Check if two arrays can be made equal by reversing subarrays multiple times

Given two arrays A[] and B[], the task is to check if array B can be made equal to A by reversing the subarrays of B by any number of times.

Examples:

Input:  A[] =  {1 2 3}, B[] =  {3 1 2}
Output: Yes
Explanation:
Reverse subarrays in array B as shown below:
Reverse subarray [3, 1], B becomes [1, 3, 2]
Reverse subarray [3, 2], B becomes [1, 2, 3] = A
There are multiple ways to convert B to A.
Input:  A[] = {1 2 3}, B[] = {3 4 1 }
Output: No

Approach: Since we have to make array B equal to array A only by reversing any sub array any number of times, Therefore it is possible only when both the arrays are anagrams. To check if both arrays are anagrams, we will sort both the arrays and check if at any index elements are not same then we will return false, otherwise we will return true at the end.

Below implementation of the above approach:

C++

// C++ implementation to check if
// two arrays can be made equal

#include <bits/stdc++.h>
using namespace std;

// Function to check if array B
// can be made equal to array A
bool canMadeEqual(int A[],
                  int B[], int n)
{
    // sort both the arrays
    sort(A, A + n);
    sort(B, B + n);

    // Check if both the arrays
    // are equal or not
    for (int i = 0; i < n; i++)
        if (A[i] != B[i])
            return false;
    return true;
}

// Driver Code
int main()
{
    int A[] = { 1, 2, 3 };
    int B[] = { 1, 3, 2 };
    int n = sizeof(A) / sizeof(A[0]);
    if (canMadeEqual(A, B, n))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

C

// C implementation to check if
// two arrays can be made equal
#include<stdio.h>
#include<math.h>

int sort(int a[],int n)
{
    int i, j, tmp;
    for(i = 0; i < n; i++)
    {
        for(j = i + 1; j < n; j++)
        {
            if(a[j] <a[i])
            {
                tmp = a[i];
                a[i] = a[j];
                a[j] = tmp;
            }
        }
    }
    return 0;
}

// Function to check if array B
// can be made equal to array A
int canMadeEqual(int A[], int B[], int n)
{
    int i;

    // Sort both arrays
    sort(A, n);
    sort(B, n);

    // Check both arrays equal or not
    for(i = 0; i < n; i++)
    {
        if (A[i] != B[i])
        {
            return(0);
        }
    }
    return (1);
}

// Driver Code
int main()
{
    int A[] = { 1, 2, 3 };
    int n;
    int B[] = { 1, 3, 2 };

    n = sizeof(A) / sizeof(A[0]);
    if(canMadeEqual(A, B, n))
    {
        printf("Yes");
    }
    else
    {
        printf("No");
    }
    return 0;
}

// This code is contributed by adityakumar27200

Java

// Java implementation to check if
// two arrays can be made equal
import java.util.*;

class GFG{

// Function to check if array B
// can be made equal to array A
public static boolean canMadeEqual(int[] A,
                                   int[] B,
                                   int n)
{

    // Sort both the arrays
    Arrays.sort(A);
    Arrays.sort(B);

    // Check if both the arrays
    // are equal or not
    for(int i = 0; i < n; i++)
    {
        if (A[i] != B[i])
        {
            return false;
        }
    }
    return true;
}

// Driver code
public static void main(String[] args)
{
    int A[] = { 1, 2, 3 };
    int B[] = { 1, 3, 2 };
    int n = A.length;

    if (canMadeEqual(A, B, n))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 implementation to check if
# two arrays can be made equal

# Function to check if array B
# can be made equal to array A
def canMadeEqual(A, B, n):

  # Sort both the arrays
  A.sort()
  B.sort()

  # Check if both the arrays
  # are equal or not
  for i in range(n):
    if (A[i] != B[i]):
      return False

  return True

# Driver Code
if __name__ == "__main__":

  A = [ 1, 2, 3 ]
  B = [ 1, 3, 2 ]
  n = len(A)

  if (canMadeEqual(A, B, n)):
    print( "Yes")
  else:
    print("No")

# This code is contributed by chitranayal

C#

// C# implementation to check if
// two arrays can be made equal
using System;

class GFG{

// Function to check if array B
// can be made equal to array A
static bool canMadeEqual(int[] A,
                         int[] B,
                         int n)
{

    // Sort both the arrays
    Array.Sort(A);
    Array.Sort(B);

    // Check if both the arrays
    // are equal or not
    for(int i = 0; i < n; i++)
    {
        if (A[i] != B[i])
        {
            return false;
        }
    }
    return true;
}

// Driver code
public static void Main()
{
    int[] A = { 1, 2, 3 };
    int[] B = { 1, 3, 2 };
    int n = A.Length;

    if (canMadeEqual(A, B, n))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}

// This code is contributed by adityakumar27200

Javascript

<script>


// Javascript implementation to check if
// two arrays can be made equal

// Function to check if array B
// can be made equal to array A
function canMadeEqual(A, B, n)
{
    // sort both the arrays
    A.sort();
    B.sort();

    // Check if both the arrays
    // are equal or not
    for (var i = 0; i < n; i++)
        if (A[i] != B[i])
            return false;
    return true;
}

// Driver Code
var A = [ 1, 2, 3 ];
var B = [ 1, 3, 2 ];
var n = A.length;
if (canMadeEqual(A, B, n))
    document.write( "Yes");
else
    document.write("No");

// This code is contributed by rutvik_56.
</script>

Output

Yes

Time complexity: O(N*Log N)
Auxiliary Space: O(1)