从 A[]中选择 X 元素，从 B[]中选择满足给定条件的 Y 元素
原文:https://www . geesforgeks . org/choose-x-elements-from-a-和-y-elements-from-b-满足给定条件/
给定两个数组 A[] 和 B[] 以及两个整数 X 和 Y ，任务是从 A[] 和 Y 元素中选择 X 元素，使得从 B[] 中选择的所有元素少于从B[ T21 中选择的所有元素
输入: A[] = {1，1，1，1，1}，B[] = {3，1}，X = 3，Y = 1 输出:是 从 A[]中选择{1，1，1}，从 B[]中选择{3}。 输入: A[] = {5，4}，B[] = {2，3，2，2}，X = 2，Y = 1 输出:否
方法:为了满足给定的条件，最小的 X 元素必须从A【】中选择，最大的 Y 元素必须从B【】中选择。这可以通过对两个数组进行排序，然后从 A[] 中选择 X th 最小的元素，说 xSmall 和YthT21 从 B[] 中选择最大的元素来完成。 这是因为如果 xSmall 小于 yLarge ，那么所有小于它的元素肯定会小于 yLarge ，所有大于 yLarge 的元素肯定会大于 xSmall 。 以下是上述方法的实施:****
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to that returns true if
// it possible to choose the elements
bool isPossible(int A[], int B[], int n,
int m, int x, int y)
{
// If elements can't be chosen
if (x > n || y > m)
return false;
// Sort both the arrays
sort(A, A + n);
sort(B, B + m);
// If xth smallest element of A[]
// is smaller than the yth
// greatest element of B[]
if (A[x - 1] < B[m - y])
return true;
else
return false;
}
// Driver code
int main()
{
int A[] = { 1, 1, 1, 1, 1 };
int B[] = { 2, 2 };
int n = sizeof(A) / sizeof(int);
int m = sizeof(B) / sizeof(int);
int x = 3, y = 1;
if (isPossible(A, B, n, m, x, y))
cout << "Yes";
else
cout << "No";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
import java.util.*;
class GFG
{
// Function to that returns true if
// it possible to choose the elements
static boolean isPossible(int A[], int B[], int n,
int m, int x, int y)
{
// If elements can't be chosen
if (x > n || y > m)
return false;
// Sort both the arrays
Arrays.sort(A);
Arrays.sort(B);
// If xth smallest element of A[]
// is smaller than the yth
// greatest element of B[]
if (A[x - 1] < B[m - y])
return true;
else
return false;
}
// Driver code
public static void main (String[] args)
{
int A[] = { 1, 1, 1, 1, 1 };
int B[] = { 2, 2 };
int n = A.length;
int m = B.length;;
int x = 3, y = 1;
if (isPossible(A, B, n, m, x, y))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by AnkitRai01
Python 3
# Python3 implementation of the approach
# Function to that returns true if
# it possible to choose the elements
def isPossible(A, B, n, m, x, y) :
# If elements can't be chosen
if (x > n or y > m) :
return False
# Sort both the arrays
A.sort()
B.sort()
# If xth smallest element of A[]
# is smaller than the yth
# greatest element of B[]
if (A[x - 1] < B[m - y]) :
return True
else :
return False
# Driver code
A = [ 1, 1, 1, 1, 1 ]
B = [ 2, 2 ]
n = len(A)
m = len(B)
x = 3
y = 1
if (isPossible(A, B, n, m, x, y)):
print("Yes")
else:
print("No")
# This code is contributed by
# divyamohan123
C
// C# implementation of the above approach
using System;
class GFG
{
// Function to that returns true if
// it possible to choose the elements
static bool isPossible(int []A, int []B, int n,
int m, int x, int y)
{
// If elements can't be chosen
if (x > n || y > m)
return false;
// Sort both the arrays
Array.Sort(A);
Array.Sort(B);
// If xth smallest element of A[]
// is smaller than the yth
// greatest element of B[]
if (A[x - 1] < B[m - y])
return true;
else
return false;
}
// Driver code
public static void Main (String[] args)
{
int []A = { 1, 1, 1, 1, 1 };
int []B = { 2, 2 };
int n = A.Length;
int m = B.Length;;
int x = 3, y = 1;
if (isPossible(A, B, n, m, x, y))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by Rajput-Ji
java 描述语言
<script>
// javascript implementation of the above approach
// Function to that returns true if
// it possible to choose the elements
function isPossible(A , B , n, m , x , y)
{
// If elements can't be chosen
if (x > n || y > m)
return false;
// Sort both the arrays
A.sort();
B.sort();
// If xth smallest element of A
// is smaller than the yth
// greatest element of B
if (A[x - 1] < B[m - y])
return true;
else
return false;
}
// Driver code
var A = [ 1, 1, 1, 1, 1 ];
var B = [ 2, 2 ];
var n = A.length;
var m = B.length;;
var x = 3, y = 1;
if (isPossible(A, B, n, m, x, y))
document.write("Yes");
else
document.write("No");
// This code is contributed by 29AjayKumar
</script>
Output:
Yes
时间复杂度: O(N*log(N))
辅助空间: O(1)
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