# Count of elements that can be deleted without disturbing the mean of the initial array

Given an array arr[] of N integers, the task is to find the count of elements from the array such that after removing them individually (only a single element can be deleted) from the array will not disturb the initial mean of the array.

Examples:

Input: arr[] = {1, 2, 3, 4, 5}
Output: 1
3 is the only element removing which
will not affect the mean of the array.
i.e. (1 + 2 + 4 + 5) / 4 = 12 / 4 = 3
which is the mean of the original array.

Input: arr[] = {5, 4, 3, 6}
Output: 0

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

Approach:

• Find the initial mean and the sum of the array elements and store them in the variables mean and sum respectively.
• Now, initialise count = 0 and for every element arr[i] if (sum – arr[i]) / (N – 1) = mean then increment the count.
• Print the count in the end.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Funcion to find the elements ` `// which do not change the mean on removal ` `int` `countElements(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// To store the sum of the array elements ` `    ``int` `sum = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``sum += arr[i]; ` ` `  `    ``// To store the initial mean ` `    ``float` `mean = (``float``)sum / n; ` ` `  `    ``// to store the count of required elements ` `    ``int` `cnt = 0; ` `    ``// Iterate over the array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Finding the new mean ` `        ``float` `newMean = (``float``)(sum - arr[i]) / (n - 1); ` ` `  `        ``// If the new mean equals to the initial mean ` `        ``if` `(newMean == mean) ` `            ``cnt++; ` `    ``} ` ` `  `    ``return` `cnt; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3, 4, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << countElements(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG  ` `{ ` ` `  `// Funcion to find the elements ` `// which do not change the mean on removal ` `static` `int` `countElements(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// To store the sum of the array elements ` `    ``int` `sum = ``0``; ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``sum += arr[i]; ` ` `  `    ``// To store the initial mean ` `    ``float` `mean = (``float``)sum / n; ` ` `  `    ``// to store the count of required elements ` `    ``int` `cnt = ``0``; ` `     `  `    ``// Iterate over the array ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` ` `  `        ``// Finding the new mean ` `        ``float` `newMean = (``float``)(sum - arr[i]) /  ` `                                      ``(n - ``1``); ` ` `  `        ``// If the new mean equals to  ` `        ``// the initial mean ` `        ``if` `(newMean == mean) ` `            ``cnt++; ` `    ``} ` `    ``return` `cnt; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5` `}; ` `    ``int` `n = arr.length; ` `    ``System.out.println(countElements(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992  `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Funcion to find the elements ` `# which do not change the mean on removal ` `def` `countElements(arr, n): ` ` `  `    ``# To store the Sum of the array elements ` `    ``Sum` `=` `0` `    ``for` `i ``in` `range``(n): ` `        ``Sum` `+``=` `arr[i] ` ` `  `    ``# To store the initial mean ` `    ``mean ``=` `Sum` `/` `n ` ` `  `    ``# to store the count of required elements ` `    ``cnt ``=` `0` `     `  `    ``# Iterate over the array ` `    ``for` `i ``in` `range``(n): ` `         `  `        ``# Finding the new mean ` `        ``newMean ``=` `(``Sum` `-` `arr[i]) ``/` `(n ``-` `1``) ` ` `  `        ``# If the new mean equals to ` `        ``# the initial mean ` `        ``if` `(newMean ``=``=` `mean): ` `            ``cnt ``+``=` `1` ` `  `    ``return` `cnt ` ` `  `# Driver code ` `arr ``=` `[``1``, ``2``, ``3``, ``4``, ``5``] ` `n ``=` `len``(arr) ` ` `  `print``(countElements(arr, n)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` `class` `GFG  ` `{ ` ` `  `// Funcion to find the elements ` `// which do not change the mean on removal ` `static` `int` `countElements(``int` `[]arr, ``int` `n) ` `{ ` ` `  `    ``// To store the sum of the array elements ` `    ``int` `sum = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``sum += arr[i]; ` ` `  `    ``// To store the initial mean ` `    ``float` `mean = (``float``)sum / n; ` ` `  `    ``// to store the count of required elements ` `    ``int` `cnt = 0; ` `     `  `    ``// Iterate over the array ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` ` `  `        ``// Finding the new mean ` `        ``float` `newMean = (``float``)(sum - arr[i]) /  ` `                                      ``(n - 1); ` ` `  `        ``// If the new mean equals to  ` `        ``// the initial mean ` `        ``if` `(newMean == mean) ` `            ``cnt++; ` `    ``} ` `    ``return` `cnt; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``int` `[]arr = { 1, 2, 3, 4, 5 }; ` `    ``int` `n = arr.Length; ` `    ``Console.WriteLine(countElements(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```1
```

Time Complexity: O(N)

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