Average numbers in array
Last Updated :
17 Feb, 2023
Given a sequence of positive integers a1, a2, …, an. Find all such indexes i such that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Examples :
Input : 5
1 2 3 4 5
Output : 1 no. of elements
2 index of element
Average of 1, 2, 4 & 5 is 3 so the
output is single index i.e. 3.
Input : 4
50 50 50 50
Output : 4 no. of elements
0 1 2 3 index of element
Average of 50, 50, 50 & 50 is 50 and
all the indexes has the same i.e. 50
so the output is indexes 1, 2, 3 & 4.
C++
#include <bits/stdc++.h>
using namespace std;
void averageNumbers( int arr[], int n, int sum)
{
int cnt = 0;
sum /= ( double )n;
cout << count(arr, arr + n, sum)
<< endl;
for ( int i = 0; i < n; i++) {
if (( double )arr[i] == sum) {
cout << i << " " ;
cnt++;
}
}
}
int main()
{
int n;
int arr[] = { 1, 2, 3, 4, 5 };
n = sizeof (arr) / sizeof (arr[0]);
double sum = 0;
int cnt = 0;
for ( int i = 0; i < n; i++) {
sum += ( double )arr[i];
}
averageNumbers(arr, n, sum);
return 0;
}
|
C
#include <stdio.h>
void averageNumbers( int arr[], int n, int sum)
{
int cnt = 0;
sum /= ( double )n;
for ( int i = 0; i < n; i++) {
if (( double )arr[i] == sum) {
cnt++;
}
}
printf ( "%d\n" ,cnt);
for ( int i = 0; i < n; i++) {
if (( double )arr[i] == sum) {
printf ( "%d " ,i);
cnt++;
}
}
}
int main()
{
int n;
int arr[] = { 1, 2, 3, 4, 5 };
n = sizeof (arr) / sizeof (arr[0]);
double sum = 0;
int cnt = 0;
for ( int i = 0; i < n; i++) {
sum += ( double )arr[i];
}
averageNumbers(arr, n, sum);
return 0;
}
|
Java
public class GFG {
static void averageNumbers( int arr[], int n, int sum) {
int cnt = 0 ;
sum /= ( double ) n;
System.out.println(count(arr, sum));
for ( int i = 0 ; i < n; i++) {
if (( double ) arr[i] == sum) {
System.out.print(i + " " );
cnt++;
}
}
}
static int count( int [] array, int sum) {
int count = 0 ;
for ( int i = 0 ; i < array.length; i++) {
if (array[i] == sum) {
count++;
}
}
return count;
}
public static void main(String[] args) {
int n;
int arr[] = { 1 , 2 , 3 , 4 , 5 };
n = arr.length;
int sum = 0 ;
int cnt = 0 ;
for ( int i = 0 ; i < n; i++) {
sum += ( double ) arr[i];
}
averageNumbers(arr, n, sum);
}
}
|
Python3
def averageNumbers(arr, n, sum ):
cnt = 0
sum / = n
print (count(arr, sum ))
for i in range ( 0 , n):
if (arr[i] = = sum ):
print (i, " " )
cnt + = 1
def count(array, sum ):
count = 0
for i in range ( 0 , len (array)):
if (array[i] = = sum ):
count + = 1
return count
if __name__ = = '__main__' :
n = 0
arr = [ 1 , 2 , 3 , 4 , 5 ]
n = len (arr)
sum = 0
cnt = 0
for i in range ( 0 , n):
sum + = arr[i]
averageNumbers(arr, n, sum )
|
C#
using System;
public class GFG {
static void averageNumbers( int []arr, int n, int sum) {
int cnt = 0;
sum /= n;
Console.WriteLine(count(arr, sum));
for ( int i = 0; i < n; i++) {
if (( double ) arr[i] == sum) {
Console.Write(i + " " );
cnt++;
}
}
}
static int count( int [] array, int sum) {
int count = 0;
for ( int i = 0; i < array.Length; i++) {
if (array[i] == sum) {
count++;
}
}
return count;
}
public static void Main() {
int n;
int []arr = {1, 2, 3, 4, 5};
n = arr.Length;
int sum = 0;
for ( int i = 0; i < n; i++) {
sum += arr[i];
}
averageNumbers(arr, n, sum);
}
}
|
PHP
<?php
function coun_t( $arr , $sum )
{
$cnt = 0;
for ( $i = 0; $i < count ( $arr ); $i ++)
{
if ( $arr [ $i ] == $sum )
{
$cnt ++;
}
}
return $cnt ;
}
function averageNumbers( $arr , $n , $sum )
{
$cnt = 0;
$sum /= $n ;
echo coun_t( $arr , $sum ) . "\n" ;
for ( $i = 0; $i < $n ; $i ++)
{
if ( $arr [ $i ] == $sum )
{
echo $i . " " ;
$cnt ++;
}
}
}
$n = 0;
$arr = array ( 1, 2, 3, 4, 5 );
$n = count ( $arr );
$sum = 0;
$cnt = 0;
for ( $i = 0; $i < $n ; $i ++)
{
$sum += $arr [ $i ];
}
averageNumbers( $arr , $n , $sum );
?>
|
Javascript
<script>
function averageNumbers(arr, n, sum) {
let cnt = 0;
sum /= n;
document.write(count(arr, sum) + "</br>" );
for (let i = 0; i < n; i++) {
if (arr[i] == sum) {
document.write(i + " " );
cnt++;
}
}
}
function count(array, sum) {
let count = 0;
for (let i = 0; i < array.length; i++) {
if (array[i] == sum) {
count++;
}
}
return count;
}
let n;
let arr = [1, 2, 3, 4, 5];
n = arr.length;
let sum = 0;
for (let i = 0; i < n; i++) {
sum += arr[i];
}
averageNumbers(arr, n, sum);
</script>
|
Time Complexity: O(n), Auxiliary Space: O(1)
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