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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2201_2300.s2248_intersection_of_multiple_arrays;

// #Easy #Array #Hash_Table #Counting #2022_06_09_Time_2_ms_(99.46%)_Space_42.6_MB_(92.53%)

import java.util.ArrayList;
import java.util.List;

/**
 * 2248 - Intersection of Multiple Arrays\.
 *
 * Easy
 *
 * 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**.
**/
public class Solution {
    public List intersection(int[][] nums) {
        List 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;
    }
}