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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) {
}
}
return ans;
}
}

############

class Solution {
public List<Integer> intersection(int[][] nums) {
int[] cnt = new int[1001];
for (var arr : nums) {
for (int x : arr) {
++cnt[x];
}
}
List<Integer> ans = new ArrayList<>();
for (int x = 0; x < 1001; ++x) {
if (cnt[x] == nums.length) {
}
}
return ans;
}
}

• 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)


• class Solution {
public:
vector<int> intersection(vector<vector<int>>& nums) {
int cnt[1001]{};
for (auto& arr : nums) {
for (int& x : arr) {
++cnt[x];
}
}
vector<int> ans;
for (int x = 0; x < 1001; ++x) {
if (cnt[x] == nums.size()) {
ans.push_back(x);
}
}
return ans;
}
};

• func intersection(nums [][]int) (ans []int) {
cnt := [1001]int{}
for _, arr := range nums {
for _, x := range arr {
cnt[x]++
}
}
for x, v := range cnt {
if v == len(nums) {
ans = append(ans, x)
}
}
return
}

• function intersection(nums: number[][]): number[] {
const cnt = new Array(1001).fill(0);
for (const arr of nums) {
for (const x of arr) {
cnt[x]++;
}
}
const ans: number[] = [];
for (let x = 0; x < 1001; x++) {
if (cnt[x] === nums.length) {
ans.push(x);
}
}
return ans;
}


• class Solution {
/**
* @param Integer[][] $nums * @return Integer[] */ function intersection($nums) {
$rs = []; for ($i = 0; $i < count($nums); $i++) { for ($j = 0; $j < count($nums[$i]);$j++) {
$hashtable[$nums[$i][$j]] += 1;
if ($hashtable[$nums[$i][$j]] === count($nums)) array_push($rs, $nums[$i][$j]); } } sort($rs);
return \$rs;
}
}


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).