检查是否可以分别从两个数组中选择X和Y元素，以使X中的最大值小于Y中的最小值

给定两个分别由N和M个整数以及两个整数X和Y组成的数组arr1 []和arr2 [] ，任务是检查是否有可能从arr1 []中选择X个元素，并从arr2 []中选择Y个元素，例如这些X元素中最大的元素小于这些Y元素中的最小元素。如果可能，请打印“是” 。否则，打印“否” 。

例子：

Input: arr1[] = {1, 1, 1, 1, 1}, arr2[] = {2, 2}, X = 3, Y = 1
Output: Yes
Explanation: Every possible selection satisfies the above condition as every element of arr1[] is less than minimum element in the arr2[].

Input: arr1[] = {1, 2, 3}, arr2[] = {3, 4, 5}, X = 2, Y = 1
Output: Yes
Explanation: One possible selection is take elements at indices 0 and 1 from arr1[] and indices 0 from arr2[], i.e {1, 2} and {3}.

方法：想法是按照升序对两个数组进行排序，然后从arr1 []中选择第一个X元素，然后从arr2 []中选择最后一个Y元素。请按照以下步骤解决问题：

• 以升序对两个数组进行排序。
• 如果X大于N或Y大于M ，则打印“否”，因为不可能选择任何这样的组合。
• 否则，如果arr1 [X – 1]的值小于arr2 [M – Y] ，则打印“是” 。
• 否则，打印“否” 。如果以上条件都不满足。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to check if it is possible to
// choose X and Y elements from a[] and
// b[] such that maximum element among
// X element is less than minimum
// element among Y elements
string check(int a[], int b[], int Na,
             int Nb, int k, int m)
{
    // Check if there are atleast X
    // elements in arr1[] and atleast
    // Y elements in arr2[]
    if (Na < k || Nb < m)
        return "No";
 
    // Sort arrays in ascending order
    sort(a, a + Na);
    sort(b, b + Nb);
 
    // Check if (X - 1)-th element in arr1[]
    // is less than from M-Yth element
    // in arr2[]
    if (a[k - 1] < b[Nb - m]) {
        return "Yes";
    }
 
    // Return false
    return "No";
}
 
// Driver Code
int main()
{
    int arr1[] = { 1, 2, 3 };
    int arr2[] = { 3, 4, 5 };
    int N = sizeof(arr1) / sizeof(arr1[0]);
    int M = sizeof(arr2) / sizeof(arr2[0]);
 
    int X = 2, Y = 1;
 
    // Function Call
    cout << check(arr1, arr2, N, M, X, Y);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to check if it is possible to
// choose X and Y elements from a[] and
// b[] such that maximum element among
// X element is less than minimum
// element among Y elements
static String check(int[] a, int[] b,
                    int Na, int Nb,
                    int k, int m)
{
     
    // Check if there are atleast X
    // elements in arr1[] and atleast
    // Y elements in arr2[]
    if (Na < k || Nb < m)
        return "No";
 
    // Sort arrays in ascending order
    Arrays.sort(a);
    Arrays.sort(b);
 
    // Check if (X - 1)-th element in arr1[]
    // is less than from M-Yth element
    // in arr2[]
    if (a[k - 1] < b[Nb - m])
    {
        return "Yes";
    }
 
    // Return false
    return "No";
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr1 = { 1, 2, 3 };
    int[] arr2 = { 3, 4, 5 };
    int N = arr1.length;
    int M = arr2.length;
 
    int X = 2, Y = 1;
 
    // Function Call
    System.out.println(check(
        arr1, arr2, N, M, X, Y));
}
}
 
// This code is contributed by Dharanendra L V

Python3

# Python3 program for the above approach
 
# Function to check if it is possible to
# choose X and Y elements from a[] and
# b[] such that maximum element among
# X element is less than minimum
# element among Y elements
def check( a, b, Na, Nb, k, m):
   
  # Check if there are atleast X
  # elements in arr1[] and atleast
  # Y elements in arr2[]
  if (Na < k or Nb < m):
    return "No"
 
  # Sort arrays in ascending order
  a.sort()
  a.sort()
 
  # Check if (X - 1)-th element in arr1[]
  # is less than from M-Yth element
  # in arr2[]
  if (a[k - 1] < b[Nb - m]):
    return "Yes"
 
  # Return false
  return "No"
 
# Driver Code
arr1 = [ 1, 2, 3 ]
arr2 = [ 3, 4, 5 ]
N = len(arr1)
M = len(arr2)
X = 2
Y = 1
 
# Function Call
print(check(arr1, arr2, N, M, X, Y))
 
# This code is contributed by rohitsongh07052.

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if it is possible to
// choose X and Y elements from a[] and
// b[] such that maximum element among
// X element is less than minimum
// element among Y elements
static string check(int[] a, int[] b,
                    int Na, int Nb,
                    int k, int m)
{
     
    // Check if there are atleast X
    // elements in arr1[] and atleast
    // Y elements in arr2[]
    if (Na < k || Nb < m)
        return "No";
 
    // Sort arrays in ascending order
    Array.Sort(a);
    Array.Sort(b);
 
    // Check if (X - 1)-th element in arr1[]
    // is less than from M-Yth element
    // in arr2[]
    if (a[k - 1] < b[Nb - m])
    {
        return "Yes";
    }
 
    // Return false
    return "No";
}
 
// Driver Code
static public void Main()
{
    int[] arr1 = { 1, 2, 3 };
    int[] arr2 = { 3, 4, 5 };
    int N = arr1.Length;
    int M = arr2.Length;
 
    int X = 2, Y = 1;
 
    // Function Call
    Console.WriteLine(check(
        arr1, arr2, N, M, X, Y));
}
}
 
// This code is contributed by Dharanendra L V

Javascript

// javascript program for the above approach   
// Function to check if it is possible to
    // choose X and Y elements from a and
    // b such that maximum element among
    // X element is less than minimum
    // element among Y elements
    function check(a,  b , Na , Nb , k , m)
    {
 
        // Check if there are atleast X
        // elements in arr1 and atleast
        // Y elements in arr2
        if (Na < k || Nb < m)
            return "No";
 
        // Sort arrays in ascending order
        a.sort();
        b.sort();
 
        // Check if (X - 1)-th element in arr1
        // is less than from M-Yth element
        // in arr2
        if (a[k - 1] < b[Nb - m]) {
            return "Yes";
        }
 
        // Return false
        return "No";
    }
 
    // Driver Code
        var arr1 = [ 1, 2, 3 ];
        var arr2 = [ 3, 4, 5 ];
        var N = arr1.length;
        var M = arr2.length;
 
        var X = 2, Y = 1;
 
        // Function Call
        document.write(check(arr1, arr2, N, M, X, Y));
 
// This code is contributed by todaysgaurav

输出：

Yes

时间复杂度： O(N * log N)
辅助空间O(1)