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 = 1Output:4Explanation: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], [1,2,3,2,4]. All of them require 4 operations. It is suboptimal to change the array to, for example, [4,6,7,8,9] because it would take 5 operations. It can be shown that we cannot make the array K-increasing in less than 4 operations.10

**Example 2:**

Input:arr = [4,1,5,2,6,2], k = 2Output:0Explanation: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 = 3Output:2Explanation: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,,6,4]. Note that there can be other ways to make the array K-increasing, but none of them require less than 2 operations.5

**Constraints:**

`1 <= arr.length <= 10`

^{5}`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