小范围未排序数组中的第k个最小/最大

在未排序的数组中找到第k个最小或最大的元素，其中k <=数组的大小。假设数组元素在小范围内。

例子：

Input : arr[] = {3, 2, 9, 5, 7, 11, 13}
        k = 5
Output: 9

Input : arr[] = {16, 8, 9, 21, 43}
        k = 3
Output: 16

Input : arr[] = {50, 50, 40}
        k = 2
Output: 50

由于给定的数组元素在较小范围内，因此我们可以指示索引表执行类似于计数排序的操作。我们存储元素的计数，然后遍历count数组并打印第k个元素。

以下是上述算法的实现

C++

// C++ program of kth smallest/largest in 
// a small range unsorted array
#include 
using namespace std;
#define maxs 1000001
  
int kthSmallestLargest(int* arr, int n, int k)
{
    int max_val = *max_element(arr, arr+n);
    int hash[max_val+1] = { 0 };
  
    // Storing counts of elements
    for (int i = 0; i < n; i++) 
        hash[arr[i]]++;    
      
    // Traverse hash array build above until
    // we reach k-th smallest element.
    int count = 0;
    for (int i=0; i <= max_val; i++)
    {
        while (hash[i] > 0)
        {
           count++;
           if (count == k)
              return i;
           hash[i]--;
        }
    }
      
    return -1;
}
  
int main()
{
    int arr[] = { 11, 6, 2, 9, 4, 3, 16 };
    int n = sizeof(arr) / sizeof(arr[0]), k = 3;
    cout << "kth smallest number is: "
         << kthSmallestLargest(arr, n, k) << endl;
    return 0;
}

Java

// Java program of kth smallest/largest in 
// a small range unsorted array 
import java.util.Arrays;
  
class GFG 
{
  
    static int maxs = 1000001;
  
    static int kthSmallestLargest(int[] arr, int n, int k) 
    {
        int max_val = Arrays.stream(arr).max().getAsInt();
        int hash[] = new int[max_val + 1];
  
        // Storing counts of elements
        for (int i = 0; i < n; i++) 
        {
            hash[arr[i]]++;
        }
  
        // Traverse hash array build above until
        // we reach k-th smallest element.
        int count = 0;
        for (int i = 0; i <= max_val; i++)
        {
            while (hash[i] > 0) 
            {
                count++;
                if (count == k) 
                {
                    return i;
                }
                hash[i]--;
            }
        }
  
        return -1;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {11, 6, 2, 9, 4, 3, 16};
        int n = arr.length, k = 3;
        System.out.println("kth smallest number is: "
                + kthSmallestLargest(arr, n, k));
    }
}
  
// This code has been contributed by 29AjayKumar

Python3

# Python 3 program of kth smallest/largest 
# in a small range unsorted array
  
def kthSmallestLargest(arr, n, k):
    max_val = arr[0]
    for i in range(len(arr)):
        if (arr[i] > max_val):
            max_val = arr[i]
    hash = [0 for i in range(max_val + 1)] 
  
    # Storing counts of elements
    for i in range(n):
        hash[arr[i]] += 1
      
    # Traverse hash array build above until
    # we reach k-th smallest element.
    count = 0
    for i in range(max_val + 1):
        while (hash[i] > 0):
            count += 1
            if (count == k):
                return i
            hash[i] -= 1
          
    return -1
  
# Driver Code
if __name__ == '__main__':
    arr = [11, 6, 2, 9, 4, 3, 16]
    n = len(arr)
    k = 3
    print("kth smallest number is:",
           kthSmallestLargest(arr, n, k))
      
# This code is contributed by
# Surendra_Gangwar

C#

// C# program of kth smallest/largest in 
// a small range unsorted array 
using System;
using System.Linq;
  
class GFG 
{ 
  
    static int maxs = 1000001; 
  
    static int kthSmallestLargest(int[] arr, int n, int k) 
    { 
        int max_val = arr.Max(); 
        int []hash = new int[max_val + 1]; 
  
        // Storing counts of elements 
        for (int i = 0; i < n; i++) 
        { 
            hash[arr[i]]++; 
        } 
  
        // Traverse hash array build above until 
        // we reach k-th smallest element. 
        int count = 0; 
        for (int i = 0; i <= max_val; i++) 
        { 
            while (hash[i] > 0) 
            { 
                count++; 
                if (count == k) 
                { 
                    return i; 
                } 
                hash[i]--; 
            } 
        } 
        return -1; 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int []arr = {11, 6, 2, 9, 4, 3, 16}; 
        int n = arr.Length, k = 3; 
        Console.WriteLine("kth smallest number is: "
                + kthSmallestLargest(arr, n, k)); 
    } 
} 
  
/* This code contributed by PrinciRaj1992 */

PHP

 0)
        {
           $count++;
           if ($count == $k)
              return $i;
           $hash[$i]--;
        }
    }
       
    return -1;
}
   
  
    $arr = array ( 11, 6, 2, 9, 4, 3, 16 );
    $n = sizeof($arr) / sizeof($arr[0]);
    $k = 3;
    echo "kth smallest number is: "
         . kthSmallestLargest($arr, $n, $k)."\n";
    return 0;
?>

输出：

kth smallest number is: 4

时间复杂度： O(n + max_val)