检查我们是否可以通过交换 A[i]和 B[i]
来对两个数组进行排序
原文:https://www . geeksforgeeks . org/check-we-能否通过交换对两个数组进行排序-ai-and-bi/
给定两个数组，我们必须检查是否可以通过交换 A[i]和 B[i]以严格的升序对两个数组进行排序。
示例:
输入: A[ ]={ 1，4，3，5，7}，B[ ]={ 2，2，5，8，9} 输出:真 交换 A[1]和 B[1]后，两个数组都排序。
输入: A[ ]={ 1，4，5，5，7}，B[ ]={ 2，2，5，8，9} 输出:假 不可能让两个数组都用任意数量的互换排序。
给我们两个数组，我们可以用 B[i]交换 A[i]，这样我们就可以严格按照升序对两个数组进行排序，所以我们必须以 A[i] < A[i+1] and B[i] < B[i+1]. 的方式对数组进行排序，我们将使用贪婪的方法来解决这个问题。 我们将得到 A[i]和 B[i]的最小值和最大值，并将最小值分配给 B[i]，将最大值分配给 A[i]。 现在，我们将检查数组 A 和数组 B 是否严格递增。 让我们考虑我们的方法是不正确的，(有可能安排但我们的方法给出假的)，这意味着任何一个或多个位置被切换。
这意味着 a[i-1]不小于 a[i]或 a[i+1]不大于 a[i]。现在如果 a[i]不大于 a[i-1]，我们就不能用 b[i]来交换 a[i]，因为 b[i]总是小于 a[i]。现在让我们假设 a[i+1]不大于 a[i]，这样我们就可以将 a[i]与 b[i]交换为 a[i] > b[i]，但作为 a[i] > b[i]和 a[i+1]> b[i+1]和 a[i]>a[i+1]，所以 a[i]永远不能小于 b[i+1]，所以没有可能的交换。我们同样可以证明 b[i]。
因此，证明了当输出为是时，排列数组可能有更多可能的组合，但是当输出为否时，没有可能根据约束排列数组
下面是上述方法的实现:
C++
// C++ implementation of above approach
#include <iostream>
using namespace std;
// Function to check whether both the array can be
// sorted in (strictly increasing ) ascending order
bool IsSorted(int A[], int B[], int n)
{
// Traverse through the array
// and find out the min and max
// variable at each position
// make one array of min variables
// and another of maximum variable
for (int i = 0; i < n; i++) {
int x, y;
// Maximum and minimum variable
x = max(A[i], B[i]);
y = min(A[i], B[i]);
// Assign min value to
// B[i] and max value to A[i]
A[i] = x;
B[i] = y;
}
// Now check whether the array is
// sorted or not
for (int i = 1; i < n; i++) {
if (A[i] <= A[i - 1] || B[i] <= B[i - 1])
return false;
}
return true;
}
// Driver code
int main()
{
int A[] = { 1, 4, 3, 5, 7 };
int B[] = { 2, 2, 5, 8, 9 };
int n = sizeof(A) / sizeof(int);
cout << (IsSorted(A, B, n) ? "True" : "False");
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of above approach
import java.io.*;
class GFG
{
// Function to check whether both the array can be
// sorted in (strictly increasing ) ascending order
static boolean IsSorted(int []A, int []B, int n)
{
// Traverse through the array
// and find out the min and max
// variable at each position
// make one array of min variables
// and another of maximum variable
for (int i = 0; i < n; i++)
{
int x, y;
// Maximum and minimum variable
x = Math.max(A[i], B[i]);
y = Math.min(A[i], B[i]);
// Assign min value to
// B[i] and max value to A[i]
A[i] = x;
B[i] = y;
}
// Now check whether the array is
// sorted or not
for (int i = 1; i < n; i++)
{
if (A[i] <= A[i - 1] || B[i] <= B[i - 1])
return false;
}
return true;
}
// Driver code
public static void main (String[] args)
{
int []A = { 1, 4, 3, 5, 7 };
int []B = { 2, 2, 5, 8, 9 };
int n = A.length;
if(IsSorted(A, B, n) == true)
{
System.out.println("True");
}
else
{
System.out.println("False");
}
}
}
// This code is contributed by ajit
Python 3
# Python3 implementation of above approach
# Function to check whether both the array can be
# sorted in (strictly increasing ) ascending order
def IsSorted(A, B, n) :
# Traverse through the array
# and find out the min and max
# variable at each position
# make one array of min variables
# and another of maximum variable
for i in range(n) :
# Maximum and minimum variable
x = max(A[i], B[i]);
y = min(A[i], B[i]);
# Assign min value to
# B[i] and max value to A[i]
A[i] = x;
B[i] = y;
# Now check whether the array is
# sorted or not
for i in range(1, n) :
if (A[i] <= A[i - 1] or B[i] <= B[i - 1]) :
return False;
return True;
# Driver code
if __name__ == "__main__" :
A = [ 1, 4, 3, 5, 7 ];
B = [ 2, 2, 5, 8, 9 ];
n = len(A);
if (IsSorted(A, B, n)) :
print(True)
else :
print(False)
# This code is contributed by AnkitRai01
C
// C# implementation of above approach
using System;
class GFG
{
// Function to check whether both the array can be
// sorted in (strictly increasing ) ascending order
static bool IsSorted(int []A, int []B, int n)
{
// Traverse through the array
// and find out the min and max
// variable at each position
// make one array of min variables
// and another of maximum variable
for (int i = 0; i < n; i++) {
int x, y;
// Maximum and minimum variable
x = Math.Max(A[i], B[i]);
y = Math.Min(A[i], B[i]);
// Assign min value to
// B[i] and max value to A[i]
A[i] = x;
B[i] = y;
}
// Now check whether the array is
// sorted or not
for (int i = 1; i < n; i++) {
if (A[i] <= A[i - 1] || B[i] <= B[i - 1])
return false;
}
return true;
}
// Driver code
public static void Main()
{
int []A = { 1, 4, 3, 5, 7 };
int []B = { 2, 2, 5, 8, 9 };
int n = A.Length;
if(IsSorted(A, B, n) == true)
{
Console.Write("True");
}
else
{
Console.Write("False");
}
}
}
// This code is contributed
// by Akanksha Rai
java 描述语言
<script>
// Javascript implementation of the approach
// Function to check whether both the
// array can be sorted in (strictly
// increasing) ascending order
function IsSorted(A, B, n)
{
// Traverse through the array
// and find out the min and max
// variable at each position
// make one array of min variables
// and another of maximum variable
for(var i = 0; i < n; i++)
{
var x, y;
// Maximum and minimum variable
x = Math.max(A[i], B[i]);
y = Math.min(A[i], B[i]);
// Assign min value to
// B[i] and max value to A[i]
A[i] = x;
B[i] = y;
}
// Now check whether the array is
// sorted or not
for(var i = 1; i < n; i++)
{
if (A[i] <= A[i - 1] ||
B[i] <= B[i - 1])
return false;
}
return true;
}
// Driver Code
var A = [ 1, 4, 3, 5, 7 ];
var B = [ 2, 2, 5, 8, 9 ];
var n = A.length;
document.write(IsSorted(A, B, n) ?
"True" : "False");
// This code is contributed by SoumikMondal
</script>
Output:
True
时间复杂度: O(N)
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