给定两个数组,我们必须通过交换 A[i] 和 B[i] 来检查是否可以严格按升序对两个数组进行排序。
例子:
Input : A[ ]={ 1, 4, 3, 5, 7}, B[ ]={ 2, 2, 5, 8, 9}
Output : True
After swapping A[1] and B[1], both the arrays are sorted.
Input : A[ ]={ 1, 4, 5, 5, 7}, B[ ]={ 2, 2, 5, 8, 9}
Output : False
It is not possible to make both the arrays sorted with any number of swaps.
给定两个数组,我们可以将 A[i] 与 B[i] 交换,以便我们可以严格按照升序对数组进行排序,因此我们必须以 A[i] < A[i] 的方式对数组进行排序+1] 和 B[i] < B[i+1]。
我们将使用贪婪的方法来解决问题。
我们将获得 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。
让我们考虑一下我们的做法是错误的(有可能安排但我们的做法是错误的),这意味着任何一个或多个位置都被转换了。
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],我们不能将 a[i] 与 b[i] 切换,因为 b[i] 总是小于 a[i]。现在让我们取 a[i+1] 不大于 a[i],所以我们可以将 a[i] 与 b[i] 切换为 a[i] > b[i],但是作为 a[i] > b [i] and a[i+1]> b[i+1] and a[i]>a[i+1] 所以 a[i] 永远不会小于 b[i+1] 所以不可能转变。我们可以类似地证明 b[i]。
因此证明,当输出为YES时,排列数组可能有更多可能的组合,但当输出为NO时,没有可能根据约束排列数组的方法。
下面是上述方法的实现:
C++
// C++ implementation of above approach
#include
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
// 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
Python3
# 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
Javascript
True
时间复杂度: O(N)
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