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# Intersection of Multiple Arrays LeetCode Solution

## Problem – Intersection of Multiple Arrays LeetCode Solution

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.

### Intersection of Multiple Arrays LeetCode Solution in Java

``````
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;
}
}
``````

### Intersection of Multiple Arrays LeetCode Solution in C++

``````vector<int> intersection(vector<vector<int>>& nums) {
vector<int> cnt(1001), res;
for (auto &arr: nums)
for (int n : arr)
++cnt[n];
for (int i = 0; i < cnt.size(); ++i)
if (cnt[i] == nums.size())
res.push_back(i);
return res;
}
``````

### Intersection of Multiple Arrays LeetCode Solution in Python

``````	res=[]
arr=[]
n=len(nums)
for num in nums:
arr.extend(num)
count=Counter(arr)
for i in count:
if(count[i]==n):
res.append(i)
return sorted(res) if res else res
``````
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