Leetcode is a popular platform for practicing coding problems, offering a wide range of challenges that cover everything from data structures and algorithms to database queries and shell scripting. Among these problems is “Make Two Arrays Equal by Reversing Subarrays,” a problem that requires you to engage with array manipulation and subarray reversing.

In this extensive guide, we’ll walk you through:

- Problem Statement
- Initial Thoughts and Approach
- Further Optimizations
- Python Implementations
- Test Cases and Edge Cases
- Complexity Analysis
- Conclusion

## 1. Problem Statement

Given two integer arrays of equal length `target`

and `arr`

, you need to determine if `target`

can be made equal to `arr`

by reversing any subarray from `arr`

.

Return `True`

if `target`

can be made equal to `arr`

by reversing any subarray from `arr`

, and `False`

otherwise.

#### Example 1:

Input: `target = [1, 2, 3, 4], arr = [2, 4, 1, 3]`

Output: `True`

Explanation: You can reverse the subarray `[4, 1, 3]`

to make `arr`

equal to `target`

.

#### Example 2:

Input: `target = [7], arr = [7]`

Output: `True`

Explanation: `arr`

is already equal to `target`

.

## 2. Initial Thoughts and Approach

An intuitive approach to solving this problem is to sort both the arrays and then check for their equality. This is based on the understanding that the order of elements in an array can always be changed by reversing a subarray.

## 3. Further Optimizations

Sorting both arrays and checking their equality is the most straightforward way to solve this problem. You can’t really improve the time complexity further, as sorting itself is an O(n log n) operation.

## 4. Python Implementations

### Initial Implementation

```
def canBeEqual(target, arr):
target.sort()
arr.sort()
return target == arr
```

### “Optimized” Implementation

Since sorting is the best approach, the “optimized” implementation is identical to the initial one.

## 5. Test Cases and Edge Cases

It is important to validate your solution with various test cases:

```
# Test 1
assert canBeEqual([1, 2, 3, 4], [2, 4, 1, 3]) == True
# Test 2
assert canBeEqual([7], [7]) == True
# Test 3 (Edge Case)
assert canBeEqual([1, 1, 1, 1], [1, 1, 1, 2]) == False
# Test 4 (Edge Case)
assert canBeEqual([3, 4, 2, 5], [5, 4, 3, 2]) == True
```

## 6. Complexity Analysis

### Time Complexity

Both the initial and “optimized” solutions have a time complexity of O(n log n), where n is the length of either `target`

or `arr`

. This is because we’re sorting both arrays.

### Space Complexity

The space complexity for the sort operation is O(n), where n is the length of the array. This is especially true for Python’s built-in TimSort, which requires additional space.

## 7. Conclusion

The “Make Two Arrays Equal by Reversing Subarrays” problem on LeetCode is a relatively straightforward problem that tests your understanding of array manipulations and sorting algorithms. While it might not present a wide scope for optimization, it serves as a good reminder that sometimes the most straightforward solution is the most effective one.