计算选择差异最大的一对的方法

给定一个由n个整数组成的数组，我们需要找到no。选择差异最大的对的方式。

例子：

Input : a[] = {3, 2, 1, 1, 3}
Output : 4
Explanation:- Here, the maximum difference 
you can find is 2 which is from (1, 3).
No. of ways of choosing it:
 1) Choosing the first and third elements,
 2) Choosing the first and fourth elements,
 3) Choosing the third and fifth elements,
 4) Choosing the fourth and fifth elements.
Hence ans is 4.

Input : a[] = {2, 4, 1, 1}
Output : 2
Explanation:- Here, the maximum difference 
is 3 from (1, 4). No. of ways choosing it:
1) Choosing the second and third elements,
2) Choosing the second and fourth elements.
   Hence ans is 2.

天真的方法：一个简单的解决方案是找到最小元素和最大元素以找到最大差。然后，我们可以找到否。通过运行两个循环来选择一对的方法。在内部循环中，检查两个元素(一个在外部循环中，另一个在内部循环中)是否有最大差异，如果是，则增加计数。最后输出计数。
时间复杂度：O(n ^ 2)
辅助空间：O(1)

高效的方法：
一种有效的方法是：

• 情况一(如果所有元素都相等)：ans否。从一组n个元素中选择2个元素的方法$nC2$就是n(n-1)/ 2。
• 情况二(如果所有要素都不相等)：答案是“否”计数的乘积。最小元素(c1)的数量和数量最大元素的数量(c2)，即c1 * c2

C++

// CPP Code to find no. of Ways of choosing
// a pair with maximum difference
#include 
using namespace std;
  
int countPairs(int a[], int n)
{
    // To find minimum and maximum of
    // the array
    int mn = INT_MAX;
    int mx = INT_MIN;
    for (int i = 0; i < n; i++) {
        mn = min(mn, a[i]);
        mx = max(mx, a[i]);
    }
  
    // to find the count of minimum and
    // maximum elements
    int c1 = 0;
    int c2 = 0; // Count variables
    for (int i = 0; i < n; i++) {
        if (a[i] == mn)
            c1++;
        if (a[i] == mx)
            c2++;
    }
  
    // condition for all elements equal
    if (mn == mx)
        return n * (n - 1) / 2;
    else
        return c1 * c2;
}
  
// Driver code
int main()
{
    int a[] = { 3, 2, 1, 1, 3 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << countPairs(a, n);
    return 0;
}

Java

// Java Code to find no. of Ways of choosing
// a pair with maximum difference
import java.util.*;
  
class GFG {
  
    static int countPairs(int a[], int n)
    {
  
        // To find minimum and maximum of
        // the array
        int mn = Integer.MAX_VALUE;
        int mx = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            mn = Math.min(mn, a[i]);
            mx = Math.max(mx, a[i]);
        }
  
        // to find the count of minimum and
        // maximum elements
        int c1 = 0;
        int c2 = 0; // Count variables
        for (int i = 0; i < n; i++) {
            if (a[i] == mn)
                c1++;
            if (a[i] == mx)
                c2++;
        }
  
        // condition for all elements equal
        if (mn == mx)
            return n * (n - 1) / 2;
        else
            return c1 * c2;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int a[] = { 3, 2, 1, 1, 3 };
        int n = a.length;
        System.out.print(countPairs(a, n));
    }
}
  
// This code is contributed by Anant Agarwal.

Python3

# Python Code to find no.
# of Ways of choosing
# a pair with maximum difference
  
def countPairs(a, n):
  
    # To find minimum and maximum of 
    # the array 
    mn = [phone]
    mx = -2147483648
    for i in range(n):
        mn = min(mn, a[i])
        mx = max(mx, a[i])
      
       
    # to find the count of minimum and 
    # maximum elements
    c1 = 0
    c2 = 0 # Count variables
    for i in range(n):
        if (a[i] == mn):
            c1+= 1
        if (a[i] == mx):
            c2+= 1
      
   
    # condition for all elements equal
    if (mn == mx): 
        return  n*(n - 1) // 2
    else:
        return c1 * c2
  
# Driver code
  
a = [ 3, 2, 1, 1, 3]
n = len(a)
  
print(countPairs(a, n))
  
# This code is contributed
# by Anant Agarwal.

C#

// C# Code to find no. of Ways of choosing
// a pair with maximum difference
using System;
  
class GFG {
  
    static int countPairs(int[] a, int n)
    {
  
        // To find minimum and maximum of
        // the array
        int mn = int.MaxValue;
        int mx = int.MinValue;
        for (int i = 0; i < n; i++) {
            mn = Math.Min(mn, a[i]);
            mx = Math.Max(mx, a[i]);
        }
  
        // to find the count of minimum and
        // maximum elements
        int c1 = 0;
        int c2 = 0; // Count variables
        for (int i = 0; i < n; i++) {
            if (a[i] == mn)
                c1++;
            if (a[i] == mx)
                c2++;
        }
  
        // condition for all elements equal
        if (mn == mx)
            return n * (n - 1) / 2;
        else
            return c1 * c2;
    }
  
    // Driver code
    public static void Main()
    {
          
        int[] a = { 3, 2, 1, 1, 3 };
        int n = a.Length;
          
        Console.WriteLine(countPairs(a, n));
    }
}
  
// This code is contributed by vt_m.

输出：

4

时间复杂度：找到最小和最大值的时间复杂度为O(n) ，找到最小和最大计数的时间复杂度为O(n)
总时间复杂度：O(n)
辅助空间：O(1)