从所有元素中减去最大值 K 次后找到最终数组

给定一个包含N个整数的数组arr[]和一个整数K，任务是对该数组执行以下操作K次。操作如下：

从数组中找到最大元素（比如M ）。
现在用Ma _i替换每个元素_。其中 1 ≤ i ≤ N。

例子：

Input: N = 6, K = 3, arr[] = {5, 38, 4, 96, 103, 41}
Output: 98 65 99 7 0 62
Explanation:
1st operation => Maximum array element is 103. Subtract 103 from all elements. arr[] = {98, 65, 99, 7, 0, 62}.
1st operation => Maximum array element is 99. Subtract 99 from all elements. arr[] = {1, 34, 0, 92, 99, 37}
1st operation => Maximum array element is 99. Subtract 99 from all elements. arr[] = {98, 65, 99, 7, 0, 62}.
This will the last state.

Input: N = 5, K = 1, arr[] = {8, 4, 3, 10, 15}
Output: 7 11 12 5 0

编程需要懂一点英语

Naive Approach：简单的方法是将上述操作执行K次。每次在数组中找到最大元素，然后更新与最大值不同的所有元素。

时间复杂度： O(N*K)。
辅助空间： O(1)。

有效方法：此问题的有效解决方案基于以下观察：

If K is odd then the final answer will be equivalent to the answer after applying the operation only one time and if K is even then the final array will be equivalent to the array after applying operations only two time.

This can be proved by as per the following:

Say maximum = M, and initial array is arr[].
After the operations are performed 1st time:
=> The maximum element of arr[] becomes 0.
=> All other elements are M – arr[i].
=> Say the new array is a1[]. So, a1[i] = M – arr[i].

After the operation are performed 2nd time:
=> Say maximum of a1[] is M1.
=> The maximum element of a1[] becomes 0.
=> All other elements are M1 – a1[i].
=> Say the new array is a2[]. So, a2[i] = M1 – a1[i].
=> Maximum of a2[] will be M1, because the 0 present in a1[] changes to M1 and all other elements are less than M1.

After the operation are performed 3rd time:
=> The maximum is M1.
=> The maximum changes to 0.
=> All other elements are M1 – a2[i] = M1 – (M1 – a1[i]) = a1[i].
=> So the new array is same as a1[].

After the operation are performed 4th time:
=> Say maximum of the array is M1.
=> The maximum element changes to be 0.
=> All other elements are M1 – a1[i].
=> So the new array is the same as a2[]

From the above it is clear that the alternate states of the array are same.

编程需要懂一点英语

按照以下步骤实施观察：

取一个变量max并存储arr的最大元素。
如果K是奇数
- 遍历数组并从最大元素中减去每个元素。
别的
- 遍历arr并从最大元素中减去每个元素，然后
  将最大元素存储在max1变量中。
- 再次遍历arr并从最大元素中减去每个元素。
打印最终数组。

下面是上述方法的实现：

C++

// C++ code for the approach
  
#include 
using namespace std;
  
// Function for converting the array
void find(int n, int k, int arr[])
{
    // Find the maximum element in array
    int max = INT_MIN;
    for (int i = 0; i < n; i++) {
  
        if (arr[i] > max) {
            max = arr[i];
        }
    }
  
    // If k is odd
    if (k % 2 != 0) {
        for (int i = 0; i < n; i++) {
            cout << max - arr[i] << " ";
        }
    }
    // If k is even
    else {
  
        // Subtract the max from every
        // element of array and store
        // the next maximum element in max1
        int max1 = INT_MIN;
        for (int i = 0; i < n; i++) {
            arr[i] = max - arr[i];
            if (arr[i] > max1) {
                max1 = arr[i];
            }
        }
  
        // Print the output
        for (int i = 0; i < n; i++) {
            cout << max1 - arr[i] << " ";
        }
    }
}
  
// Driver code
int main()
{
    int N = 6, K = 3;
    int arr[] = { 5, 38, 4, 96, 103, 41 };
  
    // Function call
    find(N, K, arr);
    return 0;
}

输出

98 65 99 7 0 62

时间复杂度： O(N)
辅助空间： O(1)