LeetCode in Kotlin
1636. Sort Array by Increasing Frequency
Easy
Given an array of integers nums, sort the array in increasing order based on the frequency of the values. If multiple values have the same frequency, sort them in decreasing order.
Return the sorted array.
Example 1:
Input: nums = [1,1,2,2,2,3]
Output: [3,1,1,2,2,2]
Explanation: ‘3’ has a frequency of 1, ‘1’ has a frequency of 2, and ‘2’ has a frequency of 3.
Example 2:
Input: nums = [2,3,1,3,2]
Output: [1,3,3,2,2]
Explanation: ‘2’ and ‘3’ both have a frequency of 2, so they are sorted in decreasing order.
Example 3:
Input: nums = [-1,1,-6,4,5,-6,1,4,1]
Output: [5,-1,4,4,-6,-6,1,1,1]
Constraints:
1 <= nums.length <= 100-100 <= nums[i] <= 100
Solution
import java.util.Collections
import java.util.TreeMap
class Solution {
fun frequencySort(nums: IntArray): IntArray {
val count: MutableMap<Int, Int> = HashMap()
for (num in nums) {
count[num] = count.getOrDefault(num, 0) + 1
}
val map = TreeMap<Int, MutableList<Int>>()
for ((key, freq) in count) {
map.putIfAbsent(freq, ArrayList())
val list = map[freq]!!
list.add(key)
map[freq] = list
}
val result = IntArray(nums.size)
var i = 0
for (entry in map.entries) {
val list = entry.value
list.sortWith(Collections.reverseOrder<Any>())
var k = entry.key
var j = 0
while (j < list.size) {
while (k-- > 0) {
result[i++] = list[j]
}
j++
k = entry.key
}
}
return result
}
}