How do you solve Find the Difference of Two Arrays on LeetCode?

Find the Difference of Two Arrays (LeetCode #2215) 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 sets can significantly reduce the time complexity for membership checks.

What is the time complexity of Find the Difference of Two Arrays?

The optimal solution for Find the Difference of Two Arrays runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse each array once to create sets and then perform set operations which are efficient.

What is the brute force solution for Find the Difference of Two Arrays?

The brute force approach for Find the Difference of Two Arrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each element of the first array against all elements of the second array. This is straightforward but inefficient as it requires nested loops. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Difference of Two Arrays?

Find the Difference of Two Arrays uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find the Difference of Two Arrays?

Always think about the data structures you can use to optimize your solution. Discuss your thought process and approach with the interviewer before coding. Test your solution with edge cases, such as empty arrays or arrays with all elements the same.

What are common mistakes when solving Find the Difference of Two Arrays?

Not considering duplicates in the input arrays, leading to incorrect results. Using nested loops without realizing the inefficiency, especially for larger inputs.

#2215

Find the Difference of Two Arrays

Easy

ArrayHash TableHash 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 sets allows us to efficiently find distinct elements and check for membership in constant time, significantly improving performance.

⚙️

Algorithm

3 steps

1Step 1: Convert nums1 and nums2 into sets to eliminate duplicates.
2Step 2: Use set difference to find elements in nums1 not in nums2 and vice versa.
3Step 3: Return the results as lists.

solution.py4 lines

1def findDifference(nums1, nums2):
2    set1 = set(nums1)
3    set2 = set(nums2)
4    return [list(set1 - set2), list(set2 - set1)]

ℹ

Complexity note: The time complexity is O(n) because we traverse each array once to create sets and then perform set operations which are efficient.

1Using sets can significantly reduce the time complexity for membership checks.
2Distinct elements can be easily managed with sets, avoiding duplicates automatically.

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