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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).

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