LeetCode in Kotlin
350. Intersection of Two Arrays II
Easy
Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.
Example 1:
Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]
Example 2:
Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]
Explanation: [9,4] is also accepted.
Constraints:
1 <= nums1.length, nums2.length <= 10000 <= nums1[i], nums2[i] <= 1000
Follow up:
- What if the given array is already sorted? How would you optimize your algorithm?
- What if
nums1’s size is small compared tonums2’s size? Which algorithm is better? - What if elements of
nums2are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?
Solution
class Solution {
fun intersect(nums1: IntArray, nums2: IntArray): IntArray {
val a: MutableMap<Int, Int> = mutableMapOf()
for (i in 0 until nums1.size) {
a[nums1[i]] = a.getOrDefault(nums1[i], 0) + 1
}
var s = MutableList<Int>(0) { 0 }
for (i in 0 until nums2.size) {
if (a.getOrDefault(nums2[i], 0) > 0) {
s.add(nums2[i])
a[nums2[i]] = a.getValue(nums2[i]) - 1
}
}
return s.toIntArray()
}
}