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2956. Find Common Elements Between Two Arrays

Description

You are given two integer arrays nums1 and nums2 of sizes n and m, respectively. Calculate the following values:

  • answer1 : the number of indices i such that nums1[i] exists in nums2.
  • answer2 : the number of indices i such that nums2[i] exists in nums1.

Return [answer1,answer2].

 

Example 1:

Input: nums1 = [2,3,2], nums2 = [1,2]

Output: [2,1]

Explanation:

Example 2:

Input: nums1 = [4,3,2,3,1], nums2 = [2,2,5,2,3,6]

Output: [3,4]

Explanation:

The elements at indices 1, 2, and 3 in nums1 exist in nums2 as well. So answer1 is 3.

The elements at indices 0, 1, 3, and 4 in nums2 exist in nums1. So answer2 is 4.

Example 3:

Input: nums1 = [3,4,2,3], nums2 = [1,5]

Output: [0,0]

Explanation:

No numbers are common between nums1 and nums2, so answer is [0,0].

 

Constraints:

  • n == nums1.length
  • m == nums2.length
  • 1 <= n, m <= 100
  • 1 <= nums1[i], nums2[i] <= 100

Solutions

Solution 1: Hash Table or Array

We can use two hash tables or arrays \(s1\) and \(s2\) to record the elements that appear in the two arrays respectively.

Next, we create an array \(ans\) of length \(2\), where \(ans[0]\) represents the number of elements in \(nums1\) that appear in \(s2\), and \(ans[1]\) represents the number of elements in \(nums2\) that appear in \(s1\).

Then, we traverse each element \(x\) in the array \(nums1\). If \(x\) has appeared in \(s2\), we increment \(ans[0]\). After that, we traverse each element \(x\) in the array \(nums2\). If \(x\) has appeared in \(s1\), we increment \(ans[1]\).

Finally, we return the array \(ans\).

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

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class Solution:
    def findIntersectionValues(self, nums1: List[int], nums2: List[int]) -> List[int]:
        s1, s2 = set(nums1), set(nums2)
        return [sum(x in s2 for x in nums1), sum(x in s1 for x in nums2)]

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class Solution {
    public int[] findIntersectionValues(int[] nums1, int[] nums2) {
        int[] s1 = new int[101];
        int[] s2 = new int[101];
        for (int x : nums1) {
            s1[x] = 1;
        }
        for (int x : nums2) {
            s2[x] = 1;
        }
        int[] ans = new int[2];
        for (int x : nums1) {
            ans[0] += s2[x];
        }
        for (int x : nums2) {
            ans[1] += s1[x];
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {
        int s1[101]{};
        int s2[101]{};
        for (int& x : nums1) {
            s1[x] = 1;
        }
        for (int& x : nums2) {
            s2[x] = 1;
        }
        vector<int> ans(2);
        for (int& x : nums1) {
            ans[0] += s2[x];
        }
        for (int& x : nums2) {
            ans[1] += s1[x];
        }
        return ans;
    }
};

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func findIntersectionValues(nums1 []int, nums2 []int) []int {
    s1 := [101]int{}
    s2 := [101]int{}
    for _, x := range nums1 {
        s1[x] = 1
    }
    for _, x := range nums2 {
        s2[x] = 1
    }
    ans := make([]int, 2)
    for _, x := range nums1 {
        ans[0] += s2[x]
    }
    for _, x := range nums2 {
        ans[1] += s1[x]
    }
    return ans
}

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function findIntersectionValues(nums1: number[], nums2: number[]): number[] {
    const s1: number[] = Array(101).fill(0);
    const s2: number[] = Array(101).fill(0);
    for (const x of nums1) {
        s1[x] = 1;
    }
    for (const x of nums2) {
        s2[x] = 1;
    }
    const ans: number[] = Array(2).fill(0);
    for (const x of nums1) {
        ans[0] += s2[x];
    }
    for (const x of nums2) {
        ans[1] += s1[x];
    }
    return ans;
}

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