计算数组中满足给定条件的有效对

给定一个整数数组arr[] ，任务是计算arr中有效元素对的数量。一对(arr[x], arr[y])被认为是无效的，如果

• arr[x] < arr[y]
• abs(arr[x] – arr[y]) 是奇数

注意：对（ARR [X]，ARR [Y]）和（ARR [Y]，ARR [X]）是两个不同的双当x = y和改编[I]的所有可能值的值的i是！ ≤ 120 。
例子：

Input: arr[] = {16, 17, 18} 
Output: 2 
Only valid pair is (18, 16)
Input: arr[] = {16, 16} 
Output: 2 
Valid pairs are (16, 16) and (16, 16) 
 

方法：我们可以处理表示数组中元素arr[i]计数的(arr[i], count)对，而不是处理所有元素。由于只有120 个可能的值，我们对每个元素组进行频率计数，以降低整体复杂性。
对于每对(arr[x], countX)和(arr[y], countY) ，如果满足条件，则总有效对将为countX * countY 。
如果arr[x] = arr[y] ，那么我们高估了一些对。在这种情况下，有效的对将是countA * (countA – 1)，因为没有元素可以与自己配对。
下面是上述方法的实现：

C++

//C++ implementation of the approach
#include
using namespace std;
 
// Function to return total valid pairs
int ValidPairs(int arr[],int n)
{
  
    // Initialize count of all the elements
    int count[121]={0};
  
    // frequency count of all the elements
    for(int i=0;i

---

Java

//Java implementation of the approach
 
class GFG{
// Function to return total valid pairs
static int ValidPairs(int arr[],int n)
{
 
    // Initialize count of all the elements
    int[] count=new int[121];
 
    // frequency count of all the elements
    for(int i=0;i

---

Python3

# Python3 implementation of the approach
 
# Function to return total valid pairs
def ValidPairs(arr):
 
    # Initialize count of all the elements
    count = [0] * 121
 
    # frequency count of all the elements
    for ele in arr:
        count[ele] += 1
 
    ans = 0
    for eleX, countX in enumerate(count):
        for eleY, countY in enumerate(count):
            if eleX < eleY:
                continue
            if (abs(eleX - eleY) % 2 == 1):
                continue
 
            # Add total valid pairs
            ans += countX * countY
            if eleX == eleY:
 
                # Exclude pairs made with a single element
                # i.e. (x, x)
                ans -= countX
 
    return ans
 
# Driver Code
arr = [16, 17, 18]
 
# Function call to print required answer
print(ValidPairs(arr))

C#

//C# implementation of the approach
using System;
 
class GFG{
// Function to return total valid pairs
static int ValidPairs(int[] arr,int n)
{
 
    // Initialize count of all the elements
    int[] count=new int[121];
 
    // frequency count of all the elements
    for(int i=0;i

---

Javascript

// javascript implementation of the approach    
// Function to return total valid pairs
    function ValidPairs(arr , n)
    {
 
        // Initialize count of all the elements
        var count = Array(121).fill(0);
 
        // frequency count of all the elements
        for (i = 0; i < n; i++)
            count[arr[i]] += 1;
 
        var ans = 0;
        for (i = 0; i < n; i++)
            for (j = 0; j < n; j++) {
                if (arr[i] < arr[j])
                    continue;
                if (Math.abs(arr[i] - arr[j]) % 2 == 1)
                    continue;
 
                // Add total valid pairs
                ans += count[arr[i]] * count[arr[j]];
                if (arr[i] == arr[j])
 
                    // Exclude pairs made with a single element
                    // i.e. (x, x)
                    ans -= count[arr[i]];
            }
        return ans;
    }
 
    // Driver Code
     
        var arr = [ 16, 17, 18 ];
        var n = arr.length;
 
        // Function call to prvar required answer
        document.write(ValidPairs(arr, n));
 
// This code is contributed by umadevi9616

输出：

1

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。