具有一个元素为 K 乘以另一个元素的不同对的计数

给定一个数组arr[]和一个整数K，找出一个元素是另一个元素的K倍的最大对数，即arr[i]=K*arr[j] 。

例子：

Input: arr[] = {1, 2, 1, 2, 4} K = 2
Output: 2
Explanation: There are two possible ways to construct pairs: ({1, 2}, {1, 2}) and ({1, 2}, {2, 4}).

Input: a = {5, 4, 3, 2, 1} K = 2
Output: 1
Explanation: We can construct either set {1, 2} or set {2, 4}.

方法：对给定的数组arr[]进行排序，并检查数组arr [] 的所有可能对，并检查给定的(i, j) arr[i]=2*arr[j]。请按照以下步骤解决问题：

• 使用 C++ STL 中的 sort函数对数组arr[]进行排序。
• 初始化一个向量，用于保持已使用元素的计数。
• 将变量ans初始化为0以存储所有可能对的计数。
• 使用变量i遍历范围[0, N-1]并执行以下步骤：  
  • 使用变量j遍历范围[l, N-1]并执行以下操作：
    • 如果used[j]和used[i]的值为 false并且arr[j]=K*arr[i] ，则将used[i]和used[j]的值设置为true并增加ans 1并打破循环。
• 最后，完成上述步骤后，打印ans 的值。

下面是上述方法的实现：

C++

// C++ program for the above approach.
#include 
using namespace std;
 
// Function to find the maximum number
// of pairs.
int maxPairs(vector a, int k)
{
    // Sort the array.
    sort(a.begin(), a.end());
    int n = a.size(), ans = 0;
 
    // mark as used
    vector used(n);
 
    // iterate over all elements
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
 
            // if condition is satisfied,
            // pair the elements
            if (!used[j]
                && a[j] == k * a[i]
                && !used[i]) {
                used[j] = used[i] = true;
                ans++;
                break;
            }
        }
    }
 
    return ans;
}
 
// Driver Code
int32_t main()
{
    vector a{ 1, 2, 1, 2, 4 };
    int k = 2;
    cout << maxPairs(a, k);
    return 0;
}

Java

// Java program for the above approach.
 
 
import java.util.Arrays;
 
class GFG {
 
    // Function to find the maximum number
    // of pairs.
public static int maxPairs(int[] a, int k)
{
    // Sort the array.
    Arrays.sort(a);
    int n = a.length, ans = 0;
 
    // mark as used
    boolean[] used = new boolean[n];
 
 
    // iterate over all elements
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
 
            // if condition is satisfied,
            // pair the elements
            if (!used[j]
                && a[j] == k * a[i]
                && !used[i]) {
                used[i] =  true;
                used[j] = used[i];
                ans++;
                break;
            }
        }
    }
 
    return ans;
}
 
    // Driver Code
    public static void main(String args[])
    {
        int[] a = {1, 2, 1, 2, 4};
        int k = 2;
        System.out.println(maxPairs(a, k));
    }
}
 
// This code is contributed by _saurabh_jaiswal.

Python3

# Python Program for the above approach
 
# Function to find the maximum number
# of pairs.
def maxPairs(a, k):
 
    # Sort the array.
    a.sort()
    n = len(a)
    ans = 0
 
    # mark as used
    used = [False] * n
 
    # iterate over all elements
    for i in range(0, n):
        for j in range(i + 1, n):
 
            # if condition is satisfied,
            # pair the elements
            if (used[j] == False and a[j] == k * a[i] and used[i] == False):
                used[j] = used[j] = True
                ans += 1
                break
 
    return ans
 
# Driver Code
a = [1, 2, 1, 2, 4]
k = 2
print(maxPairs(a, k))
 
# This code is contributed by _saurabh_jaiswal

C#

// C# program for the above approach.
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the maximum number
// of pairs.
static int maxPairs(List a, int k)
{
   
    // Sort the array.
    a.Sort();
    int n = a.Count, ans = 0;
 
    // mark as used
    int [] Ar = new int[n];
    List used = new List(Ar);
 
    // iterate over all elements
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
 
            // if condition is satisfied,
            // pair the elements
            if (used[j]==0
                && a[j] == k * a[i]
                && used[i]==0) {
                used[j] = used[i] = 1;
                ans++;
                break;
            }
        }
    }
 
    return ans;
}
 
// Driver Code
  public static void Main(){
    List a = new List(){ 1, 2, 1, 2, 4 };
    int k = 2;
    Console.Write(maxPairs(a, k));
  }
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript

        // JavaScript Program for the above approach
 
        // Function to find the maximum number
        // of pairs.
        function maxPairs(a, k)
        {
         
            // Sort the array.
            a.sort(function (x, y) { return x - y });
            let n = a.length, ans = 0;
 
            // mark as used
            let used = new Array(n).fill(false);
 
            // iterate over all elements
            for (let i = 0; i < n; i++) {
                for (let j = i + 1; j < n; j++) {
 
                    // if condition is satisfied,
                    // pair the elements
                    if (used[j] == false && a[j] == k * a[i] && used[i] == false) {
                        used[j] = used[j] = true;
                        ans++;
                        break;
                    }
                }
            }
 
            return ans;
        }
 
        // Driver Code
        let a = [1, 2, 1, 2, 4];
        let k = 2;
        document.write(maxPairs(a, k));
         
    // This code is contributed by Potta Lokesh
    

输出

2

时间复杂度： O(N^2)
空间复杂度： O(N)