350. Intersection of Two Arrays II (Easy)

https://leetcode.com/problems/intersection-of-two-arrays-ii/

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

Example 1:

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

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]

Note:

• Each element in the result should appear as many times as it shows in both arrays.
• The result can be in any order.

Follow up:

• What if the given array is already sorted? How would you optimize your algorithm?
• What if nums1's size is small compared to nums2's size? Which algorithm is better?
• 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?

Solutions

1.

class Solution {

    // Use map to keep the count of each element in nums1, then compare with nums2.

    public int[] intersect(int[] nums1, int[] nums2) {
        if (nums1 == null || nums1.length == 0 || nums2 == null || nums1.length == 0) {
            return new int[0];
        }

        List<Integer> tmp = new ArrayList<>();

        Map<Integer, Integer> freq = new HashMap<>();
        for (int i = 0; i < nums1.length; i++) {
            if (!freq.containsKey(nums1[i])) {
                freq.put(nums1[i], 0);
            }

            freq.put(nums1[i], freq.get(nums1[i]) + 1);
        }

        for (int i = 0; i < nums2.length; i++) {
            if (!freq.containsKey(nums2[i]) || freq.get(nums2[i]) == 0) {
                continue;
            }

            freq.put(nums2[i], freq.get(nums2[i]) - 1);
            tmp.add(nums2[i]);
        }

        int[] ans = new int[tmp.size()];
        for (int i = 0; i < tmp.size(); i++) {
            ans[i] = tmp.get(i);
        }

        return ans;
    }
}

2.

class Solution {

    public int[] intersect(int[] nums1, int[] nums2) {

        if (nums1 == null || nums1.length == 0 || nums2 == null || nums1.length == 0) {
            return new int[0];
        }

        Arrays.sort(nums1);
        Arrays.sort(nums2);

        int i = 0;
        int j = 0;

        ArrayList<Integer> list = new ArrayList<>();

        while (i < nums1.length && j < nums2.length) {
            /*
            // Do not separate the condition statements since i, j may be changed.
            // It will affect the followup code if not move to the next iteration.

            if (nums1[i] > nums2[j]) {
                j++;
            }

            if (nums1[i] < nums2[j]) {
                i++;
            }

            if (nums1[i] == nums2[j]) {
                list.add(nums1[i]);

                i++;
                j++;
            }
            */

            if (nums1[i] > nums2[j]) {
                j++;
            } else if (nums1[i] < nums2[j]) {
                i++;
            } else { // nums1[i] == nums2[j]
                list.add(nums1[i]);

                i++;
                j++;
            }
        }

        int[] ans = new int[list.size()];
        for (i = 0; i < list.size(); ++i) {
            ans[i] = list.get(i);
        }

        return ans;
    }
}

Incorrect Solutions

References