Find mean of K adjacent elements on each sides for each Array element

Given a circular array arr[] of N numbers, and an integer K. The task is to print the average for 2K+1 numbers for each elements (K from left, K from right, and the element self).

Examples:

Input: arr []= { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, K = 3
Output: {4.85714, 4.57143, 4.28571, 4.0, 5.0, 6.0, 5.71429, 5.42857, 5.14286}
Explanation: For each value the averages are: 
for 1 – right part:9, left part:24 & result:4.85714
for 2 – right part:12, left part:18 & result:4.57143
for 3 – right part:15, left part:12 & result:4.28571
for 4 – right part:18, left part:6 & result:4
for 5 – right part:21, left part:9 & result:5
for 6 – right part:24, left part:12 & result:6
for 7 – right part:18, left part:15 & result:5.71429
for 8 – right part:12, left part:18 & result:5.42857
for 9 – right part:6, left part:21 & result:5.14286

Input: arr[] = {2, 2, 2, 2, 2}, K = 3
Output: {2, 2, 2, 2, 2}

Naive Approach: The simplest approach to solve the problem is to traverse the required number of elements for each element of the array. Follow the steps mentioned below:

• Traverse the array and for each element do the following:
  • traverse the next K and previous K elements and calculate the mean for these elements.
• Print the answer for each element.

Below is the implementation of the above approach:

C++

// C++ code to implement the above approach
#include <iostream>
using namespace std;

// Function to calculate average
void average(int arr[], int N, int K)
{
    // Iterate over all elements
    for (int i = 0; i < N; i++) {
        int leftSum = 0;
        int rightSum = 0;
        // find the right sum
        for (int j = 1; j <= K; j++) {
            rightSum += arr[(i + j) % N];
        }
        // Find the leftSum
        for (int j = 1; j <= K; j++) {
            leftSum += arr[(i - j < 0
                                ? i - j + N
                                : i - j)
                           % N];
        }

        // Print mean for each element
        cout << ((leftSum + rightSum + arr[i])
                 * 1.0)
                    / (2 * K + 1)
             << " ";
    }
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    int K = 3;
    int N = sizeof(arr) / sizeof(int);

    average(arr, N, K);
    return 0;
}

Java

// Java code to implement the above approach
import java.util.*;
public class GFG{

// Function to calculate average
static void average(int[] arr, int N, int K)
{

    // Iterate over all elements
    for(int i = 0; i < N; i++)
    {
        int leftSum = 0;
        int rightSum = 0;

        // Find the right sum
        for(int j = 1; j <= K; j++)
        {
            rightSum += arr[(i + j) % N];
        }

        // Find the leftSum
        for(int j = 1; j <= K; j++)
        {
            leftSum += arr[(i - j < 0 ? i - j + N : i - j) % N];
        }

        // Print mean for each element
        System.out.print( ((leftSum + rightSum + arr[i]) * 1.0) /
                              (2 * K + 1) + " ");
    }
}

// Driver code
public static void main(String args[])
{
    int[] arr = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    int K = 3;
    int N = arr.length;

    average(arr, N, K);
}
}

// This code is contributed by Samim Hossain Mondal.

Python3

# Python code for the above approach

# Function to calculate average
def average(arr, N, K):

    # Iterate over all elements
    for i in range(N):
        leftSum = 0
        rightSum = 0

        # find the right sum
        for j in range(1, K + 1):
            rightSum += arr[(i + j) % N]

        # Find the leftSum
        for j in range(1, K + 1):
            leftSum += arr[((i - j + N) if (i - j < 0) else (i - j)) % N]


        # Print mean for each element
        print(round(((leftSum + rightSum + arr[i]) * 1.0) / (2 * K + 1), 5), end=" ")

# Driver code
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
K = 3
N = len(arr)

average(arr, N, K)

# This code is contributed by Saurabh Jaiswal

C#

// C# code to implement the above approach
using System;

class GFG{

// Function to calculate average
static void average(int[] arr, int N, int K)
{

    // Iterate over all elements
    for(int i = 0; i < N; i++)
    {
        int leftSum = 0;
        int rightSum = 0;

        // Find the right sum
        for(int j = 1; j <= K; j++)
        {
            rightSum += arr[(i + j) % N];
        }

        // Find the leftSum
        for(int j = 1; j <= K; j++)
        {
            leftSum += arr[(i - j < 0 ? i - j + N : i - j) % N];
        }

        // Print mean for each element
        Console.Write( ((leftSum + rightSum + arr[i]) * 1.0) /
                              (2 * K + 1) + " ");
    }
}

// Driver code
public static void Main()
{
    int[] arr = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    int K = 3;
    int N = arr.Length;

    average(arr, N, K);
}
}

// This code is contributed by ukasp

Javascript

  <script>
      // JavaScript code for the above approach

      // Function to calculate average
      function average(arr, N, K)
      {

          // Iterate over all elements
          for (let i = 0; i < N; i++)
          {
              let leftSum = 0;
              let rightSum = 0;

              // find the right sum
              for (let j = 1; j <= K; j++) {
                  rightSum += arr[(i + j) % N];
              }

              // Find the leftSum
              for (let j = 1; j <= K; j++) {
                  leftSum += arr[(i - j < 0
                      ? i - j + N
                      : i - j)
                      % N];
              }

              // Print mean for each element
              document.write((((leftSum + rightSum + arr[i])
                  * 1.0)
                  / (2 * K + 1)).toPrecision(6)
                  + " ");
          }
      }

      // Driver code

      let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];
      let K = 3;
      let N = arr.length;

      average(arr, N, K);

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

Output

4.85714 4.57143 4.28571 4 5 6 5.71429 5.42857 5.14286 

Time Complexity: O(N*K), as we are using nested loops to traverse N*K times.

Auxiliary Space: O(1), as we are not using any extra space.

Efficient Approach: To reduce the time complexity the concept of prefix sum can be used where prefix sum array(say preSum[]) is calculated like preSum[i] = arr[0] + . . . + arr[i]. And using the preSum[] array the mean can be calculated easily for each elements without traversing (2K + 1) elements in each iterations. Condition for calculating the sum of previous K elements (leftSum) and K next elements (rightSum) for each of the element is given below:

Calculation for sum of K-elements in right:

• If there are K-elements present in right:
                        rightSum = preSum[i + k] – preSum[i];
• If no elements are in right:
                        rightSum = preSum[k – 1];
• If some elements are in right and some needs circular traversal:
                        eleInRight = n – i – 1;
                        rightSum = presum[n – 1] – presum[i]  + presum[k – eleInRight – 1];

Calculation for sum of K-elements in left:

• If there are more than  K-elements present in left:
                     leftSum = preSum[i – 1] – preSum[i – k – 1];
• If only k-elements are present in left:
                    leftSum = preSum[i – 1];
• if no elements are in left:
                   leftSum = preSum[n – 1] – preSum[n – 1 – k];
• If some elements are in left and some needs circular traversal:
                  eleInLeft = i
                 leftSum = preSum[i – 1] + preSum[n – 1]  – preSum[n – 1 – (k – eleInLeft)];

Follow the steps mentioned below to implement it:

• Iterate the array create the prefix sum array.
• For each element get the left and right sum as show in the observation and calculate the average.

Below is the implementation of the above approach:

C++

// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate the average
void average(int arr[], int N, int K)
{
    int presum[N];
    presum[0] = arr[0];
    for (int i = 1; i < N; i++) {
        presum[i] = presum[i - 1] + arr[i];
    }

    for (int i = 0; i < N; i++) {

        // Right part
        int rightSum = 0;

        // When all k-elements are
        // in right
        if (i + K < N)
            rightSum = presum[i + K]
                       - presum[i];
        else {
            int eleInRight = N - i - 1;

            // When some are in right and
            // some needs circular traversal
            if (eleInRight > 0) {
                rightSum = presum[N - 1]
                           - presum[i]
                           + presum[K - eleInRight - 1];
            }
            else {
                rightSum = presum[K - eleInRight - 1];
            }
        }

        // Left part
        int leftSum = 0;

        // When more than k-elements
        // are in left
        if (i - K > 0)
            leftSum = presum[i - 1]
                      - presum[i - K - 1];

        // When exact k-elements are in left
        else if (i - K == 0) {
            leftSum = presum[i - 1];
        }

        // When some are in left some
        // needs circular traversal
        else {
            int eleInLeft = i;
            if (eleInLeft > 0) {
                leftSum = presum[i - 1]
                          + presum[N - 1]
                          - presum[N - 1 - (K - eleInLeft)];
            }
            else {
                leftSum = presum[N - 1]
                          - presum[N - 1 - K];
            }
        }
        cout << ((arr[i] + leftSum + rightSum)
                 * 1.0)
                    / (2 * K + 1)
             << " ";
    }
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    int K = 3;
    int N = sizeof(arr) / sizeof(int);

    average(arr, N, K);
    return 0;
}

Java

// Java code to implement the above approach
import java.util.*;
public class GFG
{

// Function to calculate the average
static void average(int arr[], int N, int K)
{
    int presum[] = new int[N];
    presum[0] = arr[0];
    for (int i = 1; i < N; i++) {
        presum[i] = presum[i - 1] + arr[i];
    }

    for (int i = 0; i < N; i++) {

        // Right part
        int rightSum = 0;

        // When all k-elements are
        // in right
        if (i + K < N)
            rightSum = presum[i + K]
                       - presum[i];
        else {
            int eleInRight = N - i - 1;

            // When some are in right and
            // some needs circular traversal
            if (eleInRight > 0) {
                rightSum = presum[N - 1]
                           - presum[i]
                           + presum[K - eleInRight - 1];
            }
            else {
                rightSum = presum[K - eleInRight - 1];
            }
        }

        // Left part
        int leftSum = 0;

        // When more than k-elements
        // are in left
        if (i - K > 0)
            leftSum = presum[i - 1]
                      - presum[i - K - 1];

        // When exact k-elements are in left
        else if (i - K == 0) {
            leftSum = presum[i - 1];
        }

        // When some are in left some
        // needs circular traversal
        else {
            int eleInLeft = i;
            if (eleInLeft > 0) {
                leftSum = presum[i - 1]
                          + presum[N - 1]
                          - presum[N - 1 - (K - eleInLeft)];
            }
            else {
                leftSum = presum[N - 1]
                          - presum[N - 1 - K];
            }
        }
        System.out.print(((arr[i] + leftSum + rightSum)
                 * 1.0)
                    / (2 * K + 1)
             + " ");
    }
}

// Driver code
public static void main(String args[])
{
    int[] arr = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    int K = 3;
    int N = arr.length;

    average(arr, N, K);
}
}

// This code is contributed by Samim Hossain Mondal.

Python3

# Python code to implement the above approach

# Function to calculate the average
def average(arr, N, K):

    presum = [0]*N
    presum[0] = arr[0]
    for i in range(1,N):
        presum[i] = presum[i - 1] + arr[i]

    for i in range(0,N):

        # Right part
        rightSum = 0

        # When all k-elements are
        # in right
        if (i + K < N):
            rightSum = presum[i + K] - presum[i]
        else:
            eleInRight = N - i - 1

            # When some are in right and
            # some needs circular traversal
            if (eleInRight > 0):
                rightSum = presum[N - 1] - presum[i] + presum[K - eleInRight - 1]

            else:
                rightSum = presum[K - eleInRight - 1]

        # Left part
        leftSum = 0

        # When more than k-elements
        # are in left
        if (i - K > 0):
            leftSum = presum[i - 1] - presum[i - K - 1]

        # When exact k-elements are in left
        elif (i - K == 0):
            leftSum = presum[i - 1]

        # When some are in left some
        # needs circular traversal
        else:
            eleInLeft = i
            if (eleInLeft > 0):
                leftSum = presum[i - 1] + presum[N - 1] - presum[N - 1 - (K - eleInLeft)]

            else:
                leftSum = presum[N - 1] - presum[N - 1 - K]

        print("{:.5f}".format(((arr[i] + leftSum + rightSum) * 1.0) / (2 * K + 1)),end = " ")

# Driver code
arr =  [1, 2, 3, 4, 5, 6, 7, 8, 9]
K = 3
N = len(arr)

average(arr, N, K)

# This code is contributed by Shubham Singh

C#

// C# code to implement the above approach
using System;
public class GFG{

  // Function to calculate the average
  static void average(int[] arr, int N, int K)
  {
    int[] presum = new int[N];
    presum[0] = arr[0];
    for (int i = 1; i < N; i++) {
      presum[i] = presum[i - 1] + arr[i];
    }

    for (int i = 0; i < N; i++) {

      // Right part
      int rightSum = 0;

      // When all k-elements are
      // in right
      if (i + K < N)
        rightSum = presum[i + K]
        - presum[i];
      else {
        int eleInRight = N - i - 1;

        // When some are in right and
        // some needs circular traversal
        if (eleInRight > 0) {
          rightSum = presum[N - 1]
            - presum[i]
            + presum[K - eleInRight - 1];
        }
        else {
          rightSum = presum[K - eleInRight - 1];
        }
      }

      // Left part
      int leftSum = 0;

      // When more than k-elements
      // are in left
      if (i - K > 0)
        leftSum = presum[i - 1]
        - presum[i - K - 1];

      // When exact k-elements are in left
      else if (i - K == 0) {
        leftSum = presum[i - 1];
      }

      // When some are in left some
      // needs circular traversal
      else {
        int eleInLeft = i;
        if (eleInLeft > 0) {
          leftSum = presum[i - 1]
            + presum[N - 1]
            - presum[N - 1 - (K - eleInLeft)];
        }
        else {
          leftSum = presum[N - 1]
            - presum[N - 1 - K];
        }
      }
      Console.Write(((arr[i] + leftSum + rightSum)
                     * 1.0) / (2 * K + 1) + " ");
    }
  }

  // Driver code
  public static void Main()
  {
    int[] arr = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    int K = 3;
    int N = arr.Length;

    average(arr, N, K);
  }
}

// This code is contributed by Shubham Singh

Javascript

<script>
    // JavaScript code to implement the above approach

    // Function to calculate the average
    const average = (arr, N, K) => {
        let presum = new Array(N).fill(0);
        presum[0] = arr[0];
        for (let i = 1; i < N; i++) {
            presum[i] = presum[i - 1] + arr[i];
        }

        for (let i = 0; i < N; i++) {

            // Right part
            let rightSum = 0;

            // When all k-elements are
            // in right
            if (i + K < N)
                rightSum = presum[i + K]
                    - presum[i];
            else {
                let eleInRight = N - i - 1;

                // When some are in right and
                // some needs circular traversal
                if (eleInRight > 0) {
                    rightSum = presum[N - 1]
                        - presum[i]
                        + presum[K - eleInRight - 1];
                }
                else {
                    rightSum = presum[K - eleInRight - 1];
                }
            }

            // Left part
            let leftSum = 0;

            // When more than k-elements
            // are in left
            if (i - K > 0)
                leftSum = presum[i - 1]
                    - presum[i - K - 1];

            // When exact k-elements are in left
            else if (i - K == 0) {
                leftSum = presum[i - 1];
            }

            // When some are in left some
            // needs circular traversal
            else {
                let eleInLeft = i;
                if (eleInLeft > 0) {
                    leftSum = presum[i - 1]
                        + presum[N - 1]
                        - presum[N - 1 - (K - eleInLeft)];
                }
                else {
                    leftSum = presum[N - 1]
                        - presum[N - 1 - K];
                }
            }
            document.write(`${((arr[i] + leftSum + rightSum)
                * 1.0)
                / (2 * K + 1)} `);
        }
    }

    // Driver code
    let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];
    let K = 3;
    let N = arr.length;

    average(arr, N, K);

    // This code is contributed by rakeshsahni

</script>

Output

4.85714 4.57143 4.28571 4 5 6 5.71429 5.42857 5.14286 

Time Complexity: O(N), as we are using a loop to traverse N times.

Auxiliary Space: O(N), as we are using extra space for presum.