花花酱 LeetCode 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 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

Solution: Longest increasing subsequence

if k = 1, we need to modify the following arrays
1. [a[0], a[1], a[2], …]
if k = 2, we need to modify the following arrays
1. [a[0], a[2], a[4], …]
2. [a[1], a[3], a[5], …]
if k = 3, we need to modify the following arrays
1. [a[0], a[3], a[6], …]
2. [a[1], a[4], a[7], …]
3. [a[2], a[5], a[8], …]
…

These arrays are independent of each other, we just need to find LIS of it, # ops = len(arr) – LIS(arr).
Ans = sum(len(arr_i) – LIS(arr_i)) 1 <= i <= k

Reference: 花花酱 LeetCode 300. Longest Increasing Subsequence

Time complexity: O(k * (n/k)* log(n/k)) = O(n * log(n/k))
Space complexity: O(n/k)