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###### tags: `Leetcode`
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# Leetcode 1460. Make Two Arrays Equal by Reversing Subarrays

[link](https://leetcode.com/problems/make-two-arrays-equal-by-reversing-subarrays/)

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You are given two integer arrays of equal length target and arr. In one step, you can select any non-empty subarray 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 subarray [2,4,1], arr becomes [1,4,2,3]
2- Reverse subarray [4,2], arr becomes [1,2,4,3]
3- Reverse subarray [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: true
Explanation: arr is equal to target without any reverses.

#### Example 3:

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

#### Constraints:

- target.length == arr.length
- 1 <= target.length <= 1000
- 1 <= target[i] <= 1000
- 1 <= arr[i] <= 1000

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題意: 判斷arr數組能否通過逆序操作變成target數組

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We can sort each array and iterate each element in those two arrays. If we find the element that does not appear in another array, return false.

#### Solution 1

class Solution:
    def canBeEqual(self, target: List[int], arr: List[int]) -> bool:
        target.sort()
        arr.sort()
        for t, a in zip(target, arr):
            if(t != a):
                return False
        return True     

O(T): O(nlogn)
O(S): O(1)