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

## Description

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

## Solutions

• class Solution {
public int kIncreasing(int[] arr, int k) {
int n = arr.length;
int ans = 0;
for (int i = 0; i < k; ++i) {
List<Integer> t = new ArrayList<>();
for (int j = i; j < n; j += k) {
}
ans += lis(t);
}
return ans;
}

private int lis(List<Integer> arr) {
List<Integer> t = new ArrayList<>();
for (int x : arr) {
int idx = searchRight(t, x);
if (idx == t.size()) {
} else {
t.set(idx, x);
}
}
return arr.size() - t.size();
}

private int searchRight(List<Integer> arr, int x) {
int left = 0, right = arr.size();
while (left < right) {
int mid = (left + right) >> 1;
if (arr.get(mid) > x) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
}

• class Solution {
public:
int kIncreasing(vector<int>& arr, int k) {
int ans = 0, n = arr.size();
for (int i = 0; i < k; ++i) {
vector<int> t;
for (int j = i; j < n; j += k) t.push_back(arr[j]);
ans += lis(t);
}
return ans;
}

int lis(vector<int>& arr) {
vector<int> t;
for (int x : arr) {
auto it = upper_bound(t.begin(), t.end(), x);
if (it == t.end())
t.push_back(x);
else
*it = x;
}
return arr.size() - t.size();
}
};

• class Solution:
def kIncreasing(self, arr: List[int], k: int) -> int:
def lis(arr):
t = []
for x in arr:
idx = bisect_right(t, x)
if idx == len(t):
t.append(x)
else:
t[idx] = x
return len(arr) - len(t)

return sum(lis(arr[i::k]) for i in range(k))


• func kIncreasing(arr []int, k int) int {
searchRight := func(arr []int, x int) int {
left, right := 0, len(arr)
for left < right {
mid := (left + right) >> 1
if arr[mid] > x {
right = mid
} else {
left = mid + 1
}
}
return left
}

lis := func(arr []int) int {
var t []int
for _, x := range arr {
idx := searchRight(t, x)
if idx == len(t) {
t = append(t, x)
} else {
t[idx] = x
}
}
return len(arr) - len(t)
}

n := len(arr)
ans := 0
for i := 0; i < k; i++ {
var t []int
for j := i; j < n; j += k {
t = append(t, arr[j])
}
ans += lis(t)
}
return ans
}