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[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.

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[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;
    }
}

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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