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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) { 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; } } 
  • 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 }

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