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++
#include <bits/stdc++.h>
using namespace std;
int findMissing( int arr[], int n, int k, int avg)
{
int sum = 0;
for ( int i = 0; i < n; i++) {
sum += arr[i];
}
int num = (avg * (n + k)) - sum;
int den = k;
if (num % den != 0)
return -1;
return (num / den);
}
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
class GFG
{
static int findMissing( int arr[], int n,
int k, int avg)
{
int sum = 0 ;
for ( int i = 0 ; i < n; i++)
{
sum += arr[i];
}
int num = (avg * (n + k)) - sum;
int den = k;
if (num % den != 0 )
return - 1 ;
return ( int )(num / den);
}
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));
}
}
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Python3
def findMissing(arr, n, k, avg):
sum = 0 ;
for i in range (n):
sum + = arr[i];
num = (avg * (n + k)) - sum ;
den = k;
if (num % den ! = 0 ):
return - 1 ;
return ( int )(num / den);
k = 3 ; avg = 4 ;
arr = [ 2 , 7 , 3 ] ;
n = len (arr);
print (findMissing(arr, n, k, avg));
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C#
using System;
class GFG
{
static int findMissing( int []arr, int n,
int k, int avg)
{
int sum = 0;
for ( int i = 0; i < n; i++)
{
sum += arr[i];
}
int num = (avg * (n + k)) - sum;
int den = k;
if (num % den != 0)
return -1;
return ( int )(num / den);
}
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));
}
}
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Javascript
<script>
function findMissing(arr, n, k, avg)
{
var sum = 0;
for ( var i = 0; i < n; i++) {
sum += arr[i];
}
var num = (avg * (n + k)) - sum;
var den = k;
if (num % den != 0)
return -1;
return (Math.floor(num / den));
}
var k = 3;
var avg = 4;
var arr = [ 2, 7, 3 ];
var n = arr.length;
document.write(findMissing(arr, n, k, avg));
</script>
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Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(1)