排列两个数组，使得每对数组的和大于或等于 K
原文:https://www . geesforgeks . org/permute-两个数组-每对求和-大于等于-k/
给定两个大小相等的数组 n 和一个整数 k 。任务是对两个数组进行置换，使得它们对应元素的和大于或等于 k，即 a[i] + b[i] > = k。如果存在这样的置换，任务是打印“是”，否则打印“否”。 例:
Input : a[] = {2, 1, 3},
b[] = { 7, 8, 9 },
k = 10\.
Output : Yes
Permutation a[] = { 1, 2, 3 } and b[] = { 9, 8, 7 }
satisfied the condition a[i] + b[i] >= K.
Input : a[] = {1, 2, 2, 1},
b[] = { 3, 3, 3, 4 },
k = 5\.
Output : No
其思想是按升序对一个数组进行排序，按降序对另一个数组进行排序，如果任何索引不满足条件 a[i] + b[i] >= K，则打印“否”，否则打印“是”。 如果排序数组的条件失败，则不存在满足不等式的数组置换。证明， 假设 a 排序 [] 按升序排序 a[]，而 b 排序 [] 按降序排序 b[]。 让新的置换 b[]通过交换 b 的任意两个索引 I，j 来创建排序 []，
- 情况 1: i < j，b【I】处的元素现在是 b 排序【j】。 现在，a 排序【I】+b排序【j】排序\【I】>b排序【j】作为 b【】是按降序排序的，我们知道 a 排序【I】+b排序【I】<K。
- 情况 2: i > j，b【I】处的元素现在是 b 排序【j】。 现在，a 排序【j】+b排序【I】排序\【I】>a排序【j】作为一个[]按照递增顺序排序，我们知道 a 排序【I】+b排序【I】<k。
下面是这个方法的实现:
C++
// C++ program to check whether permutation of two
// arrays satisfy the condition a[i] + b[i] >= k.
#include<bits/stdc++.h>
using namespace std;
// Check whether any permutation exists which
// satisfy the condition.
bool isPossible(int a[], int b[], int n, int k)
{
// Sort the array a[] in decreasing order.
sort(a, a + n);
// Sort the array b[] in increasing order.
sort(b, b + n, greater<int>());
// Checking condition on each index.
for (int i = 0; i < n; i++)
if (a[i] + b[i] < k)
return false;
return true;
}
// Driven Program
int main()
{
int a[] = { 2, 1, 3 };
int b[] = { 7, 8, 9 };
int k = 10;
int n = sizeof(a)/sizeof(a[0]);
isPossible(a, b, n, k) ? cout << "Yes" :
cout << "No";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to check whether
// permutation of two arrays satisfy
// the condition a[i] + b[i] >= k.
import java.util.*;
class GFG
{
// Check whether any permutation
// exists which satisfy the condition.
static boolean isPossible(Integer a[], int b[],
int n, int k)
{
// Sort the array a[] in decreasing order.
Arrays.sort(a, Collections.reverseOrder());
// Sort the array b[] in increasing order.
Arrays.sort(b);
// Checking condition on each index.
for (int i = 0; i < n; i++)
if (a[i] + b[i] < k)
return false;
return true;
}
// Driver code
public static void main(String[] args) {
Integer a[] = {2, 1, 3};
int b[] = {7, 8, 9};
int k = 10;
int n = a.length;
if (isPossible(a, b, n, k))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by Anant Agarwal.
Python 3
# Python program to check
# whether permutation of two
# arrays satisfy the condition
# a[i] + b[i] >= k.
# Check whether any
# permutation exists which
# satisfy the condition.
def isPossible(a,b,n,k):
# Sort the array a[]
# in decreasing order.
a.sort(reverse=True)
# Sort the array b[]
# in increasing order.
b.sort()
# Checking condition
# on each index.
for i in range(n):
if (a[i] + b[i] < k):
return False
return True
# Driver code
a = [ 2, 1, 3]
b = [7, 8, 9]
k = 10
n =len(a)
if(isPossible(a, b, n, k)):
print("Yes")
else:
print("No")
# This code is contributed
# by Anant Agarwal.
C
// C# program to check whether
// permutation of two arrays satisfy
// the condition a[i] + b[i] >= k.
using System;
class GFG
{
// Check whether any permutation
// exists which satisfy the condition.
static bool isPossible(int []a, int []b,
int n, int k)
{
// Sort the array a[]
// in decreasing order.
Array.Sort(a);
// Sort the array b[]
// in increasing order.
Array.Reverse(b);
// Checking condition on each index.
for (int i = 0; i < n; i++)
if (a[i] + b[i] < k)
return false;
return true;
}
// Driver code
public static void Main()
{
int []a = {2, 1, 3};
int []b = {7, 8, 9};
int k = 10;
int n = a.Length;
if (isPossible(a, b, n, k))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by anuj_67.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to check whether
// permutation of two arrays satisfy
// the condition a[i] + b[i] >= k.
// Check whether any permutation
// exists which satisfy the condition.
function isPossible( $a, $b, $n, $k)
{
// Sort the array a[] in
// decreasing order.
sort($a);
// Sort the array b[] in
// increasing order.
rsort($b);
// Checking condition on each
// index.
for ( $i = 0; $i < $n; $i++)
if ($a[$i] + $b[$i] < $k)
return false;
return true;
}
// Driven Program
$a = array( 2, 1, 3 );
$b = array( 7, 8, 9 );
$k = 10;
$n = count($a);
if(isPossible($a, $b, $n, $k))
echo "Yes" ;
else
echo "No";
// This code is contributed by
// anuj_67.
?>
java 描述语言
<script>
// JavaScript program to check whether
// permutation of two arrays satisfy
// the condition a[i] + b[i] >= k.
// Check whether any permutation
// exists which satisfy the condition.
function isPossible(a, b, n, k)
{
// Sort the array a[]
// in decreasing order.
a.sort(function(a, b){return a - b});
// Sort the array b[]
// in increasing order.
b.reverse();
// Checking condition on each index.
for (let i = 0; i < n; i++)
if (a[i] + b[i] < k)
return false;
return true;
}
let a = [2, 1, 3];
let b = [7, 8, 9];
let k = 10;
let n = a.length;
if (isPossible(a, b, n, k))
document.write("Yes");
else
document.write("No");
</script>
输出:
Yes
时间复杂度: O(n log n)。 本文由 Anuj Chauhan 供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用write.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 contribute@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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