# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// 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 `

Output:

```4
```

Time Complexity: O(1)

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