How do you solve Single Number on LeetCode?

Single Number (LeetCode #136) 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 operation allows us to cancel out duplicate numbers efficiently.

What is the brute force solution for Single Number?

The brute force approach for Single Number is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each element against every other element to find the unique one. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

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

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

What are the interview tips for Single Number?

Always consider bit manipulation techniques for problems involving unique elements. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice dry-running your code with examples to ensure you understand how it works.

What are common mistakes when solving Single Number?

Not recognizing the properties of XOR and trying to use other methods. Overcomplicating the problem by using additional data structures.

#136

Single Number

Easy

ArrayBit ManipulationBit ManipulationArray

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 the XOR operator is efficient because it has properties that allow us to cancel out duplicate numbers. This way, we can find the unique number in linear time with constant space.

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Algorithm

3 steps

1Step 1: Initialize a variable result to 0.
2Step 2: Loop through each number in the array and apply the XOR operation with result.
3Step 3: The final value of result will be the unique number.

solution.py5 lines

1def singleNumber(nums):
2    result = 0
3    for num in nums:
4        result ^= num
5    return result

ℹ

Complexity note: The time complexity is O(n) because we traverse the array once, and the space complexity is O(1) since we only use a single variable to store the result.

1XOR operation allows us to cancel out duplicate numbers efficiently.
2The unique number will remain after all duplicates are canceled out.

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