LeetCode in Kotlin
2366. Minimum Replacements to Sort the Array
Hard
You are given a 0-indexed integer array nums. In one operation you can replace any element of the array with any two elements that sum to it.
- For example, consider
nums = [5,6,7]. In one operation, we can replacenums[1]with2and4and convertnumsto[5,2,4,7].
Return the minimum number of operations to make an array that is sorted in non-decreasing order.
Example 1:
Input: nums = [3,9,3]
Output: 2
Explanation: Here are the steps to sort the array in non-decreasing order:
-
From [3,9,3], replace the 9 with 3 and 6 so the array becomes [3,3,6,3]
-
From [3,3,6,3], replace the 6 with 3 and 3 so the array becomes [3,3,3,3,3]
There are 2 steps to sort the array in non-decreasing order. Therefore, we return 2.
Example 2:
Input: nums = [1,2,3,4,5]
Output: 0
Explanation: The array is already in non-decreasing order. Therefore, we return 0.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 109
Solution
class Solution {
fun minimumReplacement(nums: IntArray): Long {
var limit = nums[nums.size - 1]
var ans: Long = 0
for (i in nums.size - 2 downTo 0) {
var replacements = nums[i] / limit - 1
if (nums[i] % limit != 0) {
replacements++
}
ans += replacements.toLong()
limit = nums[i] / (replacements + 1)
}
return ans
}
}