Leetcode 2956: Find Common Elements Between Two Arrays

Input: Two integer arrays nums1 and nums2.

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

Constraints:

• 1 <= n, m <= 100

• 1 <= nums1[i], nums2[i] <= 100

Output: Return an array of two integers: [answer1, answer2]. answer1 is the number of elements from nums1 that are present in nums2, and answer2 is the number of elements from nums2 that are present in nums1.

Example: [2, 1]

Constraints:

Goal: To find the count of matching elements between the two arrays in both directions.

Steps:

• Convert nums1 and nums2 into sets for efficient lookup.

• Iterate through nums1 and count how many of its elements exist in nums2.

• Similarly, iterate through nums2 and count how many of its elements exist in nums1.

Goal: The arrays have a maximum size of 100 elements, and each element is between 1 and 100.

Steps:

• n == nums1.length

• m == nums2.length

• 1 <= nums1[i], nums2[i] <= 100

Assumptions:

• Both input arrays will have at least one element.

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

• Explanation: The elements in nums1 that exist in nums2 are 2 and 2, so answer1 is 2. The element 2 in nums2 exists in nums1, so answer2 is 1.

Approach: The approach is to count the number of matching elements between nums1 and nums2 by using set lookups for efficient matching.

Observations:

• Using sets will allow for O(1) average-time lookups.

• We can use two sets to check how many elements from nums1 exist in nums2 and vice versa.

Steps:

• Convert both arrays into sets.

• For each element in nums1, check if it exists in nums2.

• For each element in nums2, check if it exists in nums1.

Empty Inputs:

• If one or both arrays are empty, the result will be [0, 0].

Large Inputs:

• Ensure the solution runs efficiently when n and m are at their maximum size of 100.

Special Values:

• Arrays with no common elements should return [0, 0].

Constraints:

• n and m are guaranteed to be between 1 and 100.

vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {
    set<int> n1, n2;
    for(int x: nums1) n1.insert(x);
    for(int x: nums2) n2.insert(x);
    int cnt1 = 0;
    for(int x: nums1) if(n2.count(x)) cnt1++;
    int cnt2 = 0;
    for(int x: nums2) if(n1.count(x)) cnt2++;        
    return vector<int>{cnt1, cnt2};
}

1 : Function Definition

vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {

Defines the function 'findIntersectionValues', which takes two vectors of integers, 'nums1' and 'nums2', and returns a vector of integers with the counts of common elements from each array.

2 : Declare Sets

    set<int> n1, n2;

Declares two sets, 'n1' and 'n2', to store the unique elements from 'nums1' and 'nums2', respectively. Sets automatically handle duplicates.

3 : Populate Set n1

    for(int x: nums1) n1.insert(x);

Iterates over 'nums1' and inserts each element into the set 'n1'. This ensures all elements in 'nums1' are unique.

4 : Populate Set n2

    for(int x: nums2) n2.insert(x);

Iterates over 'nums2' and inserts each element into the set 'n2'. This ensures all elements in 'nums2' are unique.

5 : Initialize Count Variables

    int cnt1 = 0;

Initializes 'cnt1' to zero. This variable will keep track of the number of common elements between 'nums1' and 'nums2'.

6 : Count Common Elements from nums1

    for(int x: nums1) if(n2.count(x)) cnt1++;

Iterates over 'nums1' and checks if each element exists in 'n2'. If it does, it increments 'cnt1', counting how many elements in 'nums1' are also in 'nums2'.

7 : Initialize Second Count Variable

    int cnt2 = 0;

Initializes 'cnt2' to zero. This variable will keep track of the number of common elements between 'nums2' and 'nums1'.

8 : Count Common Elements from nums2

    for(int x: nums2) if(n1.count(x)) cnt2++;

Iterates over 'nums2' and checks if each element exists in 'n1'. If it does, it increments 'cnt2', counting how many elements in 'nums2' are also in 'nums1'.

9 : Return Result

    return vector<int>{cnt1, cnt2};

Returns a vector containing the counts of common elements: 'cnt1' (common elements from 'nums1') and 'cnt2' (common elements from 'nums2').

Best Case: O(n + m)

Average Case: O(n + m)

Worst Case: O(n + m)

Description: Converting arrays into sets takes O(n) and O(m) time, and each lookup is O(1).

Best Case: O(n + m)

Worst Case: O(n + m)

Description: The space complexity is O(n + m) due to the storage of the sets.

Leetcode 2956: Find Common Elements Between Two Arrays

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