Array 中存在 K 次且不存在双精度的元素的数量

Array 中存在 K 次且不存在双精度的元素的计数

给定一个包含 N 个整数的数组arr[] ，任务是找出数组中存在K 次且它们的 double 不存在于数组中的元素的计数。

例子：

Input: arr[] = {10, 6, 12, 8, 10, 8}, K = 2
Output: 2
Explanation: 10 is a valid number since it appears exactly two times and 20 does not appear in array.
8 is a valid number since it appears two times and 16 does not appear in array.

Input: arr[] = {1, 3, 5, 3}, K = 3
Output: 0
Explanation: No element in the given array satisfy the condition.

Input: arr[] = {1, 2, 2, 3, 3, 4}, K = 2
Output: 1
Explanation: Only 3 is valid element.
Though 2 is present twice but its double is also present.

方法：可以使用hashmap解决任务按照以下步骤解决问题：

• 将每个数字的出现存储在哈希图中
• 对于每个元素，如果频率为K ，请检查其双精度值是否存在于哈希图中。如果不存在，请将其存储为有效数字之一。

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 count valid numbers
int find(int arr[], int N, int K)
{
    // Store ans
    int ans = 0;
    unordered_map mp;
    for (int i = 0; i < N; i++) {
 
        // Storing frequency of elements
        mp[arr[i]]++;
    }
 
    for (auto it : mp) {
        // Simply checking the
        // given condition in question
        if (it.second == K) {
            if (mp.find(it.first * 2)
                == mp.end()) {
                ans++;
            }
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 10, 6, 12, 8, 10, 8 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 2;
    cout << find(arr, N, K);
    return 0;
}

Java

// JAVA program for the above approach
import java.util.*;
class GFG
{
 
  // Function to find the count valid numbers
  public static int find(int[] arr, int N, int K)
  {
 
    // Store ans
    int ans = 0;
    HashMap mp = new HashMap<>();
 
    for (int i = 0; i < N; i++) {
 
      // Storing frequency of elements
      if (mp.containsKey(arr[i])) {
        mp.put(arr[i], mp.get(arr[i]) + 1);
      }
      else {
        mp.put(arr[i], 1);
      }
    }
 
    for (Map.Entry i :
         mp.entrySet()) {
 
      // Simply checking the
      // given condition in question
      if (i.getValue() == K) {
        if (mp.containsKey(i.getKey() * 2)
            == false) {
          ans++;
        }
      }
    }
    return ans;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int[] arr = new int[] { 10, 6, 12, 8, 10, 8 };
    int N = arr.length;
    int K = 2;
    System.out.print(find(arr, N, K));
  }
}
 
// This code is contributed by Taranpreet

Python3

# Python program for the above approach
 
# Function to find the count valid numbers
def find(arr, N, K):
    # Store ans
    ans = 0
    mp = {}
    for i in range(N):
        # Storing frequency of elements
        if arr[i] not in mp:
          mp[arr[i]] = 1
        else:
          mp[arr[i]]+=1
     
    for i in mp:
        # Simply checking the
        # given condition in question
        if (mp[i] == K):
            if ((i * 2) not in mp):
                ans += 1
    return ans
 
# Driver Code
arr = [ 10, 6, 12, 8, 10, 8 ]
N = len(arr)
K = 2
print(find(arr, N, K))
 
# This code is contributed by rohitsingh07052

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
  // Function to find the count valid numbers
  public static int find(int[] arr, int N, int K)
  {
 
    // Store ans
    int ans = 0;
    Dictionary mp =
      new Dictionary();
 
    for (int i = 0; i < N; i++) {
 
      // Storing frequency of elements
      if (!mp.ContainsKey(arr[i])) {
        mp.Add(arr[i], 1);
      }
      else {
        mp[arr[i]] = mp[arr[i]] + 1;
      }
    }
 
    foreach(KeyValuePair x in mp)
    {
 
      // Simply checking the
      // given condition in question
      if (x.Value == K) {
        if (mp.ContainsKey(x.Key * 2) == false) {
          ans++;
        }
      }
    }
    return ans;
  }
 
  // Driver Code
  public static void Main()
  {
    int[] arr = new int[] { 10, 6, 12, 8, 10, 8 };
    int N = arr.Length;
    int K = 2;
    Console.Write(find(arr, N, K));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

     // JavaScript code for the above approach
 
     // Function to find the count valid numbers
     function find(arr, N, K)
     {
      
         // Store ans
         let ans = 0;
         let mp = new Map();
         for (let i = 0; i < N; i++) {
 
             // Storing frequency of elements
             if (mp.has(arr[i])) {
                 mp.set(arr[i], mp.get(arr[i]) + 1)
             }
             else {
                 mp.set(arr[i], 1)
             }
         }
 
         for (let [key, val] of mp) {
 
             // Simply checking the
             // given condition in question
             if (val == K) {
                 if (mp.has(key * 2)
                     == false) {
                     ans++;
                 }
             }
         }
         return ans;
     }
 
     // Driver Code
     let arr = [10, 6, 12, 8, 10, 8];
     let N = arr.length;
     let K = 2;
 
     document.write(find(arr, N, K));
      
    // This code is contributed by Potta Lokesh
 

输出

2

时间复杂度： O(N)
辅助空间： 在）