计算给定数组中不相等的元素对

给定一个包含N 个元素的数组arr[] 。任务是计算索引(i, j)的总数，使得arr[i] != arr[j]和i < j 。
例子：

Input: arr[] = {1, 1, 2} 
Output: 2 
(1, 2) and (1, 2) are the only valid pairs.
Input: arr[] = {1, 2, 3} 
Output: 3
Input: arr[] = {1, 1, 1} 
Output: 0 
 

方法：初始化一个计数变量cnt = 0并运行两个嵌套循环来检查每个可能的对当前对是否有效。如果有效，则增加计数变量。最后，打印有效对的计数。
下面是上述方法的实现：

C++

// C++ implementation of the approach
 
#include 
using namespace std;
 
// Function to return the
// count of valid pairs
int countPairs(int arr[], int n)
{
 
    // To store the required count
    int cnt = 0;
 
    // For each index pair (i, j)
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
 
            // If current pair is valid
            // then increment the count
            if (arr[i] != arr[j])
                cnt++;
        }
    }
    return cnt;
}
 
// Driven code
int main()
{
    int arr[] = { 1, 1, 2 };
    int n = sizeof(arr) / sizeof(int);
 
    cout << countPairs(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
    // Function to return the
    // count of valid pairs
    static int countPairs(int arr[], int n)
    {
     
        // To store the required count
        int cnt = 0;
     
        // For each index pair (i, j)
        for (int i = 0; i < n; i++)
        {
            for (int j = i + 1; j < n; j++)
            {
     
                // If current pair is valid
                // then increment the count
                if (arr[i] != arr[j])
                    cnt++;
            }
        }
        return cnt;
    }
     
    // Driven code
    public static void main (String[] args)
    {
        int arr[] = { 1, 1, 2 };
        int n = arr.length;
     
        System.out.println(countPairs(arr, n));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
 
# Function to return the
# count of valid pairs
def countPairs(arr, n):
 
    # To store the required count
    cnt = 0;
 
    # For each index pair (i, j)
    for i in range(n):
        for j in range(i + 1, n):
 
            # If current pair is valid
            # then increment the count
            if (arr[i] != arr[j]):
                cnt += 1;
 
    return cnt;
 
# Driver code
if __name__ == '__main__':
    arr = [ 1, 1, 2 ];
    n = len(arr);
 
    print(countPairs(arr, n));
     
# This code is contributed by 29AjayKumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function to return the
    // count of valid pairs
    static int countPairs(int []arr, int n)
    {
     
        // To store the required count
        int cnt = 0;
     
        // For each index pair (i, j)
        for (int i = 0; i < n; i++)
        {
            for (int j = i + 1; j < n; j++)
            {
     
                // If current pair is valid
                // then increment the count
                if (arr[i] != arr[j])
                    cnt++;
            }
        }
        return cnt;
    }
     
    // Driven code
    public static void Main()
    {
        int []arr = { 1, 1, 2 };
        int n = arr.Length;
     
        Console.WriteLine(countPairs(arr, n));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

 
// Javascript implementation of the approach
 
// Function to return the
// count of valid pairs
function countPairs(arr, n)
{
 
    // To store the required count
    var cnt = 0;
 
    // For each index pair (i, j)
    for (var i = 0; i < n; i++) {
        for (var j = i + 1; j < n; j++) {
 
            // If current pair is valid
            // then increment the count
            if (arr[i] != arr[j])
                cnt++;
        }
    }
    return cnt;
}
 
// Driven code
var arr = [ 1, 1, 2 ];
var n = arr.length;
document.write(countPairs(arr, n));
 

输出：

2

时间复杂度： O(N ² )

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