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:Â
Input: arr[] = {2, 7, 3}, K = 3, avg = 4Â
Output: 4Â
The original array was {2, 7, 3, 4, 4, 4}Â
where 4 which occurred thrice was deleted.Â
(2 + 7 + 3 + 4 + 4 + 4) / 6 = 4Input: arr[] = {5, 2, 3}, K = 4, avg = 7;Â
Output: -1Â
The required element is 9.75 which is not an integer.Â
Approach:Â Â
- 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:Â Â
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;Â
// Function to return the missing elementint findMissing(int arr[], int n, int k, int avg){Â
    // Find the sum of the array elements    int sum = 0;    for (int i = 0; i < n; i++) {        sum += arr[i];    }Â
    // The numerator and the denominator    // of the equation    int num = (avg * (n + k)) - sum;    int den = k;Â
    // If not divisible then X is    // not an integer    // it is a floating point number    if (num % den != 0)        return -1;Â
    // Return X    return (num / den);}Â
// Driver codeint main(){Â Â Â Â int k = 3, avg = 4;Â Â Â Â int arr[] = { 2, 7, 3 };Â Â Â Â int n = sizeof(arr) / sizeof(int);Â
    cout << findMissing(arr, n, k, avg);Â
    return 0;} |
Java
// Java implementation of the approachclass 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));     }}Â
// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approachÂ
# Function to return the missing elementdef findMissing(arr, n, k, avg):Â
    # 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;Â
    # If not divisible then X is    # not an integer    # it is a floating point number    if (num % den != 0):        return -1;Â
    # Return X    return (int)(num / den);Â
# Driver codek = 3; avg = 4;arr = [2, 7, 3] ;n = len(arr);Â
print(findMissing(arr, n, k, avg));Â
# This code is contributed by 29AjayKumar |
C#
// C# implementation of above approachusing 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));     }}Â
// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the// above approachÂ
// Function to return the missing elementfunction findMissing(arr, n, k, avg){      // Find the sum of the array elements    var sum = 0;    for (var i = 0; i < n; i++) {        sum += arr[i];    }      // The numerator and the denominator    // of the equation    var num = (avg * (n + k)) - sum;    var 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 (Math.floor(num / den));}Â
// Driver codevar k = 3;var avg = 4;var arr = [ 2, 7, 3 ];var n = arr.length;document.write(findMissing(arr, n, k, avg));Â
// This code is contributed by ShubhamSingh10</script> |
Output:Â
4
Â
Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(1)
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