Intersection of Two Arrays II
Created: February 18, 2019 by [lek-tin]
Last updated: February 18, 2019
Given two arrays, write a function to compute their intersection.
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]
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?
Solution:
Java
// Time: O(m+n)
class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
if (nums1.length < nums2.length) {
return intersect(nums2, nums1);
}
List<Integer> intersection = new ArrayList<>();
HashMap<Integer, Integer> map1 = new HashMap<>();
HashMap<Integer, Integer> map2 = new HashMap<>();
for (int num: nums1) {
if (map1.get(num) == null) {
map1.put(num, 1);
} else {
map1.put(num, map1.get(num)+1);
}
}
for (int num: nums2) {
if (map2.get(num) == null) {
map2.put(num, 1);
} else {
map2.put(num, map2.get(num)+1);
}
}
for (Map.Entry<Integer, Integer> entry: map1.entrySet()) {
int numOfElements1 = entry.getValue();
int numOfElements2 = map2.get(entry.getKey()) == null ? 0 : map2.get(entry.getKey());
int overlapped = Math.min(numOfElements1, numOfElements2);
for (int i = 0; i < overlapped; i++) {
intersection.add(entry.getKey());
}
}
int[] res = new int[intersection.size()];
for (int i = 0; i < intersection.size(); i++) {
res[i] = intersection.get(i);
}
return res;
}
}
Python
from collections import Counter
class Solution:
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
counter = Counter(nums1)
res = []
for num in nums2:
if counter[num] > 0:
res.append(num)
counter[num] -= 1
return res