Intersection of Two Arrays II

Solution

Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.

 

 

Example

Example 1:

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

Example 2:

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]
Explanation: [9,4] is also accepted.

 

 

Explaination

public int[] intersect(int[] nums1, int[] nums2) {
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        int n = nums1.length, m = nums2.length;
        int i = 0, j = 0;
        List<Integer> list = new ArrayList<>();
        while(i < n && j < m){
            int a = nums1[i], b= nums2[j];
            if(a == b){
                list.add(a);
                i++;
                j++;
            }else if(a < b){
                i++;
            }else{
                j++;
            }
        }
        return list.stream().mapToInt(Integer::intValue).toArray();
    }

The way of solving this problem is to use two pointer interation. First the both arrays have to be sorted using Arrays.sort() methods. A List is then initialized to store the final results.
Using a while loop, I interate through both arrays nums1 and nums2. When nums1[i] < nums2[i], I need to increment j, and vice versa. Only when nums1[i] == nums2[i], the value can be stored in the list as a final result. Lastly, return an Array that is converted from the List using a stream.