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    LeetCode link: 349. Intersection of Two Arrays, difficulty: Easy.

    Given two integer arrays nums1 and nums2, return an array of their intersection.
    Each element in the result must be unique and you may return the result in any order.

    Array Intersection: The intersection of two arrays is defined as the set of elements that are present in both arrays.

    Example 1:

    Input: nums1 = [1,2,2,1], nums2 = [2,2]

    Output: [2]

    Example 2:

    Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]

    Output: [9,4]

    Explanation: [4,9] is also accepted.

    Constraints:

    • 1 <= nums1.length, nums2.length <= 1000
    • 0 <= nums1[i], nums2[i] <= 1000

    Intuition
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    1. Convert one of the arrays to a set. The elements are unique in a set.
    2. When traversing the other array, if an element is found to already exist in the set, it means that the element belongs to the intersection, and the element should be added to the results.
    3. The results is also of set type because duplicate removal is required.

    Complexity

    Time complexity

    O(N)

    Space complexity

    O(N)

    Solution in languages: Java Python C++ JavaScript C# Go Ruby

    Java # 

    class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        var results = new HashSet<Integer>();
        var num1Set = new HashSet<Integer>();

        for (var num : nums1) {
            num1Set.add(num);
        }

        for (var num : nums2) {
            if (num1Set.contains(num)) {
                results.add(num);
            }
        }

        return results.stream().mapToInt(num -> num).toArray();
    }
}

    Python # 

    class Solution:
    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
        set_of_nums1 = set(nums1)
        results = set()

        for num in nums2:
            if num in set_of_nums1:
                results.add(num)

        return list(results)

    C++ # 

    class Solution {
public:
    vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
        unordered_set<int> results;
        unordered_set<int> set_of_nums1(nums1.begin(), nums1.end());

        for (auto num : nums2) {
            if (set_of_nums1.contains(num)) {
                results.insert(num);
            }
        }

        return vector<int>(results.begin(), results.end());
    }
};

    JavaScript # 

    var intersection = function (nums1, nums2) {
  let results = new Set()
  let num1Set = new Set(nums1)

  for (const num of nums2) {
    if (num1Set.has(num)) {
      results.add(num)
    }
  }

  return Array.from(results)
};

    C# # 

    public class Solution
{
    public int[] Intersection(int[] nums1, int[] nums2)
    {
        var results = new HashSet<int>();
        var num1Set = new HashSet<int>();

        foreach (int num in nums1)
            num1Set.Add(num);

        foreach (int num in nums2)
        {
            if (num1Set.Contains(num))
            {
                results.Add(num);
            }
        }

        return results.ToArray();
    }
}

    Go # 

    func intersection(nums1 []int, nums2 []int) []int {
    results := map[int]bool{}
    num1Set := map[int]bool{}

    for _, num := range nums1 {
        num1Set[num] = true
    }

    for _, num := range nums2 {
        if _, ok := num1Set[num]; ok {
            results[num] = true
        }
    }

    return slices.Collect(maps.Keys(results))
}

    Ruby # 

    def intersection(nums1, nums2)
  nums1_set = Set.new(nums1)
  results = Set.new

  nums2.each do |num|
    if nums1_set.include?(num)
      results << num
    end
  end

  results.to_a
end
    Solution in languages: Java Python C++ JavaScript C# Go Ruby

    Other languages

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