The video discusses Leetcode problem 1460, which involves determining whether two arrays can be made equal by reversing subarrays, emphasizing the importance of checking if both arrays contain the same elements regardless of order, similar to the “valid anagram” problem. The host presents various solutions, including sorting the arrays and using hashmaps for counting element occurrences, ultimately demonstrating an optimized approach that checks counts in a single pass.

In the video, the host introduces the problem of making two arrays equal by reversing subarrays, as presented in Leetcode problem 1460. The task involves determining whether one array can be transformed into another of the same length by reversing any subarray. The host provides an example to illustrate the concept, explaining how one can manipulate the second array to achieve the target arrangement. This sets the stage for a deeper exploration of the problem’s solution.

The host emphasizes that while the problem may initially appear complex, the essence lies in checking whether both arrays contain the same elements, regardless of their order. This leads to the realization that the solution can be approached similarly to the “valid anagram” problem, highlighting that the key is to ensure that the counts of each element match between the two arrays. The goal is not to minimize the number of operations but simply to ascertain the possibility of transformation.

The discussion transitions into potential solutions, starting with a straightforward approach of sorting both arrays. The host notes that while this method is simple, it has a time complexity of O(n log n). A more efficient solution involves counting the occurrences of each element using a hashmap (or a dictionary in Python), allowing for a linear time complexity solution of O(n). The video outlines how to build these hashmaps and compare their contents to determine if the arrays can be made equal.

As the explanation progresses, the host provides a practical coding demonstration using Python’s `Counter`

class to create hashmaps for both arrays. This one-liner solution effectively checks for equality in element counts. However, the video also addresses the possibility of being asked to implement the counting mechanism manually during an interview, prompting a more detailed explanation of how to construct the hashmaps without relying on built-in functions.

Finally, the host optimizes the solution further by combining the counting process into a single hashmap that increments for the first array and decrements for the second. This method eliminates the need for a separate length check and simplifies the validation process to checking if all counts equal zero. The video concludes by running the final solution, demonstrating its effectiveness, and encourages viewers to explore more coding challenges with enthusiasm.