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

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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

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

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)

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

  • Create two hash tables, freqA and freqB, to store the frequency of elements in arrays A and B, respectively.
  • Iterate over arrays A and B from index 0 to index n-1, and increment the frequency of the corresponding element in the respective hash table.
  • Iterate over the elements in freqA.
    • For each element in freqA, check if its frequency in freqB is the same as its frequency in freqA.
      • If any frequency does not match, return false.
    • If all frequencies match, return true.

Below is the implementation of the above approach:

C++

#include <iostream>
#include <unordered_map>
using namespace std;
 
bool checkReverseSubarrays(int A[], int B[], int n) {
    // Create hash tables to store the frequency of elements in A and B
    unordered_map<int, int> freqA, freqB;
    for (int i = 0; i < n; i++) {
        freqA[A[i]]++;
        freqB[B[i]]++;
    }
 
    // Check if the frequencies of elements in A and B match
    for (auto it = freqA.begin(); it != freqA.end(); it++) {
        if (freqB[it->first] != it->second) {
            return false;
        }
    }
 
    // If frequencies match, return true
    return true;
}
 
int main() {
    int A[] = {1, 2, 3};
    int B[] = {1, 3, 2};
    int n = sizeof(A) / sizeof(A[0]);
    if (checkReverseSubarrays(A, B, n)) {
        cout << "Yes" << endl;
    } else {
        cout << "No" << endl;
    }
 
    
}
// This code is contributed by rudra1807raj

                    

Java

import java.util.HashMap;
import java.util.Map;
 
public class Main {
    // Function to check if two arrays are reversals of each other
    public static boolean checkReverseSubarrays(int[] A, int[] B, int n) {
        // Create hash maps to store the frequency of elements in A and B
        Map<Integer, Integer> freqA = new HashMap<>();
        Map<Integer, Integer> freqB = new HashMap<>();
 
        // Count the frequency of elements in arrays A and B
        for (int i = 0; i < n; i++) {
            freqA.put(A[i], freqA.getOrDefault(A[i], 0) + 1);
            freqB.put(B[i], freqB.getOrDefault(B[i], 0) + 1);
        }
 
        // Check if the frequencies of elements in A and B match
        for (Map.Entry<Integer, Integer> entry : freqA.entrySet()) {
            int element = entry.getKey();
            int frequency = entry.getValue();
 
            if (!freqB.containsKey(element) || freqB.get(element) != frequency) {
                return false;
            }
        }
 
        // If frequencies match, return true
        return true;
    }
 
    public static void main(String[] args) {
        int[] A = {1, 2, 3};
        int[] B = {1, 3, 2};
        int n = A.length;
 
        if (checkReverseSubarrays(A, B, n)) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}

                    

Python3

def check_reverse_subarrays(A, B):
    # Create dictionaries to store the frequency of elements in A and B
    freqA = {}
    freqB = {}
 
    # Count the frequency of elements in arrays A and B
    for num in A:
        freqA[num] = freqA.get(num, 0) + 1
    for num in B:
        freqB[num] = freqB.get(num, 0) + 1
 
    # Check if the frequencies of elements in A and B match
    for num, freq in freqA.items():
        if num not in freqB or freqB[num] != freq:
            return False
 
    # If frequencies match, return true
    return True
 
# Main function
if __name__ == "__main__":
    A = [1, 2, 3]
    B = [1, 3, 2]
 
    if check_reverse_subarrays(A, B):
        print("Yes")
    else:
        print("No")

                    

C#

using System;
using System.Collections.Generic;
 
namespace ReverseSubarraysCheck
{
    class Program
    {
        static bool CheckReverseSubarrays(int[] A, int[] B, int n)
        {
            // Create dictionaries to store the frequency of elements in A and B
            Dictionary<int, int> freqA = new Dictionary<int, int>();
            Dictionary<int, int> freqB = new Dictionary<int, int>();
 
            for (int i = 0; i < n; i++)
            {
                if (freqA.ContainsKey(A[i]))
                {
                    freqA[A[i]]++;
                }
                else
                {
                    freqA[A[i]] = 1;
                }
 
                if (freqB.ContainsKey(B[i]))
                {
                    freqB[B[i]]++;
                }
                else
                {
                    freqB[B[i]] = 1;
                }
            }
 
            // Check if the frequencies of elements in A and B match
            foreach (var kvp in freqA)
            {
                if (!freqB.TryGetValue(kvp.Key, out int freqBValue) || freqBValue != kvp.Value)
                {
                    return false;
                }
            }
 
            // If frequencies match, return true
            return true;
        }
 
        static void Main(string[] args)
        {
            int[] A = { 1, 2, 3 };
            int[] B = { 1, 3, 2 };
            int n = A.Length;
             
            if (CheckReverseSubarrays(A, B, n))
            {
                Console.WriteLine("Yes");
            }
            else
            {
                Console.WriteLine("No");
            }
        }
    }
}

                    

Javascript

function checkReverseSubarrays(A, B) {
    // Create objects to store the frequency of elements in A and B
    let freqA = {};
    let freqB = {};
 
    // Count the frequency of elements in arrays A and B
    for (let num of A) {
        freqA[num] = (freqA[num] || 0) + 1;
    }
    for (let num of B) {
        freqB[num] = (freqB[num] || 0) + 1;
    }
 
    // Check if the frequencies of elements in A and B match
    for (let num in freqA) {
        if (!freqB.hasOwnProperty(num) || freqB[num] !== freqA[num]) {
            return false;
        }
    }
 
    // If frequencies match, return true
    return true;
}
 
// Main function
const A = [1, 2, 3];
const B = [1, 3, 2];
 
if (checkReverseSubarrays(A, B)) {
    console.log("Yes");
} else {
    console.log("No");
}

                    

Output

Yes

Time Complexity: O(n), where n is the length of the arrays A and B, 
Auxiliary Space: O(n) as we are using two hash tables to store the frequency of elements in A and B.



Last Updated : 06 Oct, 2023
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