可以删除而不影响初始数组均值的元素数

给定一个由N个整数组成的数组arr [] ，任务是从数组中查找元素的数量，以使得从数组中单独删除它们(仅可以删除单个元素)后，不会干扰数组的初始均值。

例子：

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

方法：

• 查找数组元素的初始均值和总和，并将它们分别存储在变量均值和总和中。
• 现在，初始化计数= 0，并且对于每个元素arr [i]如果(sum – arr [i])/(N – 1)=均值，则递增计数。
• 最后打印计数。

下面是上述方法的实现：

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

输出：

1

时间复杂度： O(N)