# LeetCode 1460. Make Two Arrays Equal by Reversing Sub-arrays

## Description

https://leetcode.com/problems/make-two-arrays-equal-by-reversing-sub-arrays/

Given two integer arrays of equal length `target` and `arr`.

In one step, you can select any non-empty sub-array of `arr` and reverse it. You are allowed to make any number of steps.

Return True if you can make `arr` equal to `target`, or False otherwise.

Example 1:

```Input: target = [1,2,3,4], arr = [2,4,1,3]
Output: true
Explanation: You can follow the next steps to convert arr to target:
1- Reverse sub-array [2,4,1], arr becomes [1,4,2,3]
2- Reverse sub-array [4,2], arr becomes [1,2,4,3]
3- Reverse sub-array [4,3], arr becomes [1,2,3,4]
There are multiple ways to convert arr to target, this is not the only way to do so.
```

Example 2:

```Input: target = , arr = 
Output: true
Explanation: arr is equal to target without any reverses.
```

Example 3:

```Input: target = [1,12], arr = [12,1]
Output: true
```

Example 4:

```Input: target = [3,7,9], arr = [3,7,11]
Output: false
Explanation: arr doesn't have value 9 and it can never be converted to target.
```

Example 5:

```Input: target = [1,1,1,1,1], arr = [1,1,1,1,1]
Output: true
```

Constraints:

• `target.length == arr.length`
• `1 <= target.length <= 1000`
• `1 <= target[i] <= 1000`
• `1 <= arr[i] <= 1000`

## Explanation

Just check if the occurrence of two list are the same.

## Python Solution

``````class Solution:
def canBeEqual(self, target: List[int], arr: List[int]) -> bool:

counter1 = {}
counter2 = {}

for i in target:
counter1[i] = counter1.get(i, 0) + 1

for i in arr:
counter2[i] = counter2.get(i, 0) + 1

return counter1 == counter2``````
• Time Complexity: O(N).
• Space Complexity: O(N).