# 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[0]);        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)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.