计算等级不超过K的数组元素

给定一个由N个整数和一个整数K组成的数组arr [] ，任务是找到秩最多为K的数组元素的数量。

Equal array elements will have equal ranks. Any array element numerically smaller than its immediate greater array element will be ranked one greater than the total number of array elements greater than it.

例子：

Input: N = 4, arr[] = {100, 50, 50, 25}, K = 3
Output: 3
Explanation: Rank of the players will be {1, 2, 2, 4}.
There are 3 array elements whose ranks are less than or equal to 3.
Therefore, the answer is 3.

Input: N = 5, arr[] = {2, 2, 3, 4, 5}, K = 4
Output: 5
Explanation: Rank of the players will be {4, 4, 3, 2, 1}.
There are 5 array elements whose ranks are les than or equal to 4.
Therefore, the answer is 5.

天真的方法：最简单的方法是在数组中找到当前的最大元素，并将其与先前的最大元素进行比较。如果它们相等，则前一个元素和当前元素的等级必须相等。否则，将当前最大元素的等级分配为比先前获得的最大数量大一的等级。重复此过程，直到给定数组为空或秩大于K ，以较早者为准。遍历后，打印已删除元素的总数。

时间复杂度： O(N ² )
辅助空间： O(N)

高效方法：想法是使用排序算法。在以给定数组以非递增顺序排序之后，并且对于每个元素，如果当前元素和前一个元素不相等，则当前元素的等级必须比前一个元素大1。否则，它与上一个相同。请按照以下步骤解决问题：

• 以给定的arr []升序排序。
• 用1初始化变量rank和position ，以分别存储当前数组元素的rank和位置。
• 将排序后的数组从i =(N – 1)遍历到0，然后执行以下操作：
  • 当相邻数组元素不相等或i等于(N – 1)时，使用位置更新等级。
  • 返回位置–如果等级大于K则返回1 。
  • 每次遍历的增量位置。
• 如果所有数组元素的秩均不超过K ，则打印N。

下面是上述方法的实现：

C++14

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find count of array
// elements with rank less than or
// equal to k
int rankLessThanK(int* arr, int k, int n)
{
     
    // Initialize rank and position
    int rank = 1;
    int position = 1;
 
    // Sort the given array
    sort(arr, arr + n);
 
    // Traverse array from right to left
    for(int i = n - 1; i >= 0; i--)
    {
         
        // Update rank with position, if
        // adjacent elements are unequal
        if (i == n - 1 || arr[i] != arr[i + 1])
        {
            rank = position;
 
            // Return position - 1, if
            // rank greater than k
            if (rank > k)
                return position - 1;
        }
 
        // Increase position
        position++;
    }
    return n;
}
 
// Driver Code
int main()
{
     
    // Given array
    int arr[5] = { 2, 2, 3, 4, 5 };
 
    int N = 5;
 
    // Given K
    int K = 4;
 
    // Function Call
    cout << rankLessThanK(arr, K, N);
     
    return 0;
}
 
// This code is contributed by hemanth gadarla

Java

// Java program for the above approach
 
import java.util.*;
import java.lang.*;
class GFG {
 
    // Function to find count of array
    // elements with rank less than or
    // equal to k
    static int rankLessThanK(int[] arr,
                             int k, int n)
    {
        // Initialize rank and position
        int rank = 1;
        int position = 1;
 
        // Sort the given array
        Arrays.sort(arr);
 
        // Traverse array from right to left
        for (int i = n - 1; i >= 0; i--) {
 
            // Update rank with position, if
            // adjacent elements are unequal
            if (i == n - 1
                || arr[i] != arr[i + 1]) {
 
                rank = position;
 
                // Return position - 1, if
                // rank greater than k
                if (rank > k)
                    return position - 1;
            }
 
            // Increase position
            position++;
        }
 
        return n;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given array
        int arr[] = { 2, 2, 3, 4, 5 };
 
        int N = arr.length;
 
        // Given K
        int K = 4;
 
        // Function Call
        System.out.println(
            rankLessThanK(arr, K, N));
    }
}

Python3

# Python3 program for the
# above approach
 
# Function to find count of
# array elements with rank
# less than or equal to k
def rankLessThanK(arr, k, n):
   
    # Initialize rank and
    # position
    rank = 1;
    position = 1;
 
    # Sort the given array
    arr = sorted(arr)
 
    # Traverse array from
    # right to left
    for i in range(n - 1,
                   -1, -1):
 
        # Update rank with position,
        # if adjacent elements are
        # unequal
        if (i == n - 1 or
            arr[i] != arr[i + 1]):
            rank = position;
 
            # Return position - 1, if
            # rank greater than k
            if (rank > k):
                return position - 1;
 
        # Increase position
        position += 1;
 
    return n;
 
# Driver Code
if __name__ == '__main__':
   
    # Given array
    arr = [2, 2, 3, 4, 5];
 
    N = len(arr);
 
    # Given K
    K = 4;
 
    # Function Call
    print(rankLessThanK(arr, K, N));
 
# This code is contributed by gauravrajput1

C#

// C# program for the above approach
using System;
  
class GFG{
      
// Function to find count of array
// elements with rank less than or
// equal to k
static int rankLessThanK(int[] arr,
                         int k, int n)
{
     
    // Initialize rank and position
    int rank = 1;
    int position = 1;
 
    // Sort the given array
    Array.Sort(arr);
     
    // Traverse array from right to left
    for(int i = n - 1; i >= 0; i--)
    {
         
        // Update rank with position, if
        // adjacent elements are unequal
        if (i == n - 1 || arr[i] != arr[i + 1])
        {
            rank = position;
 
            // Return position - 1, if
            // rank greater than k
            if (rank > k)
                return position - 1;
        }
 
        // Increase position
        position++;
    }
    return n;
}      
  
// Driver Code
public static void Main()
{
     
    // Given array
    int[] arr = { 2, 2, 3, 4, 5 };
  
    int N = arr.Length;
  
    // Given K
    int K = 4;
  
    // Function Call
    Console.WriteLine(rankLessThanK(
        arr, K, N));
}
}
 
// This code is contributed by sanjoy_62

输出

5

时间复杂度： O(N * log N)
辅助空间： O(N)