Find the deleted value from the array when average of original elements is given

Given an array of length N + K. Also given the average avg of all the elements of the array. If an element that appears exactly K time got removed from the array (all the occurrences) and the resultant array is given, the task is to find the element X. Note that if X is not an integer then print -1.

Examples:

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.

Approach:

  • Find the sum of the array elements and store it in a variable sum.
  • Since X appeared K times then the sum of the original array will be sumOrg = sum + (X * K).
  • And the average is given to be avg i.e. avg = sumOrg / (N + K).
  • Now, X can be easily calculated as X = ((avg * (N + K)) – sum) / K

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

4

Time Complexity: O(1)



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