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2111. Minimum Operations to Make the Array K-Increasing

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 for k = 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 for k = 1 (because arr[0] > arr[1]) or k = 3 (because arr[0] > arr[3]).

In one operation, you can choose an index i and changearr[i] into any positive integer.

Return the minimum number of operations required to make the array K-increasing for the givenk.

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

Solutions (Python)

1. Solution

classSolution: defkIncreasing(self, arr: List[int], k: int) ->int: groups= [[] for_inrange(k)] ret=0foriinrange(len(arr)): groups[i%k].append(arr[i]) forgroupingroups: lis= [0] forxingroup: i=bisect.bisect(lis, x) ifi==len(lis): lis.append(x) else: lis[i] =xret+=len(group) -len(lis) +1returnret
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