Given two arrays, write a function to compute their intersection.

Example:
Givennums1=[1, 2, 2, 1],nums2=[2, 2], return[2].

Note:

• Each element in the result must be unique.
• The result can be in any order.

Solution 1: two pointers

    public int[] intersection(int[] nums1, int[] nums2) {
        Set<Integer> set = new HashSet<>();
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        int i = 0;
        int j = 0;
        while (i < nums1.length && j < nums2.length) {
            if (nums1[i] < nums2[j]) {
                i++;
            } else if (nums1[i] > nums2[j]) {
                j++;
            } else {
                set.add(nums1[i]);
                i++;
                j++;
            }
        }
        int[] result = new int[set.size()];
        int k = 0;
        for (Integer num : set) {
            result[k++] = num;
        }
        return result;
    }

Solution 2: hashset

        HashSet<Integer> set = new HashSet<Integer>();
        ArrayList<Integer> res = new ArrayList<Integer>();
        //Add all elements to set from array 1
        for(int i =0; i< nums1.length; i++) set.add(nums1[i]);
        for(int j = 0; j < nums2.length; j++) {
           // If present in array 2 then add to res and remove from set 
           if(set.contains(nums2[j])) {
                res.add(nums2[j]);
                set.remove(nums2[j]);
            }
        }
        // Convert ArrayList to array
        int[] arr = new int[res.size()];
        for (int i= 0; i < res.size(); i++) arr[i] = res.get(i);
        return arr;

Solution 3: binary search

    public int[] intersection(int[] nums1, int[] nums2) {
        Set<Integer> set = new HashSet<>();
        Arrays.sort(nums2);
        for (Integer num : nums1) {
            if (binarySearch(nums2, num)) {
                set.add(num);
            }
        }
        int i = 0;
        int[] result = new int[set.size()];
        for (Integer num : set) {
            result[i++] = num;
        }
        return result;
    }

    public boolean binarySearch(int[] nums, int target) {
        int low = 0;
        int high = nums.length - 1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (nums[mid] == target) {
                return true;
            }
            if (nums[mid] > target) {
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return false;
    }

What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

• If only nums2 cannot fit in memory, put all elements of nums1 into a HashMap, read chunks of array that fit into the memory, and record the intersections.

• If both nums1 and nums2 are so huge that neither fit into the memory, sort them individually (external sort), then read 2 elements from each array at a time in memory, record intersections.