LeetCode2111. Minimum Operations to Make the Array K-Increasing—

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一、题目

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文章来源地址https://www.toymoban.com/news/detail-837458.html

二、题解

class Solution {
public:
    int nums[100010];
    int ends[100010];
    int kIncreasing(vector<int>& arr, int k) {
        int n = arr.size();
        int res = 0;
        for(int i = 0;i < k;i++){
            int size = 0;
            for(int j = i;j < n;j += k){
                nums[size++] = arr[j];
            }
            res += size - lengthOfLIS(size);
        }
        return res;
    }
    int lengthOfLIS(int size) {
        int len = 0;
        for(int i = 0;i < size;i++){
            int index = binarySearch(len,nums[i]);
            if(index == -1) ends[len++] = nums[i];
            else ends[index] = nums[i];
        }
        return len;
    }
    int binarySearch(int len,int num){
        int l = 0, r = len - 1,res = -1;
        while(l <= r){
            int mid = (l + r) / 2;
            if(ends[mid] > num){
                res = mid;
                r = mid - 1;
            }
            else l = mid + 1;
        }
        return res;
    }
};