Question

2111. Minimum Operations to Make the Array K-Increasing

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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

Solution 1

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))

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();
  }
};

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
}