检查是否可以分别从两个数组中选择 X 和 Y 元素，使得 X 中的最大值小于 Y 中的最小值
原文:https://www . geesforgeks . org/check-if-x-和-y-elements-可以分别从两个数组中进行选择，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 整数组成的数组arr 1【】和arr 2【】以及两个整数 X 和 Y ，任务是检查是否有可能从arr 1【】和 Y 元素中选择 X 元素如果可能，则打印“是”。否则，打印“否”。
示例:
输入: arr1[] = {1，1，1，1，1}，arr2[] = {2，2}，X = 3，Y = 1 输出:是 解释:每个可能的选择都满足上述条件，因为 arr1[]的每个元素都小于 arr2[]中的最小元素。
输入: arr1[] = {1，2，3}，arr2[] = {3，4，5}，X = 2，Y = 1 输出:是 解释:一种可能的选择是从 arr1[]中取索引 0 和 1 处的元素，从 arr2[]中取索引 0 处的元素，即{1，2}和{3}。
方法:思路是将两个数组按照升序排序，然后从arr 1【】中选择第一个 X 元素，从arr 2【】中选择最后一个 Y 元素。按照以下步骤解决问题:
- 按升序排列两个数组。
- 如果 X 大于 N 或 Y 大于 M ，则打印“否”，因为无法选择任何此类组合。
- 否则，如果arr 1[X–1]的值小于arr 2[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 <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<script>
// 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
</script>
Output:
Yes
时间复杂度:O(N * log N) T5辅助空间** O(1)
版权属于：月萌API www.moonapi.com，转载请注明出处
