single unique number, but extending that idea to two unique numbers requires deeper reasoning.
The problem rewards those who can reason about binary representations and exploit problem constraints effectively.
What this problem teaches beyond XOR tricks
Beyond its immediate solution, this problem teaches a powerful algorithmic lesson: when input data has strong guarantees, you can often design solutions that eliminate extra memory entirely. It also demonstrates how low-level operations like bit manipulation can lead to clean, high-performance algorithms.
If you can clearly explain why XOR cancels duplicates, how a distinguishing bit separates the two unique numbers, and why the groups resolve cleanly, you demonstrate strong problem-solving fundamentals. That depth of understanding makes “Single Number III” an excellent exercise in efficient reasoning with bitwise operations.
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