Replace all elements of given Array with average of previous K and next K elements
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++
#include <bits/stdc++.h>
using namespace std;
void findAverage( int arr[], int N, int K)
{
int start, end;
for ( int i = 0; i < N; i++) {
int sum = 0;
start = max(i - K, 0);
end = min(i + K, N - 1);
int cnt = end - start;
for ( int j = start; j <= end; j++) {
if (j == i) {
continue ;
}
sum += arr[j];
}
cout << sum / cnt << ' ' ;
}
}
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
class GFG {
static void findAverage( int [] arr, int N, int K)
{
int start, end;
for ( int i = 0 ; i < N; i++) {
int sum = 0 ;
start = Math.max(i - K, 0 );
end = Math.min(i + K, N - 1 );
int cnt = end - start;
for ( int j = start; j <= end; j++) {
if (j == i) {
continue ;
}
sum += arr[j];
}
System.out.print(sum / cnt + " " );
}
}
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);
}
}
|
Python3
def findAverage(arr, N, K):
start = None
end = None
for i in range (N):
sum = 0
start = max (i - K, 0 )
end = min (i + K, N - 1 )
cnt = end - start
for j in range (start, end + 1 ):
if j = = i:
continue
sum + = arr[j]
print (( sum / / cnt), end = " " )
arr = [ 9 , 7 , 3 , 9 , 1 , 8 , 11 ]
N = len (arr)
K = 2
findAverage(arr, N, K)
|
C#
using System;
class GFG
{
static void findAverage( int []arr, int N, int K)
{
int start, end;
for ( int i = 0; i < N; i++) {
int sum = 0;
start = Math.Max(i - K, 0);
end = Math.Min(i + K, N - 1);
int cnt = end - start;
for ( int j = start; j <= end; j++) {
if (j == i) {
continue ;
}
sum += arr[j];
}
Console.Write(sum / cnt + " " );
}
}
public static void Main()
{
int []arr = { 9, 7, 3, 9, 1, 8, 11 };
int N = arr.Length;
int K = 2;
findAverage(arr, N, K);
}
}
|
Javascript
<script>
function findAverage(arr, N, K)
{
let start, end;
for (let i = 0; i < N; i++)
{
let sum = 0;
start = Math.max(i - K, 0);
end = Math.min(i + K, N - 1);
let cnt = end - start;
for (let j = start; j <= end; j++) {
if (j == i) {
continue ;
}
sum += arr[j];
}
document.write(Math.floor(sum / cnt) + ' ' );
}
}
let arr = [9, 7, 3, 9, 1, 8, 11];
let N = arr.length;
let K = 2;
findAverage(arr, N, K);
</script>
|
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++
#include <bits/stdc++.h>
using namespace std;
void findAverage( int arr[], int N, int K)
{
int i, sum = 0, next, prev, update;
int cnt = 0;
for (i = 0; i <= K and i < N; i++) {
sum += arr[i];
cnt += 1;
}
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;
}
}
}
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
import java.util.*;
public class GFG
{
static void findAverage( int arr[], int N, int K)
{
int i, sum = 0 , next = 0 , prev = 0 , update = 0 ;
int cnt = 0 ;
for (i = 0 ; i <= K && i < N; i++) {
sum += arr[i];
cnt += 1 ;
}
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 ;
}
}
}
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);
}
}
|
Python3
def findAverage(arr, N, K):
sum = 0 ; next = 0 ; prev = 0 ; update = 0 ;
cnt = 0 ;
for i in range ( 0 , K + 1 , 1 ):
if (i > = N):
break
sum + = arr[i];
cnt + = 1 ;
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 ;
if __name__ = = '__main__' :
arr = [ 9 , 7 , 3 , 9 , 1 , 8 , 11 ];
N = len (arr);
K = 2 ;
findAverage(arr, N, K);
|
C#
using System;
public class GFG
{
static void findAverage( int []arr, int N, int K)
{
int i, sum = 0, next = 0, prev = 0, update = 0;
int cnt = 0;
for (i = 0; i <= K && i < N; i++) {
sum += arr[i];
cnt += 1;
}
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;
}
}
}
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);
}
}
|
Javascript
<script>
const findAverage = (arr, N, K) => {
let i, sum = 0, next, prev, update;
let cnt = 0;
for (i = 0; i <= K && i < N; i++) {
sum += arr[i];
cnt += 1;
}
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;
}
}
}
let arr = [9, 7, 3, 9, 1, 8, 11];
let N = arr.length;
let K = 2;
findAverage(arr, N, K);
</script>
|
Time complexity: O(N)
Auxiliary Space: O(1)
Last Updated :
27 Jan, 2022
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