- Difficulty: Hard
- Tags: LeetCode, Hard, Array, Binary Search, leetcode-2111, O(nlog(n/k)), O(n/k)
Problem
You are given a 0-indexed array arr
consisting of n
positive integers, and a positive integer k
.
The array arr
is called K-increasing if arr[i-k] <= arr[i]
holds for every index i
, where k <= i <= n-1
.
- For example,
arr = [4, 1, 5, 2, 6, 2]
is K-increasing fork = 2
because:arr[0] <= arr[2] (4 <= 5)
arr[1] <= arr[3] (1 <= 2)
arr[2] <= arr[4] (5 <= 6)
arr[3] <= arr[5] (2 <= 2)
- However, the same
arr
is not K-increasing fork = 1
(becausearr[0] > arr[1]
) ork = 3
(becausearr[0] > arr[3]
).
In one operation, you can choose an index i
and change arr[i]
into any positive integer.
Return the minimum number of operations required to make the array K-increasing for the given k
.
Example 1:
Input: arr = [5,4,3,2,1], k = 1 Output: 4 Explanation: For k = 1, the resultant array has to be non-decreasing. Some of the K-increasing arrays that can be formed are [5,6,7,8,9], [1,1,1,1,1], [2,2,3,4,4]. All of them require 4 operations. It is suboptimal to change the array to, for example, [6,7,8,9,10] because it would take 5 operations. It can be shown that we cannot make the array K-increasing in less than 4 operations.
Example 2:
Input: arr = [4,1,5,2,6,2], k = 2 Output: 0 Explanation: This is the same example as the one in the problem description. Here, for every index i where 2 <= i <= 5, arr[i-2] <= arr[i]. Since the given array is already K-increasing, we do not need to perform any operations.
Example 3:
Input: arr = [4,1,5,2,6,2], k = 3 Output: 2 Explanation: Indices 3 and 5 are the only ones not satisfying arr[i-3] <= arr[i] for 3 <= i <= 5. One of the ways we can make the array K-increasing is by changing arr[3] to 4 and arr[5] to 5. The array will now be [4,1,5,4,6,5]. Note that there can be other ways to make the array K-increasing, but none of them require less than 2 operations.
Constraints:
1 <= arr.length <= 105
1 <= arr[i], k <= arr.length