LeetCode in Kotlin

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.

Solution

class Solution {
    fun intersection(nums: Array<IntArray>): List<Int> {
        val ans: MutableList<Int> = ArrayList()
        val count = IntArray(1001)
        for (arr in nums) {
            for (i in arr) {
                ++count[i]
            }
        }
        for (i in count.indices) {
            if (count[i] == nums.size) {
                ans.add(i)
            }
        }
        return ans
    }
}

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