# Solving Leetcode Interviews in Seconds with AI: Make Two Arrays Equal by Reversing Subarrays


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1460" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the approach and Python code:

*   **Check for Permutation:** The core idea is that if `arr` can be transformed into `target` through reversals, it implies `arr` and `target` must contain the same elements (i.e., they are permutations of each other). We can check this by comparing the sorted arrays or using frequency counts.
*   **Reversals Preserve Element Counts:** Reversing a subarray doesn't change the overall count of each element in the array. Therefore, if the element counts in `arr` and `target` differ, no amount of reversals will make them equal.

*   **Runtime & Storage Complexity**: O(n log n) due to sorting. O(1) storage (or O(n) depending on sort implementation).

	
	# Code
	```python
	def canBeEqual(target, arr):
    """
    Checks if arr can be transformed into target by reversing subarrays.

    Args:
        target: The target array.
        arr: The array to transform.

    Returns:
        True if arr can be transformed into target, False otherwise.
    """

    if len(target) != len(arr):
        return False

    target.sort()
    arr.sort()

    return target == arr
	```