2248 - Intersection of Multiple Arrays (Easy)

Problem Link

https://leetcode.com/problems/intersection-of-multiple-arrays/

Problem Statement

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.

Approach 1: Brute Force

Observation: Each element of the final output would have a frequency of $nums.length$.

We use hash map to store the frequency for each integer. Iterate the map and look for the those with $occurrence ==nums.length$.

Written by @wingkwong

class Solution {
public:
    vector<int> intersection(vector<vector<int>>& nums) {
        int n = nums.size();
        unordered_map<int, int> m;
        vector<int> ans;
        // count each integer
        for (auto x : nums)  for (auto y : x) m[y]++;
        // if the count is equal to n, then take this integer
        for (auto x : m) if (x.second == n) ans.push_back(x.first);
        // sort in ascending order
        sort(ans.begin(), ans.end());
        return ans;
    }
};