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 > 1Input: 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 anomaliesstatic 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 codeint 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
// Java implementation of the approachclass 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
# Python3 implementation of the approach# Function to return the # number of anomaliesdef 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 codearr = [1, 3, 5]n = len(arr)k = 1print(countAnomalies(arr, n, k))# This code is contributed # by mohit kumar |
C#
// C# implementation of the approachusing 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// 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?> |
Javascript
<script>// Javascript implementation of the approach// Function to return the number of anomaliesfunction 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 codevar arr = [ 1, 3, 5 ];var n = arr.length;var k = 1;document.write(countAnomalies(arr, n, k));// This code is contributed by umadevi9616</script> |
Output:
2
Time Complexity : O(n)
Auxiliary Space : O(1)
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