计算第一个数组中绝对差大于 K 的元素与第二个数组中的元素

给定两个数组 arr1[] 和 arr2[]以及一个整数 K ，我们的任务是找到第一个数组中的元素个数，对于元素x，在 arr1[] 中，在 arr2[] 中至少存在一个元素y使得 x 和 y 的绝对差大于整数K 。

例子：

Input: arr1 = {3, 1, 4}, arr2 = {5, 1, 2}, K = 2 
Output: 2 
Explanation: 
Such elements are 1 and 4. 
For 1, arr2[] has 5 and abs(1 – 5) = 4 which is greater than 2. 
For 4, arr2[] has 1 and abs(4 – 1) = 3 which again is greater than 2.

Input: arr1 = {1, 2}, arr2 = {4, 6}, K = 3 
Output: 2 
Explanation: 
Such elements are 1 and 2. 
For 1, arr2[] has 6 and abs(1 – 6) = 5 which is greater than 3. 
For 2, arr2[] has 6 and abs(2 – 6) = 4 which is greater than 3. 
 

朴素的方法：迭代 arr1[] 中的每个元素，并检查 arr2 中是否存在一个元素，使得它们的绝对差大于值 K。

时间复杂度： O(N * M)其中 N 和 M 分别是数组 1 和 2 的大小。

高效的方法：为了优化上面的方法，我们必须观察到对于arr1[]中的每个元素，我们只需要arr2[]中最小和最大的元素来检查它是否远离。对于arr1中的每个元素x，如果最小或最大值与x的绝对差大于K，则该元素是远距离的。

下面是上述方法的实现：

C++

// C++ program to count elements in first Array
// with absolute difference greater than K
// with an element in second Array
 
#include 
using namespace std;
 
// Function to count the such elements
void countDist(int arr1[], int n, int arr2[],
               int m, int k)
{
    // Store count of required elements in arr1
    int count = 0;
 
    // Initialise the smallest and the largest
    // value from the second array arr2[]
    int smallest = arr2[0];
    int largest = arr2[0];
 
    // Find the smallest and
    // the largest element in arr2
    for (int i = 0; i < m; i++) {
        smallest = max(smallest, arr2[i]);
        largest = min(largest, arr1[i]);
    }
    for (int i = 0; i < n; i++) {
 
        // Check if absolute difference of smallest
        // and arr1[i] or largest and arr1[i] is > K
        // then arr[i] is a required element
        if (abs(arr1[i] - smallest) > k
            || abs(arr1[i] - largest) > k)
            count++;
    }
 
    // Print the final result
    cout << count;
}
 
// Driver code
int main()
{
    int arr1[] = { 3, 1, 4 };
    int n = sizeof(arr1) / sizeof(arr1[0]);
    int arr2[] = { 5, 1, 2 };
    int m = sizeof(arr2) / sizeof(arr2[0]);
    int k = 2;
 
    countDist(arr1, n, arr2, m, k);
 
    return 0;
}

Java

// Java program to count elements in first Array
// with absolute difference greater than K
// with an element in second Array
class GFG{
 
// Function to count the such elements
static void countDist(int arr1[], int n,
                      int arr2[], int m,
                      int k)
{
    // Store count of required elements in arr1
    int count = 0;
 
    // Initialise the smallest and the largest
    // value from the second array arr2[]
    int smallest = arr2[0];
    int largest = arr2[0];
 
    // Find the smallest and
    // the largest element in arr2
    for(int i = 0; i < m; i++)
    {
       smallest = Math.max(smallest, arr2[i]);
       largest = Math.min(largest, arr1[i]);
    }
    for(int i = 0; i < n; i++)
    {
 
       // Check if absolute difference
       // of smallest and arr1[i] or
       // largest and arr1[i] is > K
       // then arr[i] is a required element
       if (Math.abs(arr1[i] - smallest) > k ||
           Math.abs(arr1[i] - largest) > k)
           count++;
    }
 
    // Print the final result
    System.out.print(count);
}
 
// Driver code
public static void main(String[] args)
{
    int arr1[] = { 3, 1, 4 };
    int n = arr1.length;
    int arr2[] = { 5, 1, 2 };
    int m = arr2.length;
    int k = 2;
 
    countDist(arr1, n, arr2, m, k);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to count elements in the first Array
# with an absolute difference greater than K
# with an element in the second Array
 
# Function to count the such elements
def countDist(arr1, n, arr2, m, k):
     
    # Store count of required elements in arr1
    count = 0
 
    # Initialise the smallest and the largest
    # value from the second array arr2[]
    smallest = arr2[0]
    largest = arr2[0]
 
    # Find the smallest and
    # the largest element in arr2
    for i in range(m):
        smallest = max(smallest, arr2[i])
        largest = min(largest, arr1[i])
 
    for i in range(n):
 
        # Check if absolute difference of smallest
        # and arr1[i] or largest and arr1[i] is > K
        # then arr[i] is a required element
        if (abs(arr1[i] - smallest) > k
            or abs(arr1[i] - largest) > k):
            count += 1
 
    # Print final result
    print(count)
 
 
# Driver code
if __name__ == '__main__':
     
    arr1= [ 3, 1, 4 ]
    n = len(arr1)
    arr2= [ 5, 1, 2 ]
    m = len(arr2)
    k = 2
 
    countDist(arr1, n, arr2, m, k)
     
# This code is contributed by mohit kumar 29

C#

// C# program to count elements in first array
// with absolute difference greater than K
// with an element in second Array
using System;
 
class GFG{
 
// Function to count the such elements
static void countDist(int []arr1, int n,
                      int []arr2, int m,
                      int k)
{
    // Store count of required elements in arr1
    int count = 0;
 
    // Initialise the smallest and the largest
    // value from the second array arr2[]
    int smallest = arr2[0];
    int largest = arr2[0];
 
    // Find the smallest and
    // the largest element in arr2
    for(int i = 0; i < m; i++)
    {
       smallest = Math.Max(smallest, arr2[i]);
       largest = Math.Min(largest, arr1[i]);
    }
    for(int i = 0; i < n; i++)
    {
        
       // Check if absolute difference
       // of smallest and arr1[i] or
       // largest and arr1[i] is > K
       // then arr[i] is a required element
       if (Math.Abs(arr1[i] - smallest) > k ||
           Math.Abs(arr1[i] - largest) > k)
           count++;
    }
 
    // Print the readonly result
    Console.Write(count);
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr1 = { 3, 1, 4 };
    int n = arr1.Length;
    int []arr2 = { 5, 1, 2 };
    int m = arr2.Length;
    int k = 2;
 
    countDist(arr1, n, arr2, m, k);
}
}
 
// This code is contributed by gauravrajput1

Javascript

 
// Javascript program to count elements in
// first Array with absolute difference
// greater than K with an element in
// second Array   
 
// Function to count the such elements
function countDist(arr1, n, arr2, m, k)
{
     
    // Store count of required elements in arr1
    var count = 0;
 
    // Initialise the smallest and the largest
    // value from the second array arr2
    var smallest = arr2[0];
    var largest = arr2[0];
 
    // Find the smallest and
    // the largest element in arr2
    for(i = 0; i < m; i++)
    {
        smallest = Math.max(smallest, arr2[i]);
        largest = Math.min(largest, arr1[i]);
    }
    for(i = 0; i < n; i++)
    {
         
        // Check if absolute difference
        // of smallest and arr1[i] or
        // largest and arr1[i] is > K
        // then arr[i] is a required element
        if (Math.abs(arr1[i] - smallest) > k ||
            Math.abs(arr1[i] - largest) > k)
            count++;
    }
 
    // Prvar the final result
    document.write(count);
}
 
// Driver code
var arr1 = [ 3, 1, 4 ];
var n = arr1.length;
var arr2 = [ 5, 1, 2 ];
var m = arr2.length;
var k = 2;
 
countDist(arr1, n, arr2, m, k);
 
// This code is contributed by umadevi9616
 

输出：

2

时间复杂度： O(N + M)，其中 N 和 M 是给定数组的大小。
辅助空间： O(1)。