找到一个数字K，以使Array包含至少K个大于或等于K的数字

给定大小为N的非负整数的数组arr [] ，任务是找到一个整数H ，以使数组中至少K个整数大于或等于K。
例子：

Input: arr[] = [3, 0, 6, 1, 5] 
Output: 3 
Explanation: There are 3 number greater than or equal to 3 in the array i.e. 3, 6 and 5.

Input: arr[] = [9, 10, 7, 5, 0, 10, 2, 0] 
Output: 5

方法：使用散列
整数K不能大于arr []的大小。因此，请维护频率数组(哈希表)中每个元素的频率。然后从末尾遍历频率数组，并返回与条件匹配的第一个索引。

下面是上述方法的实现。

C++

// C++ program for the above approach
  
#include 
using namespace std;
  
int getNumberK(vector& S)
{
    // vector to store freq.
    vector freq(S.size() + 1,
                     0);
  
    // Filling freq vector.
    for (int i = 0; i < S.size();
         i++) {
  
        if (S[i] < S.size())
            freq[S[i]]++;
        else
            freq[S.size()]++;
    }
  
    int total = 0;
  
    // Finding K number.
    for (int i = S.size(); i >= 0;
         i--) {
  
        total += freq[i];
        if (total >= i)
            return i;
    }
  
    // No K number found.
    return 0;
}
  
// Driver code
int main()
{
    vector arr{ 3, 0, 6, 1, 5 };
    cout << getNumberK(arr) << '\n';
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
      
static int getNumberK(ArrayList S)
{
      
    // To store freq.
    int[] freq = new int[S.size() + 1];
  
    // Filling freq vector.
    for(int i = 0; i < S.size(); i++)
    {
        if (S.get(i) < S.size())
            freq[S.get(i)]++;
        else
            freq[S.size()]++;
    }
      
    int total = 0;
  
    // Finding K number.
    for(int i = S.size(); i >= 0; i--)
    {
        total += freq[i];
        if (total >= i)
            return i;
    }
  
    // No K number found.
    return 0;
}
  
// Driver code
public static void main(String[] args)
{
    ArrayList arr = 
          new ArrayList<>(Arrays.asList(3, 0, 6, 1, 5));
                                                 
    System.out.println(getNumberK(arr));
}
}
  
// This code is contributed by offbeat

Python3

# Python3 program for the above approach
def getNumberK(S): 
      
    # Vector to store freq. 
    freq = [0] * (len(S) + 1)
  
    # Filling freq vector. 
    for i in range(len(S)): 
        if (S[i] < len(S)): 
            freq[S[i]] += 1
        else:
            freq[len(S)] += 1
      
    total = 0
  
    # Finding K number. 
    for i in range(len(S), -1, -1): 
        total += freq[i] 
          
        if (total >= i):
            return i 
  
    # No K number found. 
    return 0
  
# Driver code 
arr = [ 3, 0, 6, 1, 5 ] 
  
print(getNumberK(arr))
  
# This code is contributed by code_hunt

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
  
class GFG{
      
static int getNumberK(List S)
{
      
    // To store freq.
    int[] freq = new int[S.Count + 1];
  
    // Filling freq vector.
    for(int i = 0; i < S.Count; i++)
    {
        if (S[i] < S.Count)
            freq[S[i]]++;
        else
            freq[S.Count]++;
    }
      
    int total = 0;
  
    // Finding K number.
    for(int i = S.Count; i >= 0; i--)
    {
        total += freq[i];
        if (total >= i)
            return i;
    }
  
    // No K number found.
    return 0;
}
  
// Driver code
public static void Main(String[] args)
{
    List arr = new List{ 3, 0, 6, 1, 5 };
                                                  
    Console.WriteLine(getNumberK(arr));
}
}
  
// This code is contributed by Amit Katiyar

输出：

3

时间复杂度O(N)
空间复杂度O(N)