给定原始元素的平均值时，从数组中查找已删除的值

给定长度为N + K的数组。还给出了数组所有元素的平均平均值。如果从数组中删除了恰好K时间出现的元素(所有出现的情况)并给出了结果数组，则任务是找到元素X。请注意，如果X不是整数，则输出-1 。
例子：

Input: arr[] = {2, 7, 3}, K = 3, avg = 4 
Output: 4 
The original array was {2, 7, 3, 4, 4, 4} 
where 4 which occurred thrice was deleted. 
(2 + 7 + 3 + 4 + 4 + 4) / 6 = 4

Input: arr[] = {5, 2, 3}, K = 4, avg = 7; 
Output: -1 
The required element is 9.75 which is not an integer. 

方法：

• 找到数组元素的总和并将其存储在变量sum中。
• 由于X出现了K次，则原始数组的总和将为sumOrg = sum +(X * K) 。
• 平均值为avg，即avg = sumOrg /(N + K) 。
• / K –现在，X可以很容易地为X =(SUM(平均*(N + K)))来计算

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the missing element
int findMissing(int arr[], int n, int k, int avg)
{
 
    // Find the sum of the array elements
    int sum = 0;
    for (int i = 0; i < n; i++) {
        sum += arr[i];
    }
 
    // The numerator and the denominator
    // of the equation
    int num = (avg * (n + k)) - sum;
    int den = k;
 
    // If not divisible then X is
    // not an integer
    // it is a floating point number
    if (num % den != 0)
        return -1;
 
    // Return X
    return (num / den);
}
 
// Driver code
int main()
{
    int k = 3, avg = 4;
    int arr[] = { 2, 7, 3 };
    int n = sizeof(arr) / sizeof(int);
 
    cout << findMissing(arr, n, k, avg);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
    // Function to return the missing element
    static int findMissing(int arr[], int n,
                           int k, int avg)
    {
     
        // Find the sum of the array elements
        int sum = 0;
        for (int i = 0; i < n; i++)
        {
            sum += arr[i];
        }
     
        // The numerator and the denominator
        // of the equation
        int num = (avg * (n + k)) - sum;
        int den = k;
     
        // If not divisible then X is
        // not an integer
        // it is a floating point number
        if (num % den != 0)
            return -1;
     
        // Return X
        return (int)(num / den);
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int k = 3, avg = 4;
        int arr[] = { 2, 7, 3 };
        int n = arr.length;
     
        System.out.println(findMissing(arr, n, k, avg));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
 
# Function to return the missing element
def findMissing(arr, n, k, avg):
 
    # Find the sum of the array elements
    sum = 0;
    for i in range(n):
        sum += arr[i];
     
    # The numerator and the denominator
    # of the equation
    num = (avg * (n + k)) - sum;
    den = k;
 
    # If not divisible then X is
    # not an integer
    # it is a floating ponumber
    if (num % den != 0):
        return -1;
 
    # Return X
    return (int)(num / den);
 
# Driver code
k = 3; avg = 4;
arr = [2, 7, 3] ;
n = len(arr);
 
print(findMissing(arr, n, k, avg));
 
# This code is contributed by 29AjayKumar

C#

// C# implementation of above approach
using System;
     
class GFG
{
     
    // Function to return the missing element
    static int findMissing(int []arr, int n,
                           int k, int avg)
    {
     
        // Find the sum of the array elements
        int sum = 0;
        for (int i = 0; i < n; i++)
        {
            sum += arr[i];
        }
     
        // The numerator and the denominator
        // of the equation
        int num = (avg * (n + k)) - sum;
        int den = k;
     
        // If not divisible then X is
        // not an integer
        // it is a floating point number
        if (num % den != 0)
            return -1;
     
        // Return X
        return (int)(num / den);
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        int k = 3, avg = 4;
        int []arr = { 2, 7, 3 };
        int n = arr.Length;
     
        Console.WriteLine(findMissing(arr, n, k, avg));
    }
}
 
// This code is contributed by Rajput-Ji

输出：

4

时间复杂度： O(1)
辅助空间： O(1)