Intersection of Two Arrays
Created: November 28, 2018 by [lek-tin]
Last updated: November 28, 2018
Given two arrays, write a function to compute their intersection.
Example 1
Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2]
Example 2
Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [9,4]
Note
- Each element in the result must be unique.
- The result can be in any order.
Solution
Use binary search when one of the lists is very long and the other is short
class Solution:
def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
if len(nums1) < len(nums2):
return self.intersection(nums2, nums1)
nums1.sort()
res = []
for target in set(nums2):
left, right = 0, len(nums1)-1
while left <= right:
mid = left + (right-left)//2
if nums1[mid] == target:
res.append(target)
break
elif nums1[mid] < target:
left = mid + 1
else:
right = mid - 1
return res
Solution (hashset)
Python
class Solution:
def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
lookup = {}
res = set()
for num in nums1:
lookup[num] = 1
for num in nums2:
if num in lookup:
res.add(num)
return res
Java
class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
if (nums1.length < nums2.length) {
return intersection(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());
if (Math.min(numOfElements1, numOfElements2) > 0) {
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;
}
}