数组中的异常数

给定一个由N个整数组成的数组A。异常是一个数字，其与数组中每个其他数字之间的绝对差大于K，其中k是给定的正整数。查找异常数量。

例子：

Input : arr[] = {1, 3, 5}, k = 1
Output : 3
Explanation:
All the numbers in the array are anamolies 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 anamolies.

Input : arr[] = {7, 1, 8}, k = 5
Output : 1

简单方法：
我们简单地检查每个数字是否满足给定条件，即每个其他数字的绝对差大于K或不大于K。

C++

// A simple C++ solution to count anomalies in
// an array.
#include
using namespace std;
 
int countAnomalies(int arr[], int n, int k)
{
   int res = 0;
   for (int i=0; i

---

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 code
public 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 Code
if __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 code
public 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)

输出：

1

时间复杂度： O(n * n)

使用二进制搜索的有效方法
1)对数组进行排序。
2)对于每个元素，找到大于它的最大元素，并找到大于它的最小元素。我们可以使用Binary Search在O(Log n)时间中找到这两个。如果这两个元素之间的差大于k，我们将增加结果。
以下C++代码的先决条件：C++中的lower_bound，C++中的upper_bound。

C++

#include
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 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 approach
def 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 code
if __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 code
public 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

输出：

1

时间复杂度： O(n Log n)
小k的另一种有效解决方案
1)将数组的所有值插入哈希表。
2)再次遍历数组，并为每个值arr [i]搜索从arr [i] – k到arr [i] + k(不包括arr [i])的每个值。如果找不到任何元素，则arr [i]是异常的。
时间复杂度：O(nk)