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Formatted question description: https://leetcode.ca/all/2248.html
2248. Intersection of Multiple Arrays
- Difficulty: Easy.
- Related Topics: Array, Hash Table, Counting.
- Similar Questions: Intersection of Two Arrays, Intersection of Two Arrays II, Find Smallest Common Element in All Rows, Intersection of Three Sorted Arrays, Find the Difference of Two Arrays.
Problem
Given a 2D integer array nums
where nums[i]
is a non-empty array of distinct positive integers, return the list of integers that are present in **each array of** nums
** sorted in ascending order.
**Example 1:
Input: nums = [[3,1,2,4,5],[1,2,3,4],[3,4,5,6]]
Output: [3,4]
Explanation:
The only integers present in each of nums[0] = [3,1,2,4,5], nums[1] = [1,2,3,4], and nums[2] = [3,4,5,6] are 3 and 4, so we return [3,4].
Example 2:
Input: nums = [[1,2,3],[4,5,6]]
Output: []
Explanation:
There does not exist any integer present both in nums[0] and nums[1], so we return an empty list [].
Constraints:
-
1 <= nums.length <= 1000
-
1 <= sum(nums[i].length) <= 1000
-
1 <= nums[i][j] <= 1000
-
All the values of
nums[i]
are unique.
Solution (Java, C++, Python)
-
class Solution { public List<Integer> intersection(int[][] nums) { List<Integer> ans = new ArrayList<>(); int[] count = new int[1001]; for (int[] arr : nums) { for (int i : arr) { ++count[i]; } } for (int i = 0; i < count.length; i++) { if (count[i] == nums.length) { ans.add(i); } } return ans; } }
-
Todo
-
class Solution: def intersection(self, nums: List[List[int]]) -> List[int]: cnt = [0] * 1001 for arr in nums: for x in arr: cnt[x] += 1 return [x for x, v in enumerate(cnt) if v == len(nums)] ############ # 2248. Intersection of Multiple Arrays # https://leetcode.com/problems/intersection-of-multiple-arrays/ class Solution: def intersection(self, nums: List[List[int]]) -> List[int]: n = len(nums) count = Counter() for arr in nums: for x in arr: count[x] += 1 res = [] for k, v in count.items(): if v == n: res.append(k) return sorted(res)
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).