Count of elements whose absolute difference with the sum of all the other elements is greater than k
Last Updated :
03 Mar, 2022
Given an array arr[] of N integers and an integer K, the task is to find the number of anomalies in the array. An anomaly is a number for which the absolute difference between it and all the other numbers in the array is greater than K. Find the number of anomalies.
Examples:
Input: arr[] = {1, 3, 5}, k = 1
Output: 2
1 and 3 are the anomalies.
|1 - (3 + 5)| = 7 > 1
|3 - (1 + 5)| = 3 > 1
Input: arr[] = {7, 1, 8}, k = 5
Output: 1
Approach: Find the sum of all the array elements and store it in sum, now for every element of the array arr[i] if the absolute difference of arr[i] with sum - arr[i] is > k then it is an anomaly. Count all the anomalies in the array and print the result in the end.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
// Function to return the number of anomalies
static int countAnomalies(int arr[],
int n, int k)
{
// To store the count of anomalies
int cnt = 0;
// To store the sum of the array elements
int i, sum = 0;
// Find the sum of the array elements
for (i = 0; i < n; i++)
sum += arr[i];
// Count the anomalies
for (i = 0; i < n; i++)
if (abs(arr[i] -
(sum - arr[i])) > k)
cnt++;
return cnt;
}
// Driver code
int main()
{
int arr[] = { 1, 3, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 1;
cout << countAnomalies(arr, n, k);
}
// This code is contributed
// by Code_Mech
// Java implementation of the approach
class GFG {
// Function to return the number of anomalies
static int countAnomalies(int arr[], int n, int k)
{
// To store the count of anomalies
int cnt = 0;
// To store the sum of the array elements
int i, sum = 0;
// Find the sum of the array elements
for (i = 0; i < n; i++)
sum += arr[i];
// Count the anomalies
for (i = 0; i < n; i++)
if (Math.abs(arr[i] - (sum - arr[i])) > k)
cnt++;
return cnt;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 3, 5 };
int n = arr.length;
int k = 1;
System.out.print(countAnomalies(arr, n, k));
}
}
# Python3 implementation of the approach
# Function to return the
# number of anomalies
def countAnomalies(arr, n, k):
# To store the count of anomalies
cnt = 0
# To store the Sum of
# the array elements
i, Sum = 0, 0
# Find the Sum of the array elements
for i in range(n):
Sum += arr[i]
# Count the anomalies
for i in range(n):
if (abs(arr[i] - (Sum - arr[i])) > k):
cnt += 1
return cnt
# Driver code
arr = [1, 3, 5]
n = len(arr)
k = 1
print(countAnomalies(arr, n, k))
# This code is contributed
# by mohit kumar
// C# implementation of the approach
using System;
class GFG
{
// Function to return the number of anomalies
static int countAnomalies(int[] arr, int n, int k)
{
// To store the count of anomalies
int cnt = 0;
// To store the sum of the array elements
int i, sum = 0;
// Find the sum of the array elements
for (i = 0; i < n; i++)
sum += arr[i];
// Count the anomalies
for (i = 0; i < n; i++)
if (Math.Abs(arr[i] - (sum - arr[i])) > k)
cnt++;
return cnt;
}
// Driver code
public static void Main()
{
int[] arr = { 1, 3, 5 };
int n = arr.Length;
int k = 1;
Console.WriteLine(countAnomalies(arr, n, k));
}
}
// This code is contributed by Code_Mech.
<?php
// PHP implementation of the approach
// Function to return
// the number of anomalies
function countAnomalies($arr, $n, $k)
{
// To store the count of anomalies
$cnt = 0;
// To store the sum of
// the array elements
$sum = 0;
// Find the sum of the array elements
for ($i = 0; $i < $n; $i++)
$sum += $arr[$i];
// Count the anomalies
for ($i = 0; $i < $n; $i++)
if (abs($arr[$i] -
($sum - $arr[$i])) > $k)
$cnt++;
return $cnt;
}
// Driver code
$arr = array(1, 3, 5);
$n = count($arr);
$k = 1;
echo countAnomalies($arr, $n, $k);
// This code is contributed by Ryuga
?>
<script>
// Javascript implementation of the approach
// Function to return the number of anomalies
function countAnomalies(arr, n, k)
{
// To store the count of anomalies
var cnt = 0;
// To store the sum of the array elements
var i, sum = 0;
// Find the sum of the array elements
for(i = 0; i < n; i++)
sum += arr[i];
// Count the anomalies
for(i = 0; i < n; i++)
if (Math.abs(arr[i] - (sum - arr[i])) > k)
cnt++;
return cnt;
}
// Driver code
var arr = [ 1, 3, 5 ];
var n = arr.length;
var k = 1;
document.write(countAnomalies(arr, n, k));
// This code is contributed by umadevi9616
</script>
Time Complexity : O(n)
Auxiliary Space : O(1)
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