How do you solve Sort Integers by Binary Reflection on LeetCode?

Sort Integers by Binary Reflection (LeetCode #3769) 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: Binary reflection can be computed efficiently using string manipulation.

What is the time complexity of Sort Integers by Binary Reflection?

The optimal solution for Sort Integers by Binary Reflection runs in O(n log n) time and uses O(n) space. The sorting step dominates the complexity, making it O(n log n) overall, while storing tuples requires O(n) space.

What is the brute force solution for Sort Integers by Binary Reflection?

The brute force approach for Sort Integers by Binary Reflection is Brute Force, with O(n²) time complexity and O(1) space complexity. We can compute the binary reflection for each number, then sort the array based on these values. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Sort Integers by Binary Reflection?

Sort Integers by Binary Reflection uses the following concepts: Array, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Sort Integers by Binary Reflection?

Explain your thought process clearly. Discuss edge cases, like duplicates or single-element arrays. Practice coding on a whiteboard or online editor.

What are common mistakes when solving Sort Integers by Binary Reflection?

Forgetting to reverse the binary string. Not handling ties in binary reflection correctly.

#3769

Sort Integers by Binary Reflection

Easy

ArraySortingHash 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)

We can create a list of tuples containing the binary reflection and the original number, then sort this list. This reduces the number of operations significantly.

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Algorithm

3 steps

1Step 1: Create a list of tuples where each tuple contains the binary reflection and the original number.
2Step 2: Sort this list based on the binary reflection first, and the original number second.
3Step 3: Extract the sorted original numbers from the sorted list.

solution.py2 lines

1def sortByBinaryReflection(nums):
2    return [num for _, num in sorted((int(bin(num)[:1:-1], 2), num) for num in nums)]

ℹ

Complexity note: The sorting step dominates the complexity, making it O(n log n) overall, while storing tuples requires O(n) space.

1Binary reflection can be computed efficiently using string manipulation.
2Sorting based on multiple criteria is straightforward with tuples.

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