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++
// 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)
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