Given an array A of N integers. An anomaly is a number for which the absolute difference between it and every other number in the array is greater than K where k is a given positive integer. Find the number of anomalies.
Examples:
Input : arr[] = {1, 3, 5}, k = 1
Output : 3
Explanation:
All the numbers in the array are anomalies because
For the number 1 abs(1-3)=2, abs(1-5)=4 which all are greater than 1,
For the number 3 abs(3-1)=2, abs(3-5)=2 which all are again greater than 1
For the number 5 abs(5-1)=4, abs(5-3)=2 which all are again greater than 1
So there are 3 anomalies.
Input : arr[] = {7, 1, 8}, k = 5
Output : 1
Simple Approach: We simply check for each number if it satisfies the given condition, that is, the absolute difference is greater than K or not with each of the other numbers.
C++
// A simple C++ solution to count anomalies in// an array.#include<iostream>using namespace std;int countAnomalies(int arr[], int n, int k){ int res = 0; for (int i=0; i<n; i++) { int j; for (j=0; j<n; j++) if (i != j && abs(arr[i] - arr[j]) <= k) break; if (j == n) res++; } return res;}int main(){ int arr[] = {7, 1, 8}, k = 5; int n = sizeof(arr)/sizeof(arr[0]); cout << countAnomalies(arr, n, k); return 0;} |
Java
// A simple java solution to count // anomalies in an array.class GFG {static int countAnomalies(int arr[], int n, int k){ int res = 0; for (int i = 0; i < n; i++) { int j; for (j = 0; j < n; j++) if (i != j && Math.abs(arr[i] - arr[j]) <= k) break; if (j == n) res++; } return res;}// Driver codepublic static void main(String args[]){ int arr[] = {7, 1, 8}, k = 5; int n = arr.length; System.out.println(countAnomalies(arr, n, k));}}// This code is contributed by ANKITRAI1 |
Python3
# A simple Python3 solution to # count anomalies in an array. def countAnomalies(arr, n, k): res = 0 for i in range(0, n): j = 0 while j < n: if i != j and abs(arr[i] - arr[j]) <= k: break j += 1 if j == n: res += 1 return res # Driver Codeif __name__ == "__main__": arr = [7, 1, 8] k = 5 n = len(arr) print(countAnomalies(arr, n, k)) # This code is contributed by Rituraj Jain |
C#
// A simple C# solution to count // anomalies in an array.using System;class GFG {static int countAnomalies(int[] arr, int n, int k){ int res = 0; for (int i = 0; i < n; i++) { int j; for (j = 0; j < n; j++) if (i != j && Math.Abs(arr[i] - arr[j]) <= k) break; if (j == n) res++; } return res;}// Driver codepublic static void Main(){ int[] arr = {7, 1, 8}; int k = 5; int n = arr.Length; Console.WriteLine(countAnomalies(arr, n, k));}}// This code is contributed// by Akanksha Rai(Abby_akku) |
PHP
<?php // A simple PHP solution to count // anomalies in an array.function countAnomalies(&$arr, $n, $k){ $res = 0; for ($i = 0; $i < $n; $i++) { for ($j = 0; $j < $n; $j++) if ($i != $j && abs($arr[$i] - $arr[$j]) <= $k) break; if ($j == $n) $res++; } return $res;}// Driver Code$arr = array(7, 1, 8);$k = 5;$n = sizeof($arr);echo countAnomalies($arr, $n, $k);// This code is contributed// by ChitraNayal?> |
Javascript
<script>// A simple javascript solution to count// anomalies in an array.function countAnomalies(arr,n,k){ let res = 0; for (let i = 0; i < n; i++) { let j; for (j = 0; j < n; j++) if (i != j && Math.abs(arr[i] - arr[j]) <= k) break; if (j == n) res++; } return res;}// Driver codelet arr=[7, 1, 8];let k = 5;let n = arr.length;document.write(countAnomalies(arr, n, k)); // This code is contributed by avanitrachhadiya2155</script> |
Output
1
Time Complexity: O(n * n)
Auxiliary Space: O(1)
Efficient Approach using Binary Search
- Sort the array.
- For every element, find the largest element greater than it and the smallest element greater than it. We can find these two in O(Log n) time using Binary Search. If the difference between these two elements is more than k, we increment the result.
Prerequisites for below C++ code : lower_bound in C++, upper_bound in C++.
Implementation:
C++
#include<bits/stdc++.h>using namespace std;int countAnomalies(int a[], int n, int k){ // Sort the array so that we can apply binary // search. sort(a, a+n); // One by one check every element if it is // anomaly or not using binary search. int res = 0; for (int i=0; i<n; i++) { int *u = upper_bound(a, a+n, a[i]); // If arr[i] is not largest element and // element just greater than it is within // k, then return false. if (u != a+n && ((*u) - a[i]) <= k) continue; int *s = lower_bound(a, a+n, a[i]); // If there are more than one occurrences // of arr[i], return false. if (u - s > 1) continue; // If arr[i] is not smallest element and // just smaller element is not k distance away if (s != a && (*(s - 1) - a[i]) <= k) continue; res++; } return res;}int main(){ int arr[] = {7, 1, 8}, k = 5; int n = sizeof(arr)/sizeof(arr[0]); cout << countAnomalies(arr, n, k); return 0;} |
Java
import java.util.*;class GFG { static int countAnomalies(int a[], int n, int k) { // Sort the array so that we can apply binary // search. Arrays.sort(a); // One by one check every element if it is // anomaly or not using binary search. int res = 0; for (int i = 0; i < n; i++) { int u = upper_bound(a, 0, n, a[i]); // If arr[i] is not largest element and // element just greater than it is within // k, then return false. if (u < n && a[u] - a[i] <= k) continue; int s = lower_bound(a, 0, n, a[i]); // If there are more than one occurrences // of arr[i], return false. if (u - s > 1) continue; // If arr[i] is not smallest element and // just smaller element is not k distance away if (s > 0 && a[s - 1] - a[i] <= k) continue; res++; } return res; } static int lower_bound(int[] a, int low, int high, int element) { while (low < high) { int middle = low + (high - low) / 2; if (element > a[middle]) low = middle + 1; else high = middle; } return low; } static int upper_bound(int[] a, int low, int high, int element) { while (low < high) { int middle = low + (high - low) / 2; if (a[middle] > element) high = middle; else low = middle + 1; } return low; } public static void main(String[] args) { int arr[] = { 7, 1, 8 }, k = 5; int n = arr.length; System.out.print(countAnomalies(arr, n, k)); }}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to implement# the above approachdef countAnomalies(a, n, k): # Sort the array so that # we can apply binary # search. a = sorted(a); # One by one check every # element if it is anomaly # or not using binary search. res = 0; for i in range(n): u = upper_bound(a, 0, n, a[i]); # If arr[i] is not largest # element and element just # greater than it is within # k, then return False. if (u < n and a[u] - a[i] <= k): continue; s = lower_bound(a, 0, n, a[i]); # If there are more than # one occurrences of arr[i], # return False. if (u - s > 1): continue; # If arr[i] is not smallest # element and just smaller # element is not k distance away if (s > 0 and a[s - 1] - a[i] <= k): continue; res += 1; return res;def lower_bound(a, low, high, element): while (low < high): middle = int(low + int(high - low) / 2); if (element > a[middle]): low = middle + 1; else: high = middle; return low;def upper_bound(a, low, high, element): while (low < high): middle = int(low + (high - low) / 2); if (a[middle] > element): high = middle; else: low = middle + 1; return low;# Driver codeif __name__ == '__main__': arr = [7, 1, 8] k = 5; n = len(arr); print(countAnomalies(arr, n, k));# This code is contributed by shikhasingrajput |
C#
using System;class GFG{static int countAnomalies(int []a, int n, int k) { // Sort the array so that we can // apply binary search. Array.Sort(a); // One by one check every element if it is // anomaly or not using binary search. int res = 0; for(int i = 0; i < n; i++) { int u = upper_bound(a, 0, n, a[i]); // If arr[i] is not largest element and // element just greater than it is within // k, then return false. if (u < n && a[u] - a[i] <= k) continue; int s = lower_bound(a, 0, n, a[i]); // If there are more than one occurrences // of arr[i], return false. if (u - s > 1) continue; // If arr[i] is not smallest element and // just smaller element is not k distance away if (s > 0 && a[s - 1] - a[i] <= k) continue; res++; } return res;}static int lower_bound(int[] a, int low, int high, int element){ while (low < high) { int middle = low + (high - low) / 2; if (element > a[middle]) low = middle + 1; else high = middle; } return low;}static int upper_bound(int[] a, int low, int high, int element) { while (low < high) { int middle = low + (high - low) / 2; if (a[middle] > element) high = middle; else low = middle + 1; } return low;}// Driver codepublic static void Main(String[] args){ int []arr = { 7, 1, 8 }; int k = 5; int n = arr.Length; Console.Write(countAnomalies(arr, n, k));}}// This code is contributed by Amit Katiyar |
Javascript
<script>// JavaScript program for the above approach function countAnomalies(a, n, k) { // Sort the array so that we can apply binary // search. a.sort(); // One by one check every element if it is // anomaly or not using binary search. let res = 0; for (let i = 0; i < n; i++) { let u = upper_bound(a, 0, n, a[i]); // If arr[i] is not largest element and // element just greater than it is within // k, then return false. if (u < n && a[u] - a[i] <= k) continue; let s = lower_bound(a, 0, n, a[i]); // If there are more than one occurrences // of arr[i], return false. if (u - s > 1) continue; // If arr[i] is not smallest element and // just smaller element is not k distance away if (s > 0 && a[s - 1] - a[i] <= k) continue; res++; } return res; } function lower_bound(a, low, high, element) { while (low < high) { let middle = low + Math.floor((high - low) / 2); if (element > a[middle]) low = middle + 1; else high = middle; } return low; } function upper_bound(a, low, high, element) { while (low < high) { let middle = low + Math.floor((high - low) / 2); if (a[middle] > element) high = middle; else low = middle + 1; } return low; }// Driver Code let arr = [ 7, 1, 8 ], k = 5; let n = arr.length; document.write(countAnomalies(arr, n, k));</script> |
Output
1
Time Complexity: O(n Log n)
Auxiliary Space: O(1)
Another efficient solution for small k
- Insert all values of the array in a hash table.
- Traverse array again and for every value arr[i], search for every value from arr[i] – k to arr[i] + k (excluding arr[i]). If none of the elements are found, then arr[i] is anomaly.
Time Complexity: O(n k)
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