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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 = [3,1,2,4,5], nums = [1,2,3,4], and nums = [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 and nums, 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;
for (int[] arr : nums) {
for (int i : arr) {
++count[i];
}
}
for (int i = 0; i < count.length; i++) {
if (count[i] == nums.length) {
}
}
return ans;
}
}

• Todo

• class Solution:
def intersection(self, nums: List[List[int]]) -> List[int]:
cnt =  * 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).