[LeetCode] 350. Intersection of Two Arrays II

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

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

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?

Thought process:

Iterate through both arrays. Count the number of each number and put the counts into two maps. Iterate through both maps. For numbers that appear in both maps, put min(count1, count2) of that number into result.

Solution:

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class Solution { public int[] intersect(int[] nums1, int[] nums2) { Map<Integer, Integer> map1 = countNums(nums1); Map<Integer, Integer> map2 = countNums(nums2); List<Integer> intersect = new ArrayList<>(); for (int num : map1.keySet()) { if (map2.containsKey(num)) { for (int i = 0; i < Math.min(map1.get(num), map2.get(num)); i++) { intersect.add(num); } } } int[] array = new int[intersect.size()]; for (int i = 0; i < array.length; i++) { array[i] = intersect.get(i); } return array; } private Map<Integer, Integer> countNums(int[] nums) { Map<Integer, Integer> map = new HashMap<>(); for (int num : nums) { int count = map.getOrDefault(num, 0); map.put(num, count + 1); } return map; } }

Time complexity: O(a + b).

Follow-up:

• If the given arrays are already sorted, I can use two pointers at the beginning of both arrays respectively. Iterate through both arrays together and get the common numbers.