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349. Intersection of Two Arrays

Description

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.

 

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

Solutions

Solution 1: Hash Table or Array

Thinking

The distinct intersection of two arrays. Nested loops are \(O(nm)\). A set (or a small table) answers membership in \(O(1)\).

Load one side into a set, scan the other, and remove on a hit to avoid duplicates. The code is simply the set intersection.

First, we use a hash table or an array \(s\) of length \(1001\) to record the elements that appear in the array \(nums1\). Then, we iterate through each element in the array \(nums2\). If an element \(x\) is in \(s\), we add \(x\) to the answer and remove \(x\) from \(s\).

After the iteration is finished, we return the answer array.

The time complexity is \(O(n+m)\), and the space complexity is \(O(n)\). Here, \(n\) and \(m\) are the lengths of the arrays \(nums1\) and \(nums2\), respectively.

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class Solution:
    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
        return list(set(nums1) & set(nums2))

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class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        boolean[] s = new boolean[1001];
        for (int x : nums1) {
            s[x] = true;
        }
        List<Integer> ans = new ArrayList<>();
        for (int x : nums2) {
            if (s[x]) {
                ans.add(x);
                s[x] = false;
            }
        }
        return ans.stream().mapToInt(Integer::intValue).toArray();
    }
}

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class Solution {
public:
    vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
        bool s[1001];
        memset(s, false, sizeof(s));
        for (int x : nums1) {
            s[x] = true;
        }
        vector<int> ans;
        for (int x : nums2) {
            if (s[x]) {
                ans.push_back(x);
                s[x] = false;
            }
        }
        return ans;
    }
};

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func intersection(nums1 []int, nums2 []int) (ans []int) {
    s := [1001]bool{}
    for _, x := range nums1 {
        s[x] = true
    }
    for _, x := range nums2 {
        if s[x] {
            ans = append(ans, x)
            s[x] = false
        }
    }
    return
}

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function intersection(nums1: number[], nums2: number[]): number[] {
    const s = new Set(nums1);
    return [...new Set(nums2.filter(x => s.has(x)))];
}

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/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersection = function (nums1, nums2) {
    const s = new Set(nums1);
    return [...new Set(nums2.filter(x => s.has(x)))];
};

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public class Solution {
    public int[] Intersection(int[] nums1, int[] nums2) {
        HashSet<int> s1 = new HashSet<int>(nums1);
        HashSet<int> s2 = new HashSet<int>(nums2);
        s1.IntersectWith(s2);
        int[] ans = new int[s1.Count];
        s1.CopyTo(ans);
        return ans;
    }
}

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class Solution {
    /**
     * @param Integer[] $nums1
     * @param Integer[] $nums2
     * @return Integer[]
     */
    function intersection($nums1, $nums2) {
        $s1 = array_unique($nums1);
        $s2 = array_unique($nums2);
        $ans = array_intersect($s1, $s2);
        return array_values($ans);
    }
}

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