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;
    }
};

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))

############

# 2111. Minimum Operations to Make the Array K-Increasing
# https://leetcode.com/problems/minimum-operations-to-make-the-array-k-increasing/

class Solution:
    def kIncreasing(self, arr: List[int], k: int) -> int:
        A = collections.defaultdict(list)
        
        for i, x in enumerate(arr):
            A[i % k].append(x)
        
        res = 0
        for _, s in A.items():
            lis = []
            
            for x in s:
                index = bisect.bisect(lis, x)
                
                if index < len(lis):
                    lis[index] = x
                else:
                    lis.append(x)
            
            res += len(s) - len(lis)
        
        return res

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) {
                t.add(arr[j]);
            }
            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()) {
                t.add(x);
            } 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;
    }
}

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
}

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2111 - Minimum Operations to Make the Array K-Increasing

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

Solution 1. Longest Increasing Subsequence (LIS)

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

Solution 1. Longest Increasing Subsequence (LIS)

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