将所有大小为 K 的子数组转换为单个元素所需的最低成本

先决条件：滑动窗口中位数

给定一个由N 个整数和一个整数K组成的数组 arr[] ，任务是找到使每个长度为K 的子数组的每个元素相等所需的最小成本。用另一个元素替换任何数组元素的成本是两者之间的绝对差异。

例子：

Input: A[] = {1, 2, 3, 4, 6}, K = 3
Output: 7
Explanation:
Subarray 1: Cost to convert subarray {1, 2, 3} to {2, 2, 2} = |1-2| + |2-2| + |3-2| = 2
Subarray 2: Cost to convert subarray {2, 3, 4} to {3, 3, 3} = |2-3| + |3-3| + |4-3| = 2
Subarray 3: Cost to convert subarray {3, 4, 6} to {4, 4, 4} = |3-4| + |4-4| + |6-4| = 3
Minimum Cost = 2 + 2 + 3 = 7/

Input: A[] = {2, 3, 4, 4, 1, 7, 6}, K = 4
Output: 21

方法：
要找到将子数组的每个元素转换为单个元素的最小成本，请将子数组的每个元素更改为该子数组的中值。请按照以下步骤解决问题：

• 要有效地找到每个运行子数组的中位数，请使用多重集来获取每个子数组中元素的排序顺序。 Median 将是这个multiset的中间元素。
• 对于下一个子数组，从多重集中删除前一个子数组的最左边的元素，将当前元素添加到多重集中。
• 保持指针mid有效地跟踪多重集的中间元素。
• 如果新添加的元素小于前一个中间元素，则将mid移动到其紧邻较小的元素。否则，将mid移动到它的下一个元素。
• 计算替换子数组每个元素的成本| A[i] – 中值_{each_subarray} | .
• 最后打印总成本。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
  
// Function to find the minimum
// cost to convert each element of
// every subarray of size K equal
int minimumCost(vector arr, int n,
                int k)
{
    // Stores the minimum cost
    int totalcost = 0;
    int i, j;
  
    // Stores the first K elements
    multiset mp(arr.begin(),
                     arr.begin() + k);
  
    if (k == n) {
  
        // Obtain the middle element of
        // the multiset
        auto mid = next(mp.begin(),
                        n / 2 - ((k + 1) % 2));
  
        int z = *mid;
  
        // Calculate cost for the subarray
        for (i = 0; i < n; i++)
            totalcost += abs(z - arr[i]);
  
        // Return the total cost
        return totalcost;
    }
    else {
  
        // Obtain the middle element
        // in multiset
        auto mid = next(mp.begin(),
                        k / 2 - ((k + 1) % 2));
  
        for (i = k; i < n; i++) {
  
            int zz = *mid;
            int cost = 0;
            for (j = i - k; j < i; j++) {
  
                // Cost for the previous
                // k length subarray
                cost += abs(arr[j] - zz);
            }
            totalcost += cost;
  
            // Insert current element
            // into multiset
            mp.insert(arr[i]);
  
            if (arr[i] < *mid) {
  
                // New element appears
                // to the left of mid
                mid--;
            }
  
            if (arr[i - k] <= *mid) {
  
                // New element appears
                // to the right of mid
                mid++;
            }
  
            // Remove leftmost element
            // from the window
            mp.erase(mp.lower_bound(arr[i - k]));
  
            // For last element
            if (i == n - 1) {
                zz = *mid;
                cost = 0;
  
                for (j = i - k + 1;
                     j <= i; j++) {
  
                    // Calculate cost for the subarray
                    cost += abs(zz - arr[j]);
                }
  
                totalcost += cost;
            }
        }
  
        // Return the total cost
        return totalcost;
    }
}
  
// Driver Code
int main()
{
    int N = 5, K = 3;
  
    vector A({ 1, 2, 3, 4, 6 });
  
    cout << minimumCost(A, N, K);
}

输出：

7

时间复杂度： O(NlogN)
辅助空间： O(1)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live