LeetCode in Kotlin

2215. Find the Difference of Two Arrays

Easy

Given two 0-indexed integer arrays nums1 and nums2, return a list answer of size 2 where:

• answer[0] is a list of all distinct integers in nums1 which are not present in nums2.
• answer[1] is a list of all distinct integers in nums2 which are not present in nums1.

Note that the integers in the lists may be returned in any order.

Example 1:

Input: nums1 = [1,2,3], nums2 = [2,4,6]

Output: [[1,3],[4,6]]

Explanation:

For nums1, nums1[1] = 2 is present at index 0 of nums2, whereas nums1[0] = 1 and nums1[2] = 3 are not present in nums2. Therefore, answer[0] = [1,3].

For nums2, nums2[0] = 2 is present at index 1 of nums1, whereas nums2[1] = 4 and nums2[2] = 6 are not present in nums2. Therefore, answer[1] = [4,6].

Example 2:

Input: nums1 = [1,2,3,3], nums2 = [1,1,2,2]

Output: [[3],[]]

Explanation:

For nums1, nums1[2] and nums1[3] are not present in nums2. Since nums1[2] == nums1[3], their value is only included once and answer[0] = [3].

Every integer in nums2 is present in nums1. Therefore, answer[1] = [].

Constraints:

• 1 <= nums1.length, nums2.length <= 1000
• -1000 <= nums1[i], nums2[i] <= 1000

Solution

class Solution {
    fun findDifference(nums1: IntArray, nums2: IntArray): List<List<Int>> {
        val set1 = createSet(nums1)
        val set2 = createSet(nums2)
        return listOf(getMissing(set1, set2), getMissing(set2, set1))
    }

    private fun createSet(array: IntArray): Set<Int> {
        val set: MutableSet<Int> = HashSet()
        for (x in array) {
            set.add(x)
        }
        return set
    }

    private fun getMissing(first: Set<Int>, second: Set<Int>): List<Int> {
        val list: MutableList<Int> = ArrayList()
        for (x in first) {
            if (!second.contains(x)) {
                list.add(x)
            }
        }
        return list
    }
}

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