How do you solve Intersection of Two Arrays II on LeetCode?

Intersection of Two Arrays II (LeetCode #350) 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 time complexity of Intersection of Two Arrays II?

The optimal solution for Intersection of Two Arrays II runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse both arrays once, and the space complexity is O(n) due to the HashMap storing counts.

What is the brute force solution for Intersection of Two Arrays II?

The brute force approach for Intersection of Two Arrays II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each element of the first array against all elements of the second array. If a match is found, we add it to the result. This is simple but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Intersection of Two Arrays II?

Intersection of Two Arrays II uses the following concepts: Array, Hash Table, Two Pointers, Binary Search, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Intersection of Two Arrays II?

Explain your thought process clearly while coding. Consider edge cases, such as empty arrays or arrays with no intersection. Practice writing clean and efficient code.

What are common mistakes when solving Intersection of Two Arrays II?

Not handling duplicates correctly when counting occurrences. Forgetting to check if an element exists in the HashMap.

#350

Intersection of Two Arrays II

Easy

ArrayHash TableTwo PointersBinary SearchSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Using a HashMap allows us to count the occurrences of each number in one array and then find the intersection efficiently. This reduces the number of checks needed.

⚙️

Algorithm

3 steps

1Step 1: Create a HashMap to count occurrences of each number in nums1.
2Step 2: Iterate through nums2, checking if the number exists in the HashMap.
3Step 3: If it exists, add it to the result and decrement the count in the HashMap.

solution.py10 lines

1from collections import Counter
2
3def intersect(nums1, nums2):
4    counts = Counter(nums1)
5    result = []
6    for num in nums2:
7        if counts[num] > 0:
8            result.append(num)
9            counts[num] -= 1
10    return result

ℹ

Complexity note: The time complexity is O(n) because we traverse both arrays once, and the space complexity is O(n) due to the HashMap storing counts.

1Using a HashMap can significantly reduce the time complexity.
2Understanding how to manage counts of elements is crucial for intersection problems.

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