## 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:trueExplanation: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 = [7], arr = [7]Output:trueExplanation: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:falseExplanation: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).