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📜  用前 K 和后 K 元素的平均值替换给定 Array 的所有元素

📅  最后修改于: 2022-05-13 01:56:05.292000             🧑  作者: Mango

用前 K 和后 K 元素的平均值替换给定 Array 的所有元素

给定一个包含N个正整数和一个整数K的数组arr[] 。任务是用前K个和后K个元素的平均值替换每个数组元素。此外,如果不存在K个元素,则调整使用前后可用的最大元素数。

例子:

天真的方法:最简单的方法是使用嵌套循环。外循环将从左到右遍历数组,即从i = 0i < N ,内循环将从索引i - K到除i之外的索引i + K遍历子数组并计算它们的平均值。

下面是上述方法的实现:

C++
// C++ program for the above approach
#include 
using namespace std;
 
// Function to replace all array elements
// with the average of previous and
// next K elements
void findAverage(int arr[], int N, int K)
{
    int start, end;
    for (int i = 0; i < N; i++) {
        int sum = 0;
 
        // Start limit is max(i-K, 0)
        start = max(i - K, 0);
 
        // End limit in min(i+K, N-1)
        end = min(i + K, N - 1);
        int cnt = end - start;
        for (int j = start; j <= end; j++) {
 
            // Skipping the current element
            if (j == i) {
                continue;
            }
            sum += arr[j];
        }
        cout << sum / cnt << ' ';
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 9, 7, 3, 9, 1, 8, 11 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 2;
    findAverage(arr, N, K);
    return 0;
}


Java
// Java program to implement
// the above approach
class GFG {
 
  // Function to replace all array elements
  // with the average of previous and
  // next K elements
  static void findAverage(int[] arr, int N, int K)
  {
    int start, end;
    for (int i = 0; i < N; i++) {
      int sum = 0;
 
      // Start limit is max(i-K, 0)
      start = Math.max(i - K, 0);
 
      // End limit in min(i+K, N-1)
      end = Math.min(i + K, N - 1);
      int cnt = end - start;
      for (int j = start; j <= end; j++) {
 
        // Skipping the current element
        if (j == i) {
          continue;
        }
        sum += arr[j];
      }
      System.out.print(sum / cnt + " ");
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int[] arr = { 9, 7, 3, 9, 1, 8, 11 };
    int N = arr.length;
    int K = 2;
    findAverage(arr, N, K);
  }
}
 
// This code is contributed by ukasp.


Python3
# Python code for the above approach
 
# Function to replace all array elements
# with the average of previous and
# next K elements
def findAverage(arr, N, K):
    start = None
    end = None
    for i in range(N):
        sum = 0
 
        # Start limit is max(i-K, 0)
        start = max(i - K, 0)
 
        # End limit in min(i+K, N-1)
        end = min(i + K, N - 1)
        cnt = end - start
        for j in range(start, end + 1):
 
            # Skipping the current element
            if j == i:
                continue
            sum += arr[j]
        print((sum // cnt), end= " ")
 
# Driver Code
arr = [9, 7, 3, 9, 1, 8, 11]
N = len(arr)
K = 2
findAverage(arr, N, K)
 
# This code is contributed by gfgking


C#
// C# program to implement
// the above approach
using System;
class GFG
{
 
// Function to replace all array elements
// with the average of previous and
// next K elements
static void findAverage(int []arr, int N, int K)
{
    int start, end;
    for (int i = 0; i < N; i++) {
        int sum = 0;
 
        // Start limit is max(i-K, 0)
        start = Math.Max(i - K, 0);
 
        // End limit in min(i+K, N-1)
        end = Math.Min(i + K, N - 1);
        int cnt = end - start;
        for (int j = start; j <= end; j++) {
 
            // Skipping the current element
            if (j == i) {
                continue;
            }
            sum += arr[j];
        }
        Console.Write(sum / cnt + " ");
    }
}
 
// Driver Code
public static void Main()
{
    int []arr = { 9, 7, 3, 9, 1, 8, 11 };
    int N = arr.Length;
    int K = 2;
    findAverage(arr, N, K);
 
}
}
 
// This code is contributed by Samim Hossain Mondal.


Javascript


C++
// C++ program for the above approach
#include 
using namespace std;
 
// Function to replace all array elements
// with the average of previous and
// next K elements
void findAverage(int arr[], int N, int K)
{
    int i, sum = 0, next, prev, update;
    int cnt = 0;
 
    // Calculate initial sum of K+1 elements
    for (i = 0; i <= K and i < N; i++) {
        sum += arr[i];
        cnt += 1;
    }
 
    // Using the sliding window technique
    for (i = 0; i < N; i++) {
        update = sum - arr[i];
        cout << update / (cnt - 1) << " ";
        next = i + K + 1;
        prev = i - K;
        if (next < N) {
            sum += arr[next];
            cnt += 1;
        }
        if (prev >= 0) {
            sum -= arr[prev];
          cnt-=1;
        }
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 9, 7, 3, 9, 1, 8, 11 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 2;
    findAverage(arr, N, K);
    return 0;
}


Java
// Java program for the above approach
import java.util.*;
public class GFG
{
 
  // Function to replace all array elements
  // with the average of previous and
  // next K elements
  static void findAverage(int arr[], int N, int K)
  {
    int i, sum = 0, next = 0, prev = 0, update = 0;
    int cnt = 0;
 
    // Calculate initial sum of K+1 elements
    for (i = 0; i <= K && i < N; i++) {
      sum += arr[i];
      cnt += 1;
    }
 
    // Using the sliding window technique
    for (i = 0; i < N; i++) {
      update = sum - arr[i];
      System.out.print(update / (cnt - 1) + " ");
      next = i + K + 1;
      prev = i - K;
      if (next < N) {
        sum += arr[next];
        cnt += 1;
      }
      if (prev >= 0) {
        sum -= arr[prev];
        cnt-=1;
      }
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = { 9, 7, 3, 9, 1, 8, 11 };
    int N = arr.length;
    int K = 2;
    findAverage(arr, N, K);
 
  }
}
 
// This code is contributed by Samim Hossain Mondal


Python3
# Python program for the above approach
 
# Function to replace all array elements
# with the average of previous and
# next K elements
def findAverage(arr, N, K):
    sum = 0; next = 0; prev = 0; update = 0;
    cnt = 0;
 
    # Calculate initial sum of K+1 elements
    for i in range(0, K + 1, 1):
        if(i >= N):
            break
        sum += arr[i];
        cnt += 1;
 
    # Using the sliding window technique
    for i in range(0, N):
        update = sum - arr[i];
        print(update // (cnt - 1), end=" ");
        next = i + K + 1;
        prev = i - K;
        if (next < N):
            sum += arr[next];
            cnt += 1;
 
        if (prev >= 0):
            sum -= arr[prev];
            cnt -= 1;
 
# Driver Code
if __name__ == '__main__':
    arr = [9, 7, 3, 9, 1, 8, 11];
    N = len(arr);
    K = 2;
    findAverage(arr, N, K);
 
# This code is contributed by 29AjayKumar


C#
// C# program for the above approach
using System;
 
public class GFG
{
 
  // Function to replace all array elements
  // with the average of previous and
  // next K elements
  static void findAverage(int []arr, int N, int K)
  {
    int i, sum = 0, next = 0, prev = 0, update = 0;
    int cnt = 0;
 
    // Calculate initial sum of K+1 elements
    for (i = 0; i <= K && i < N; i++) {
      sum += arr[i];
      cnt += 1;
    }
 
    // Using the sliding window technique
    for (i = 0; i < N; i++) {
      update = sum - arr[i];
      Console.Write(update / (cnt - 1) + " ");
      next = i + K + 1;
      prev = i - K;
      if (next < N) {
        sum += arr[next];
        cnt += 1;
      }
      if (prev >= 0) {
        sum -= arr[prev];
        cnt-=1;
      }
    }
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int []arr = { 9, 7, 3, 9, 1, 8, 11 };
    int N = arr.Length;
    int K = 2;
    findAverage(arr, N, K);
 
  }
}
 
// This code is contributed by 29AjayKumar


Javascript



输出
5 7 6 4 7 7 4 

时间复杂度: O(N 2 )
辅助空间: O(1)

高效的方法:这种方法使用滑动窗口的方法。按照下面提到的步骤来实现这个概念:

  • 考虑每个元素都有K个下一个和前一个元素,并采用大小为 2*K + 1 的窗口来覆盖整个范围。
  • 现在首先找到前 (K+1) 个元素的总和。
  • 遍历数组时:
    • 通过将总和除以(window-1 的大小)来计算平均值。
    • 在当前窗口的最右端之后添加下一个元素。
    • 移除当前窗口最左边的元素。这会将窗口向右移动一个位置
  • 打印结果数组。

下面是上述方法的实现:

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to replace all array elements
// with the average of previous and
// next K elements
void findAverage(int arr[], int N, int K)
{
    int i, sum = 0, next, prev, update;
    int cnt = 0;
 
    // Calculate initial sum of K+1 elements
    for (i = 0; i <= K and i < N; i++) {
        sum += arr[i];
        cnt += 1;
    }
 
    // Using the sliding window technique
    for (i = 0; i < N; i++) {
        update = sum - arr[i];
        cout << update / (cnt - 1) << " ";
        next = i + K + 1;
        prev = i - K;
        if (next < N) {
            sum += arr[next];
            cnt += 1;
        }
        if (prev >= 0) {
            sum -= arr[prev];
          cnt-=1;
        }
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 9, 7, 3, 9, 1, 8, 11 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 2;
    findAverage(arr, N, K);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
 
  // Function to replace all array elements
  // with the average of previous and
  // next K elements
  static void findAverage(int arr[], int N, int K)
  {
    int i, sum = 0, next = 0, prev = 0, update = 0;
    int cnt = 0;
 
    // Calculate initial sum of K+1 elements
    for (i = 0; i <= K && i < N; i++) {
      sum += arr[i];
      cnt += 1;
    }
 
    // Using the sliding window technique
    for (i = 0; i < N; i++) {
      update = sum - arr[i];
      System.out.print(update / (cnt - 1) + " ");
      next = i + K + 1;
      prev = i - K;
      if (next < N) {
        sum += arr[next];
        cnt += 1;
      }
      if (prev >= 0) {
        sum -= arr[prev];
        cnt-=1;
      }
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = { 9, 7, 3, 9, 1, 8, 11 };
    int N = arr.length;
    int K = 2;
    findAverage(arr, N, K);
 
  }
}
 
// This code is contributed by Samim Hossain Mondal

Python3

# Python program for the above approach
 
# Function to replace all array elements
# with the average of previous and
# next K elements
def findAverage(arr, N, K):
    sum = 0; next = 0; prev = 0; update = 0;
    cnt = 0;
 
    # Calculate initial sum of K+1 elements
    for i in range(0, K + 1, 1):
        if(i >= N):
            break
        sum += arr[i];
        cnt += 1;
 
    # Using the sliding window technique
    for i in range(0, N):
        update = sum - arr[i];
        print(update // (cnt - 1), end=" ");
        next = i + K + 1;
        prev = i - K;
        if (next < N):
            sum += arr[next];
            cnt += 1;
 
        if (prev >= 0):
            sum -= arr[prev];
            cnt -= 1;
 
# Driver Code
if __name__ == '__main__':
    arr = [9, 7, 3, 9, 1, 8, 11];
    N = len(arr);
    K = 2;
    findAverage(arr, N, K);
 
# This code is contributed by 29AjayKumar

C#

// C# program for the above approach
using System;
 
public class GFG
{
 
  // Function to replace all array elements
  // with the average of previous and
  // next K elements
  static void findAverage(int []arr, int N, int K)
  {
    int i, sum = 0, next = 0, prev = 0, update = 0;
    int cnt = 0;
 
    // Calculate initial sum of K+1 elements
    for (i = 0; i <= K && i < N; i++) {
      sum += arr[i];
      cnt += 1;
    }
 
    // Using the sliding window technique
    for (i = 0; i < N; i++) {
      update = sum - arr[i];
      Console.Write(update / (cnt - 1) + " ");
      next = i + K + 1;
      prev = i - K;
      if (next < N) {
        sum += arr[next];
        cnt += 1;
      }
      if (prev >= 0) {
        sum -= arr[prev];
        cnt-=1;
      }
    }
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int []arr = { 9, 7, 3, 9, 1, 8, 11 };
    int N = arr.Length;
    int K = 2;
    findAverage(arr, N, K);
 
  }
}
 
// This code is contributed by 29AjayKumar

Javascript



输出
5 7 6 4 7 7 4 

时间复杂度: O(N)
辅助空间: O(1)