找到一个数字K，使得恰好K个数组元素大于或等于K

给定的大小N，其中仅包含非负元素的数组a []中，任务是找到任何整数K，对其中有恰好有k数组元素是大于或等于k。如果不存在这样的K ，则打印-1 。

例子：

Input: a[] = {7, 8, 9, 0, 0, 1}
Output: 3
Explanation:
Since 3 is less than or equal to 7, 8, and 9, therefore, 3 is the answer.

Input: a[] = {0, 0}
Output: -1

方法：任务是找到K使得数组元素大于或等于K。因此， K不能超过数组a [n]中存在的最大元素。请按照以下步骤解决问题：

1. 遍历数组以找到最大的数组元素，并将其存储在变量中，例如m 。
2. 初始化计数器变量cnt，以计算大于或等于K的数组元素的数量。
3. 迭代从0到m的K可能值。为每个值迭代数组，并计算大于或等于该值的数组元素的数量。
4. 如果对于任何值，发现恰好有K个数组元素大于或等于该值，则打印该值。
5. 如果在数组完全遍历之后没有获得此类值，请打印-1 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find K for which
// there are exactly K array
// elements greater than or equal to K
int zvalue(vector& nums)
{
 
    // Finding the largest array element
    int m = *max_element(nums.begin(),
                         nums.end());
    int cnt = 0;
 
    // Possible values of K
    for (int i = 0; i <= m; i++) {
        cnt = 0;
 
        // Traverse the array
        for (int j = 0; j < nums.size(); j++) {
 
            // If current array element is
            // greater than or equal to i
            if (nums[j] >= i)
                cnt++;
        }
 
        // If i array elements are
        // greater than or equal to i
        if (cnt == i)
            return i;
    }
 
    // Otherwise
    return -1;
}
 
// Driver Code
int main()
{
    vector nums = { 7, 8, 9, 0, 0, 1 };
    cout << zvalue(nums) << endl;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to find K for which
// there are exactly K array
// elements greater than or equal to K
public static int zvalue(int[] nums)
{
     
    // Finding the largest array element
    int m = max_element(nums);
    int cnt = 0;
 
    // Possible values of K
    for(int i = 0; i <= m; i++)
    {
        cnt = 0;
 
        // Traverse the array
        for(int j = 0; j < nums.length; j++)
        {
             
            // If current array element is
            // greater than or equal to i
            if (nums[j] >= i)
                cnt++;
        }
 
        // If i array elements are
        // greater than or equal to i
        if (cnt == i)
            return i;
    }
 
    // Otherwise
    return -1;
}
 
// To find maximum Element
public static int max_element(int[] nums)
{
    int max = nums[0];
    for(int i = 1; i < nums.length; i++)
        max = Math.max(max, nums[i]);
         
    return max;
}
 
// Driver Code
public static void main(String args[])
{
    int[] nums = { 7, 8, 9, 0, 0, 1 };
     
    System.out.println(zvalue(nums));
}
}
 
// This code is contributed by hemanth gadarla

Python3

# Python3 program for the above approach
 
# Function to find K for which
# there are exactly K array
# elements greater than or equal to K
def zvalue(nums):
     
    # Finding the largest array element
    m = max(nums)
    cnt = 0
 
    # Possible values of K
    for i in range(0, m + 1, 1):
        cnt = 0
 
        # Traverse the array
        for j in range(0, len(nums), 1):
             
            # If current array element is
            # greater than or equal to i
            if (nums[j] >= i):
                cnt += 1
 
        # If i array elements are
        # greater than or equal to i
        if (cnt == i):
            return i
 
    # Otherwise
    return -1
 
# Driver Code
if __name__ == '__main__':
     
    nums =  [ 7, 8, 9, 0, 0, 1 ]
     
    print(zvalue(nums))
     
# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the
// above approach
using System;
class GFG{
 
// Function to find K for which
// there are exactly K array
// elements greater than or equal to K
public static int zvalue(int[] nums)
{
  // Finding the largest array element
  int m = max_element(nums);
  int cnt = 0;
 
  // Possible values of K
  for(int i = 0; i <= m; i++)
  {
    cnt = 0;
 
    // Traverse the array
    for(int j = 0;
            j < nums.Length; j++)
    {
      // If current array element is
      // greater than or equal to i
      if (nums[j] >= i)
        cnt++;
    }
 
    // If i array elements are
    // greater than or equal to i
    if (cnt == i)
      return i;
  }
 
  // Otherwise
  return -1;
}
 
// To find maximum Element
public static int max_element(int[] nums)
{
  int max = nums[0];
   
  for(int i = 1; i < nums.Length; i++)
    max = Math.Max(max, nums[i]);
   
  return max;
}
 
// Driver Code
public static void Main(String []args)
{
  int[] nums = {7, 8, 9, 0, 0, 1};
  Console.WriteLine(zvalue(nums));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// javascript program for the above approach
 
// Function to find K for which
// there are exactly K array
// elements greater than or equal to K
function zvalue(nums)
{
     
    // Finding the largest array element
    var m = max_element(nums);
    var cnt = 0;
 
    // Possible values of K
    for(i = 0; i <= m; i++)
    {
        cnt = 0;
 
        // Traverse the array
        for(j = 0; j < nums.length; j++)
        {
             
            // If current array element is
            // greater than or equal to i
            if (nums[j] >= i)
                cnt++;
        }
 
        // If i array elements are
        // greater than or equal to i
        if (cnt == i)
            return i;
    }
 
    // Otherwise
    return -1;
}
 
// To find maximum Element
function max_element(nums)
{
    var max = nums[0];
    for(i = 1; i < nums.length; i++)
        max = Math.max(max, nums[i]);
         
    return max;
}
 
// Driver Code
nums = [ 7, 8, 9, 0, 0, 1 ];
     
document.write(zvalue(nums));
 
// This code contributed by shikhasingrajput
 

输出：

3

时间复杂度： O(N ² )
辅助空间： O(1)