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

  • Difficulty Level : Medium
  • 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#




// 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#




// 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) 


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