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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

// 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; 
} 

Java

// 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 

Python3

# Python 3 program to print 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 

C#

      
// 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 

PHP

<?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 
?> 

Output:

1
2

Time Complexity: O(n)
This article is contributed by Sagar Shukla. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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