How do you solve Find the XOR of Numbers Which Appear Twice on LeetCode?

Find the XOR of Numbers Which Appear Twice (LeetCode #3158) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a HashMap can significantly reduce the time complexity.

What is the brute force solution for Find the XOR of Numbers Which Appear Twice?

The brute force approach for Find the XOR of Numbers Which Appear Twice is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each number in the array to see how many times it appears. If it appears twice, we include it in our XOR calculation. This is straightforward but not efficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the XOR of Numbers Which Appear Twice?

Find the XOR of Numbers Which Appear Twice uses the following concepts: Array, Hash Table, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find the XOR of Numbers Which Appear Twice?

Always clarify the problem constraints and edge cases. Explain your thought process clearly while coding. Consider both time and space complexity in your solution.

What are common mistakes when solving Find the XOR of Numbers Which Appear Twice?

Not handling edge cases where no numbers appear twice. Confusing the XOR operation with addition.

#3158

Find the XOR of Numbers Which Appear Twice

Easy

ArrayHash TableBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Using a HashMap allows us to count occurrences of each number in a single pass, making it more efficient. We can then XOR the numbers that appear twice in another pass.

⚙️

Algorithm

5 steps

1Step 1: Initialize a HashMap to count occurrences of each number.
2Step 2: Loop through the array and populate the HashMap with counts.
3Step 3: Initialize a variable `result` to 0 for the XOR of numbers that appear twice.
4Step 4: Loop through the HashMap and XOR the keys (numbers) that have a count of 2.
5Step 5: Return `result`.

solution.py9 lines

1def findXOR(nums):
2    count_map = {}
3    for num in nums:
4        count_map[num] = count_map.get(num, 0) + 1
5    result = 0
6    for num, count in count_map.items():
7        if count == 2:
8            result ^= num
9    return result

ℹ

Complexity note: The time complexity is O(n) because we traverse the array and the HashMap in linear time. The space complexity is O(n) due to the storage of counts in the HashMap.

1Using a HashMap can significantly reduce the time complexity.
2XOR operation is commutative and associative, which allows us to combine results effectively.

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