Average numbers in array

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++

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// CPP program to print all such indices such
// that the i-th element equals the arithmetic
// mean of all other elements
#include <bits/stdc++.h>
using namespace std;
  
// function to find number of elements
// satisfying condition and their indexes
void averageNumbers(int arr[], int n, int sum)
{
  
    int cnt = 0;
  
    // calculating average
    sum /= (double)n;
  
    // counting how many elements
    // satisfies the condition.
    cout << count(arr, arr + n, sum)
         << endl;
    for (int i = 0; i < n; i++) {
        if ((double)arr[i] == sum) {
  
            // output the indices.
            cout << i << " ";
            cnt++;
        }
    }
}
  
// Driver code
int main()
{
    int n;
    int arr[] = { 1, 2, 3, 4, 5 };
    n = sizeof(arr) / sizeof(arr[0]);
    double sum = 0;
    int cnt = 0;
  
    // sum of the elements of the array
    for (int i = 0; i < n; i++) {
        sum += (double)arr[i];
    }
    averageNumbers(arr, n, sum);
    return 0;
}

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Java

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// Java program to print all such indices such
// that the i-th element equals the arithmetic
// mean of all other elements
public class GFG {
  
// function to find number of elements
// satisfying condition and their indexes
    static void averageNumbers(int arr[], int n, int sum) {
  
        int cnt = 0;
  
        // calculating average
        sum /= (double) n;
  
        // counting how many elements
        // satisfies the condition.
        System.out.println(count(arr, sum));
        for (int i = 0; i < n; i++) {
            if ((double) arr[i] == sum) {
  
                // output the indices.
                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;
    }
// Driver code
  
    public static void main(String[] args) {
        int n;
        int arr[] = {1, 2, 3, 4, 5};
        n = arr.length;
        int sum = 0;
        int cnt = 0;
  
        // sum of the elements of the array
        for (int i = 0; i < n; i++) {
            sum += (double) arr[i];
        }
        averageNumbers(arr, n, sum);
  
    }
  
}
// This code is contributed by 29AjayKumar

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Python3

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# Python 3 program to pr all such indices 
# such that the i-th element equals the 
# arithmetic mean of all other elements
  
# Function to find number of elements
# satisfying condition and their indexes
def averageNumbers(arr, n, sum):
    cnt = 0
  
    # calculating average
    sum /= n
  
    # counting how many elements
    # satisfies the condition.
    print(count(arr, sum))
    for i in range(0, n):
        if (arr[i] == sum):
              
            # output the indices.
            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
          
# Driver code
if __name__ == '__main__':
    n = 0
    arr = [ 1, 2, 3, 4, 5 ]
    n = len(arr)
    sum = 0
    cnt = 0
  
    # sum of the elements of the array
    for i in range(0, n):
        sum += arr[i]
    averageNumbers(arr, n, sum)
  
# This code contributed by 29AjayKumar

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C#

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// C# program to print all such indices such
// that the i-th element equals the arithmetic
// mean of all other elements
using System; 
public class GFG {
   
// function to find number of elements
// satisfying condition and their indexes
    static void averageNumbers(int []arr, int n, int sum) {
   
        int cnt = 0;
   
        // calculating average
        sum /=  n;
   
        // counting how many elements
        // satisfies the condition.
        Console.WriteLine(count(arr, sum));
        for (int i = 0; i < n; i++) {
            if ((double) arr[i] == sum) {
   
                // output the indices.
                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;
    }
// Driver code
   
    public static void Main() {
        int n;
        int []arr = {1, 2, 3, 4, 5};
        n = arr.Length;
        int sum = 0;
   
        // sum of the elements of the array
        for (int i = 0; i < n; i++) {
            sum += arr[i];
        }
        averageNumbers(arr, n, sum);
   
    }
   
}
// This code is contributed by 29AjayKumar

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PHP

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<?php
// PHP program to print all such indices 
// such that the i-th element equals the 
// arithmetic mean of all other elements
  
// counting how many elements
// satisfies the condition.
function coun_t($arr, $sum
{
    $cnt = 0;
    for ( $i = 0; $i < count($arr); $i++) 
    {
        if ($arr[$i] == $sum)
        {
            $cnt++;
        }
    }
    return $cnt;
}
  
// function to find number of elements
// satisfying condition and their indexes
function averageNumbers($arr, $n, $sum)
{
  
    $cnt = 0;
  
    // calculating average
    $sum /= $n;
  
    // counting how many elements
    // satisfies the condition.
    echo coun_t($arr, $sum) . "\n";
          
    for ( $i = 0; $i < $n; $i++) 
    {
        if ($arr[$i] == $sum
        {
  
            // output the indices.
            echo $i . " ";
            $cnt++;
        }
    }
}
  
// Driver Code
$n = 0;
$arr = array( 1, 2, 3, 4, 5 );
$n = count($arr);
$sum = 0;
$cnt = 0;
  
// sum of the elements of the array
for ($i = 0; $i < $n; $i++) 
{
    $sum += $arr[$i];
}
averageNumbers($arr, $n, $sum);
  
// This code is contributed by
// Rajput-Ji
?>

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Output:

1
2 

Time Complexity: O(n)
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