给定两个数组arr1 []和arr2 []和一个整数K ，我们的任务是在第一个数组中找到数字元素，对于arr1 []中的元素x，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.
天真的方法：对arr1 []中的每个元素进行迭代，并检查arr2中是否存在一个元素，以使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。
高效方法：要优化上述方法，我们必须观察到arr1 []中的每个元素，仅需要arr2 []的最小和最大元素即可检查其是否远离。对于每个元素x，在arr1中，如果最小值或最大值与x的绝对差大于K，则该元素较远。
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 29AjayKumarPython3
# 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 29C#
// 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输出：
2
时间复杂度： O(N + M) ，其中N和M是给定数组的大小。