Replace all elements of given Array with average of previous K and next K elements

Last Updated : 27 Jan, 2022

Given an array arr[] containing N positive integers and an integer K. The task is to replace every array element with the average of previous K and next K elements. Also, if K elements are not present then adjust use the maximum number of elements available before and after.

Examples:

Input: arr[] = {9, 7, 3, 9, 1, 8, 11}, K=2
Output: 5 7 6 4 7 7 4
Explanation: For i = 0, average = (7 + 3)/2 = 5
For i = 1, average = (9 + 3 + 9)/3 = 7
For i = 2, average = (9 + 7 + 9 + 1)/4 = 6
For i = 3, average = (7 + 3 + 1 + 8)/4 = 4
For i = 4, average = (3 + 9 + 8 + 11)/4 = 7
For i = 5, average = (9 + 1 + 11)/3 = 7
For i = 6, average = (1 + 8)/2 = 4

Input: arr[] = {13, 26, 35, 41, 23, 18, 38}, K=3
Output: 34 28 24 25 31 34 27

Naive Approach: The simplest approach is to use nested loops. The outer loop will traverse the array from left to right, i.e. from i = 0 to i < N and an inner loop will traverse the subarray from index i - K to the index i + K except i and calculate the average of them.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
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# 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

  <script>
        // JavaScript code for the above approach

        // Function to replace all array elements
        // with the average of previous and 
        // next K elements
        function findAverage(arr, N, K)
        {
            let start, end;
            for (let i = 0; i < N; i++) 
            {
                let 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);
                let cnt = end - start;
                for (let j = start; j <= end; j++) {

                    // Skipping the current element
                    if (j == i) {
                        continue;
                    }
                    sum += arr[j];
                }
                document.write(Math.floor(sum / cnt) + ' ');
            }
        }

        // Driver Code
        let arr = [9, 7, 3, 9, 1, 8, 11];
        let N = arr.length;
        let K = 2;
        findAverage(arr, N, K);

  // This code is contributed by Potta Lokesh
    </script>

Output

5 7 6 4 7 7 4

Time complexity: O(N²)
Auxiliary Space: O(1)

Efficient approach: This approach uses the sliding window method. Follow the steps mentioned below to implement this concept:

Consider that every element has K next and previous elements and take an window of size 2*K + 1 to cover this whole range.
Now initially find the sum of first (K+1) elements.
While traversing the array:
- Calculate the average by dividing the sum with (size of window-1).
- Add the next element after the rightmost end of the current window.
- Remove the leftmost element of the current window. This will shift the window one position to right
Print the resultant array.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
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# 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

<script>
    // JavaScript program for the above approach

    // Function to replace all array elements
    // with the average of previous and
    // next K elements
    const findAverage = (arr, N, K) => {
        let i, sum = 0, next, prev, update;
        let 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];
            document.write(`${parseInt(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

    let arr = [9, 7, 3, 9, 1, 8, 11];
    let N = arr.length;
    let K = 2;
    findAverage(arr, N, K);

// This code is contributed by rakeshsahni

</script>

Output

5 7 6 4 7 7 4

Time complexity: O(N)
Auxiliary Space: O(1)

Replace every array element by sum of previous and next

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Replace all elements of given Array with average of previous K and next K elements

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