差异小于K的对

给定一个由n个整数组成的数组，我们需要找到所有差异小于k的对

例子 ：

Input : a[] = {1, 10, 4, 2}
        K = 3
Output : 2
We can make only two pairs 
with difference less than 3.
(1, 2) and (4, 2)

Input : a[] = {1, 8, 7}
        K = 7
Output : 2
Pairs with difference less than 7
are (1, 7) and (8, 7)

方法1(简单) ：运行两个嵌套循环。外循环逐个选择每个元素x。内部循环会考虑x之后的所有元素，并检查差异是否在限制范围内。

C++

// CPP code to find count of Pairs with 
// difference less than K.
#include 
using namespace std;
  
int countPairs(int a[], int n, int k)
{
    int res = 0;
    for (int i = 0; i < n; i++) 
      for (int j=i+1; j

---

Java

// java code to find count of Pairs with 
// difference less than K.
import java.io.*;
  
class GFG {
    static int countPairs(int a[], int n, int k)
    {
        int res = 0;
        for (int i = 0; i < n; i++) 
        for (int j = i + 1; j < n; j++)
            if (Math.abs(a[j] - a[i]) < k) 
                res++;
      
        return res;
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        int a[] = {1, 10, 4, 2};
        int k = 3;
        int n = a.length;
        System.out.println(countPairs(a, n, k)); 
          
    }
}
  
// This code is contributed by vt_m.

Python3

# Python3 code to find count of Pairs  
# with difference less than K.
  
def countPairs(a, n, k):
    res = 0
    for i in range(n): 
        for j in range(i + 1, n):
            if (abs(a[j] - a[i]) < k):
                res += 1
  
    return res
      
# Driver code
a = [1, 10, 4, 2]
k = 3
n = len(a)
print(countPairs(a, n, k), end = "")
  
# This code is contributed by Azkia Anam.

C#

// C# code to find count of Pairs 
//  with difference less than K.
using System;
  
class GFG {
      
    // Function to count pairs
    static int countPairs(int []a, int n, 
                          int k)
    {
        int res = 0;
        for (int i = 0; i < n; i++) 
        for (int j = i + 1; j < n; j++)
            if (Math.Abs(a[j] - a[i]) < k) 
                res++;
      
        return res;
    }
      
    // Driver code
    public static void Main () 
    {
        int []a = {1, 10, 4, 2};
        int k = 3;
        int n = a.Length;
        Console.WriteLine(countPairs(a, n, k)); 
          
    }
}
  
// This code is contributed by vt_m.

输出 ：

2

时间复杂度： O(n ² )
辅助空间： O(1)

方法2(排序)：首先，我们对数组进行排序。然后，我们从第一个元素开始，并在差小于k的情况下继续考虑对。如果发现差异大于或等于k时停止循环，则移至下一个元素。

C++

// CPP code to find count of Pairs with 
// difference less than K.
#include 
using namespace std;
  
int countPairs(int a[], int n, int k)
{
    // to sort the array.
    sort(a, a + n);
  
    int res = 0;
    for (int i = 0; i < n; i++) {
  
        // Keep incrementing result while
        // subsequent elements are within
        // limits.
        int j = i+1; 
        while (j < n && a[j] - a[i] < k) {
            res++;
            j++;
        }
    }
    return res;
}
  
// Driver code
int main()
{
    int a[] =  {1, 10, 4, 2};
    int k = 3;
    int n = sizeof(a) / sizeof(a[0]);
    cout << countPairs(a, n, k) << endl; 
    return 0;
}

Java

// Java code to find count of Pairs with 
// difference less than K.
import java.io.*;
import java.util.Arrays;
  
class GFG
{
    static int countPairs(int a[], int n, int k)
    {
        // to sort the array.
        Arrays.sort(a);
      
        int res = 0;
        for (int i = 0; i < n; i++) 
        {
      
            // Keep incrementing result while
            // subsequent elements are within
            // limits.
            int j = i + 1; 
            while (j < n && a[j] - a[i] < k) 
            {
                res++;
                j++;
            }
        }
        return res;
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        int a[] = {1, 10, 4, 2};
        int k = 3;
        int n = a.length;
        System.out.println(countPairs(a, n, k));
    }
}
  
// This code is contributed by vt_m.

Python3

# Python code to find count of Pairs  
# with difference less than K.
  
def countPairs(a, n, k):
      
    # to sort the array
    a.sort()
    res = 0
    for i in range(n): 
          
        # Keep incrementing result while
        # subsequent elements are within limits.
        j = i+1
        while (j < n and a[j] - a[i] < k):
            res += 1
            j += 1
    return res
  
# Driver code
a = [1, 10, 4, 2]
k = 3
n = len(a)
print(countPairs(a, n, k), end = "")
  
# This code is contributed by Azkia Anam.

C＃

// C# code to find count of Pairs 
// with difference less than K.
using System;
  
class GFG {
      
    // Function to count pairs
    static int countPairs(int []a, int n,
                          int k)
    {
          
        // to sort the array.
        Array.Sort(a);
      
        int res = 0;
        for (int i = 0; i < n; i++) 
        {
      
            // Keep incrementing result while
            // subsequent elements are within
            // limits.
            int j = i + 1; 
            while (j < n && a[j] - a[i] < k) 
            {
                res++;
                j++;
            }
        }
        return res;
    }
      
    // Driver code
    public static void Main () 
    {
        int []a = {1, 10, 4, 2};
        int k = 3;
        int n = a.Length;
        Console.WriteLine(countPairs(a, n, k));
    }
}
  
// This code is contributed by vt_m.

输出 ：

2

时间复杂度：O(res)其中res是输出中的对数。请注意，在最坏的情况下，这也需要O(n ² )时间，但总体上效果要好得多。