与所有其他元素的和的绝对差大于k的元素数

给定N个整数和整数K的数组arr [] ，任务是查找数组中的异常数。异常是一个数字，其与数组中所有其他数字之间的绝对差大于K。查找异常数量。

例子：

Input: arr[] = {1, 3, 5}, k = 1
Output: 2
1 and 3 are the anomalies.
|1 – (3 + 5)| = 7 > 1
|3 – (1 + 5)| = 3 > 1

Input: arr[] = {7, 1, 8}, k = 5
Output: 1

方法：找到所有数组元素的总和，然后将其存储在sum中，现在对于数组arr [i]的每个元素，如果arr [i]与sum – arr [i]的绝对差大于k ，则为异常。计算数组中的所有异常，最后打印结果。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include
using namespace std;
  
// Function to return the number of anomalies
static int countAnomalies(int arr[], 
                          int n, int k)
{
  
    // To store the count of anomalies
    int cnt = 0;
  
    // To store the sum of the array elements
    int i, sum = 0;
  
    // Find the sum of the array elements
    for (i = 0; i < n; i++)
        sum += arr[i];
  
    // Count the anomalies
    for (i = 0; i < n; i++)
        if (abs(arr[i] - 
               (sum - arr[i])) > k)
            cnt++;
  
    return cnt;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 3, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 1;
    cout << countAnomalies(arr, n, k);
}
  
// This code is contributed
// by Code_Mech

Java

// Java implementation of the approach
class GFG {
  
    // Function to return the number of anomalies
    static int countAnomalies(int arr[], int n, int k)
    {
  
        // To store the count of anomalies
        int cnt = 0;
  
        // To store the sum of the array elements
        int i, sum = 0;
  
        // Find the sum of the array elements
        for (i = 0; i < n; i++)
            sum += arr[i];
  
        // Count the anomalies
        for (i = 0; i < n; i++)
            if (Math.abs(arr[i] - (sum - arr[i])) > k)
                cnt++;
  
        return cnt;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 3, 5 };
        int n = arr.length;
        int k = 1;
        System.out.print(countAnomalies(arr, n, k));
    }
}

Python3

# Python3 implementation of the approach
  
# Function to return the 
# number of anomalies
def countAnomalies(arr, n, k):
  
    # To store the count of anomalies
    cnt = 0
  
    # To store the Sum of 
    # the array elements
    i, Sum = 0, 0
  
    # Find the Sum of the array elements
    for i in range(n):
        Sum += arr[i]
  
    # Count the anomalies
    for i in range(n):
        if (abs(arr[i] - (Sum - arr[i])) > k):
            cnt += 1
  
    return cnt
  
# Driver code
arr = [1, 3, 5]
n = len(arr)
k = 1
print(countAnomalies(arr, n, k))
  
# This code is contributed 
# by mohit kumar

C#

// C# implementation of the approach
using System;
  
class GFG
{
  
    // Function to return the number of anomalies
    static int countAnomalies(int[] arr, int n, int k)
    {
  
        // To store the count of anomalies
        int cnt = 0;
  
        // To store the sum of the array elements
        int i, sum = 0;
  
        // Find the sum of the array elements
        for (i = 0; i < n; i++)
            sum += arr[i];
  
        // Count the anomalies
        for (i = 0; i < n; i++)
            if (Math.Abs(arr[i] - (sum - arr[i])) > k)
                cnt++;
  
        return cnt;
    }
  
    // Driver code
    public static void Main()
    {
        int[] arr = { 1, 3, 5 };
        int n = arr.Length;
        int k = 1;
        Console.WriteLine(countAnomalies(arr, n, k));
    }
}
  
// This code is contributed by Code_Mech.

PHP

 $k) 
            $cnt++; 
  
    return $cnt; 
}
  
// Driver code 
$arr = array(1, 3, 5);
$n = count($arr); 
$k = 1; 
  
echo countAnomalies($arr, $n, $k); 
  
// This code is contributed by Ryuga
?>

输出：

2