检查是否可以通过反转任何子数组一次使两个数组相等

给定两个大小为N 的数组A[]和B[] ，任务是通过只反转A 的任何子数组一次来检查是否可以使A[]等于B[] 。

例子：

Input: A[] = {1, 3, 2, 4} 
B[] = {1, 2, 3, 4} 
Output: Yes 
Explanation: 
The sub-array {3, 2} can be reversed to {2, 3}, which makes A equal to B.
 

Input: A[] = {1, 4, 2, 3} 
B[] = {1, 2, 3, 4} 
Output: No 
Explanation: 
There is no sub-array of A which, when reversed, makes A equal to B.

朴素的方法：检查A[] 的所有子数组，并在反转子数组后比较两个数组。
时间复杂度： O(N ² )。

有效的方法：

• 首先，找到A和B 中不相等的子数组的开始和结束索引。
• 然后，通过反转所需的子数组，我们可以检查是否可以使A等于B。
• 起始索引是A[i] != B[i]数组中的第一个索引，结束索引是A[i] != B[i]数组中的最后一个索引。

下面是上述方法的实现。

C++

// C++ implementation to
// check whether two arrays
// can be made equal by
// reversing a sub-array
// only once
#include 
using namespace std;
 
// Function to check whether two arrays
// can be made equal by reversing
// a sub-array only once
void checkArray(int A[], int B[], int N)
{
    // Integer variable for
    // storing the required
    // starting and ending
    // indices in the array
    int start = 0;
    int end = N - 1;
 
    // Finding the smallest index
    // for which A[i] != B[i]
    // i.e the starting index
    // of the unequal sub-array
    for (int i = 0; i < N; i++) {
        if (A[i] != B[i]) {
            start = i;
            break;
        }
    }
    // Finding the largest index
    // for which A[i] != B[i]
    // i.e the ending index
    // of the unequal sub-array
    for (int i = N - 1; i >= 0; i--) {
        if (A[i] != B[i]) {
            end = i;
            break;
        }
    }
 
    // Reversing the sub-array
    // A[start], A[start+1] .. A[end]
    reverse(A + start, A + end + 1);
 
    // Checking whether on reversing
    // the sub-array A[start]...A[end]
    // makes the arrays equal
 
    for (int i = 0; i < N; i++) {
        if (A[i] != B[i]) {
            // If any element of the
            // two arrays is unequal
            // print No and return
            cout << "No" << endl;
            return;
        }
    }
    // Print Yes if arrays are
    // equal after reversing
    // the sub-array
    cout << "Yes" << endl;
}
// Driver code
int main()
{
    int A[] = { 1, 3, 2, 4 };
    int B[] = { 1, 2, 3, 4 };
    int N = sizeof(A) / sizeof(A[0]);
    checkArray(A, B, N);
 
    return 0;
}

Java

// Java implementation to
// check whether two arrays
// can be made equal by
// reversing a sub-array
// only once
import java.util.*;
class GFG{
  
// Function to check whether two arrays
// can be made equal by reversing
// a sub-array only once
static void checkArray(int A[], int B[], int N)
{
    // Integer variable for
    // storing the required
    // starting and ending
    // indices in the array
    int start = 0;
    int end = N - 1;
  
    // Finding the smallest index
    // for which A[i] != B[i]
    // i.e the starting index
    // of the unequal sub-array
    for (int i = 0; i < N; i++)
    {
        if (A[i] != B[i])
        {
            start = i;
            break;
        }
    }
    // Finding the largest index
    // for which A[i] != B[i]
    // i.e the ending index
    // of the unequal sub-array
    for (int i = N - 1; i >= 0; i--)
    {
        if (A[i] != B[i])
        {
            end = i;
            break;
        }
    }
  
    // Reversing the sub-array
    // A[start], A[start+1] .. A[end]
    Collections.reverse(Arrays.asList(A));
  
    // Checking whether on reversing
    // the sub-array A[start]...A[end]
    // makes the arrays equal
    for (int i = 0; i < N; i++)
    {
        if (A[i] != B[i])
        {
            // If any element of the
            // two arrays is unequal
            // print No and return
            System.out.println("Yes");
            return;
        }
    }
    // Print Yes if arrays are
    // equal after reversing
    // the sub-array
   System.out.println("Yes");
}
// Driver code
public static void main(String[] args)
{
    int A[] = { 1, 3, 2, 4 };
    int B[] = { 1, 2, 3, 4 };
    int N = A.length;
    checkArray(A, B, N);
}
}
 
// This Code is contributed by rock_cool

Python3

# Python3 implementation to
# check whether two arrays
# can be made equal by
# reversing a sub-array
# only once
 
# Function to check whether two arrays
# can be made equal by reversing
# a sub-array only once
def checkArray(A, B, N):
     
    # Integer variable for
    # storing the required
    # starting and ending
    # indices in the array
    start = 0
    end = N - 1
 
    # Finding the smallest index
    # for which A[i] != B[i]
    # i.e the starting index
    # of the unequal sub-array
    for i in range(N):
        if (A[i] != B[i]):
            start = i
            break
             
    # Finding the largest index
    # for which A[i] != B[i]
    # i.e the ending index
    # of the unequal sub-array
    for i in range(N - 1, -1, -1):
        if (A[i] != B[i]):
            end = i
            break
 
    # Reversing the sub-array
    # A[start], A[start+1] .. A[end]
    A[start:end + 1] = reversed(A[start:end + 1])
     
    # Checking whether on reversing
    # the sub-array A[start]...A[end]
    # makes the arrays equal
    for i in range(N):
        if (A[i] != B[i]):
             
            # If any element of the
            # two arrays is unequal
            # print No and return
            print("No")
            return
             
    # Print Yes if arrays are
    # equal after reversing
    # the sub-array
    print("Yes")
     
# Driver code
if __name__ == '__main__':
     
    A = [ 1, 3, 2, 4 ]
    B = [ 1, 2, 3, 4 ]
    N = len(A)
     
    checkArray(A, B, N)
     
# This code is contributed by mohit kumar 29

C#

// C# implementation to
// check whether two arrays
// can be made equal by
// reversing a sub-array
// only once
using System;
 
class GFG{
 
// Function to check whether two arrays
// can be made equal by reversing
// a sub-array only once
static void checkArray(int []A, int []B, int N)
{
     
    // Integer variable for
    // storing the required
    // starting and ending
    // indices in the array
    int start = 0;
    int end = N - 1;
 
    // Finding the smallest index
    // for which A[i] != B[i]
    // i.e the starting index
    // of the unequal sub-array
    for(int i = 0; i < N; i++)
    {
        if (A[i] != B[i])
        {
            start = i;
            break;
        }
    }
     
    // Finding the largest index
    // for which A[i] != B[i]
    // i.e the ending index
    // of the unequal sub-array
    for(int i = N - 1; i >= 0; i--)
    {
        if (A[i] != B[i])
        {
            end = i;
            break;
        }
    }
 
    // Reversing the sub-array
    // A[start], A[start+1] .. A[end]
    Array.Reverse(A, start, end);
 
 
    // Checking whether on reversing
    // the sub-array A[start]...A[end]
    // makes the arrays equal
    for(int i = 0; i < N; i++)
    {
        if (A[i] != B[i])
        {
             
            // If any element of the
            // two arrays is unequal
            // print No and return
            Console.Write("Yes");
            return;
        }
    }
     
    // Print Yes if arrays are
    // equal after reversing
    // the sub-array
    Console.Write("Yes");
}
 
// Driver code
public static void Main(string[] args)
{
    int []A = { 1, 3, 2, 4 };
    int []B = { 1, 2, 3, 4 };
    int N = A.Length;
    checkArray(A, B, N);
}
}
 
// This code is contributed by rutvik_56

Javascript

 
 
// Javascript implementation to
// check whether two arrays
// can be made equal by
// reversing a sub-array
// only once
 
// Function to check whether two arrays
// can be made equal by reversing
// a sub-array only once
function checkArray(A, B, N)
{
    // Integer variable for
    // storing the required
    // starting and ending
    // indices in the array
    var start = 0;
    var end = N - 1;
 
    // Finding the smallest index
    // for which A[i] != B[i]
    // i.e the starting index
    // of the unequal sub-array
    for (var i = 0; i < N; i++) {
        if (A[i] != B[i]) {
            start = i;
            break;
        }
    }
    // Finding the largest index
    // for which A[i] != B[i]
    // i.e the ending index
    // of the unequal sub-array
    for (var i = N - 1; i >= 0; i--) {
        if (A[i] != B[i]) {
            end = i;
            break;
        }
    }
 
    // Reversing the sub-array
    // A[start], A[start+1] .. A[end]
    var l = start;
    var r = end;
    var d = parseInt((r-l+2)/2);
      for(var i=0;i<d;i++)
      {
         var t = A[l+i];
         A[l+i] = A[r-i];
         A[r-i] = t;
      }
 
    // Checking whether on reversing
    // the sub-array A[start]...A[end]
    // makes the arrays equal
 
    for (var i = 0; i < N; i++) {
        if (A[i] != B[i]) {
            // If any element of the
            // two arrays is unequal
            // print No and return
            document.write( "No" );
            return;
        }
    }
    // Print Yes if arrays are
    // equal after reversing
    // the sub-array
    document.write( "Yes" );
}
// Driver code
var A = [1, 3, 2, 4];
var B = [1, 2, 3, 4];
var N = A.length;
checkArray(A, B, N);
 
 

输出：

Yes

时间复杂度： O(N)

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