Intersection of Multiple Arrays

// Accepted solution for LeetCode #2248: Intersection of Multiple Arrays
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) {
                ans.add(x);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2248: Intersection of Multiple Arrays
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
}

# Accepted solution for LeetCode #2248: Intersection of Multiple Arrays
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)]

// Accepted solution for LeetCode #2248: Intersection of Multiple Arrays
/**
 * [2248] Intersection of Multiple Arrays
 *
 * 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 = [[<u>3</u>,1,2,<u>4</u>,5],[1,2,<u>3</u>,<u>4</u>],[<u>3</u>,<u>4</u>,5,6]]
 * Output: [3,4]
 * Explanation:
 * The only integers present in each of nums[0] = [<u>3</u>,1,2,<u>4</u>,5], nums[1] = [1,2,<u>3</u>,<u>4</u>], and nums[2] = [<u>3</u>,<u>4</u>,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.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/intersection-of-multiple-arrays/
// discuss: https://leetcode.com/problems/intersection-of-multiple-arrays/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn intersection(nums: Vec<Vec<i32>>) -> Vec<i32> {
        let target_count = nums.len();
        nums.into_iter()
            .flatten()
            .fold(std::collections::HashMap::new(), |mut counts, n| {
                *counts.entry(n).or_insert(0_usize) += 1;
                counts
            })
            .into_iter()
            .filter_map(|(n, count)| (count == target_count).then_some(n))
            .collect::<std::collections::BinaryHeap<_>>()
            .into_sorted_vec()
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2248_example_1() {
        let nums = vec![vec![3, 1, 2, 4, 5], vec![1, 2, 3, 4], vec![3, 4, 5, 6]];

        let result = vec![3, 4];

        assert_eq!(Solution::intersection(nums), result);
    }

    #[test]
    fn test_2248_example_2() {
        let nums = vec![vec![1, 2, 3], vec![4, 5, 6]];

        let result: Vec<i32> = vec![];

        assert_eq!(Solution::intersection(nums), result);
    }
}

// Accepted solution for LeetCode #2248: Intersection of Multiple Arrays
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;
}