How do you solve Make Two Arrays Equal by Reversing Subarrays on LeetCode?

Make Two Arrays Equal by Reversing Subarrays (LeetCode #1460) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Both arrays must contain the same elements to be made equal.

What is the time complexity of Make Two Arrays Equal by Reversing Subarrays?

The optimal solution for Make Two Arrays Equal by Reversing Subarrays runs in O(n log n) time and uses O(n) space. Sorting both arrays takes O(n log n) time, and the space complexity is O(n) due to the sorting operation.

What is the brute force solution for Make Two Arrays Equal by Reversing Subarrays?

The brute force approach for Make Two Arrays Equal by Reversing Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying all possible subarrays and reversing them to see if we can match the target array. This is straightforward but inefficient as it may require many operations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Make Two Arrays Equal by Reversing Subarrays?

Make Two Arrays Equal by Reversing Subarrays uses the following concepts: Array, Hash Table, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Make Two Arrays Equal by Reversing Subarrays?

Always consider simpler approaches first before optimizing. Clarify the problem requirements and constraints with the interviewer. Practice recognizing patterns in problems, such as using sorting for equality checks.

What are common mistakes when solving Make Two Arrays Equal by Reversing Subarrays?

Overcomplicating the problem by trying to simulate every possible reversal. Not recognizing that sorting can simplify the comparison.

#1460

Make Two Arrays Equal by Reversing Subarrays

Easy

ArrayHash TableSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal approach leverages sorting. If both arrays contain the same elements, regardless of order, they can be made equal by reversing subarrays. Sorting allows us to easily compare the contents of both arrays.

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Algorithm

3 steps

1Step 1: Sort both target and arr.
2Step 2: Compare the sorted arrays element by element.
3Step 3: If all elements match, return true; otherwise, return false.

solution.py4 lines

1# Full working Python code
2
3def canMakeEqual(target, arr):
4    return sorted(target) == sorted(arr)

ℹ

Complexity note: Sorting both arrays takes O(n log n) time, and the space complexity is O(n) due to the sorting operation.

1Both arrays must contain the same elements to be made equal.
2Sorting provides a quick way to check for equality of contents.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.