Formatted question description: https://leetcode.ca/all/2111.html

# 2111. Minimum Operations to Make the Array K-Increasing (Hard)

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

Similar Questions:

## Solution 1. Longest Increasing Subsequence (LIS)

Split the numbers in A into k groups: [0, k, 2k, 3k, ...], [1, 1+k, 1+2k, 1+3k, ...], …

Compute the minimum operations needed to make a group non-decreasing. Assume the Longest Increasing Subsequence (LIS) of this group is of length t, and the group is of length len, then the minimum operations needed is len - t.

Computing the length of LIS is a classic problem that can be solved using binary search. See 300. Longest Increasing Subsequence (Medium)

The answer is the sum of minimum operations for all the groups.

// OJ: https://leetcode.com/problems/minimum-operations-to-make-the-array-k-increasing/
// Time: O(Nlog(N/k))
// Space: O(N/k)
class Solution {
public:
int kIncreasing(vector<int>& A, int k) {
int N = A.size(), ans = 0;
for (int i = 0; i < k; ++i) { // handle each group separately
vector<int> lis{A[i]}; // the longest non-decreasing subsequence of this group
int len = 1; // the length of this group
for (int j = i + k; j < N; j += k, ++len) {
auto i = upper_bound(begin(lis), end(lis), A[j]);
if (i == end(lis)) lis.push_back(A[j]);
else *i = A[j];
}
ans += len - lis.size();
}
return ans;
}
};


## Discuss

https://leetcode.com/problems/minimum-operations-to-make-the-array-k-increasing/discuss/1634980/C%2B%2B-Longest-Increasing-Subsequence