LeetCode in Kotlin
2592. Maximize Greatness of an Array
Medium
You are given a 0-indexed integer array nums. You are allowed to permute nums into a new array perm of your choosing.
We define the greatness of nums be the number of indices 0 <= i < nums.length for which perm[i] > nums[i].
Return the maximum possible greatness you can achieve after permuting nums.
Example 1:
Input: nums = [1,3,5,2,1,3,1]
Output: 4
Explanation: One of the optimal rearrangements is perm = [2,5,1,3,3,1,1]. At indices = 0, 1, 3, and 4, perm[i] > nums[i]. Hence, we return 4.
Example 2:
Input: nums = [1,2,3,4]
Output: 3
Explanation: We can prove the optimal perm is [2,3,4,1]. At indices = 0, 1, and 2, perm[i] > nums[i]. Hence, we return 3.
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 109
Solution
import java.util.TreeMap
class Solution {
fun maximizeGreatness(nums: IntArray): Int {
val map = TreeMap<Int, Int>()
for (num in nums) map.compute(num) { _, n -> (n ?: 0) + 1 }
var count = 0
for (num in nums) {
val entry = map.higherEntry(num)
if (entry != null && entry.key != num) {
count++
if (entry.value - 1 == 0) map.remove(entry.key)
else map[entry.key] = entry.value - 1
}
}
return count
}
}