How do you solve Single Number III on LeetCode?

Single Number III (LeetCode #260) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: XOR is a powerful tool for problems involving pairs or unique elements.

What is the brute force solution for Single Number III?

The brute force approach for Single Number III is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each element and counting its occurrences. This is straightforward but inefficient, as it requires multiple passes through the array. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Single Number III?

Single Number III uses the following concepts: Array, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Single Number III?

Explain your thought process clearly while coding. Discuss the trade-offs of different approaches before jumping into coding. Practice bit manipulation problems to become comfortable with them.

What are common mistakes when solving Single Number III?

Forgetting to handle the case where numbers are negative. Not recognizing that XOR can help eliminate duplicates.

#260

Single Number III

Medium

ArrayBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

Using bit manipulation, we can efficiently find the two unique numbers in linear time and constant space. The key is to leverage the properties of XOR to separate the two unique numbers.

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Algorithm

3 steps

1Step 1: XOR all numbers to get a combined result of the two unique numbers (let's call it xorResult).
2Step 2: Find a bit that is set (1) in xorResult. This bit will help us differentiate the two unique numbers.
3Step 3: Partition the original numbers into two groups based on the set bit and XOR each group to find the two unique numbers.

solution.py15 lines

1# Full working Python code
2
3def singleNumber(nums):
4    xorResult = 0
5    for num in nums:
6        xorResult ^= num
7    # Get the rightmost set bit
8    diff = xorResult & -xorResult
9    num1, num2 = 0, 0
10    for num in nums:
11        if num & diff:
12            num1 ^= num
13        else:
14            num2 ^= num
15    return [num1, num2]

ℹ

Complexity note: This complexity is linear because we only pass through the list a constant number of times, and we use a fixed amount of extra space.

1XOR is a powerful tool for problems involving pairs or unique elements.
2Identifying a distinguishing bit helps in segregating elements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.