How do you solve Orderly Queue on LeetCode?

Orderly Queue (LeetCode #899) 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: If k is 1, we can only rotate the string, so we need to find the smallest rotation.

What is the brute force solution for Orderly Queue?

The brute force approach for Orderly Queue is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible strings by moving the first k characters to the end of the string in every possible way. This allows us to find the lexicographically smallest string by comparing all generated strings. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Orderly Queue?

Orderly Queue uses the following concepts: Math, String, Sorting. The recommended patterns to study are: String Manipulation, Sorting.

What are the interview tips for Orderly Queue?

Clarify the value of k and how it affects the operations allowed on the string. Think about edge cases, such as strings of length 1 or strings that are already sorted. Discuss your thought process and approach before jumping into coding.

What are common mistakes when solving Orderly Queue?

Not considering the case where k = 1 and only rotations are allowed. Overlooking the need to sort the string when k > 1.

#899

Orderly Queue

Hard

MathStringSortingString ManipulationSorting

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)

When k > 1, we can sort the entire string to get the lexicographically smallest arrangement. For k = 1, we can only rotate, so we need to find the smallest rotation.

⚙️

Algorithm

2 steps

1Step 1: If k == 1, find the lexicographically smallest rotation by checking all rotations.
2Step 2: If k > 1, sort the string and return the sorted string.

solution.py4 lines

1def orderlyQueue(s, k):
2    if k == 1:
3        return min(s[i:] + s[:i] for i in range(len(s)))
4    return ''.join(sorted(s))

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the string when k > 1. The space complexity is O(n) because we create a new sorted string.

1If k is 1, we can only rotate the string, so we need to find the smallest rotation.
2If k is greater than 1, we can sort the string to get the smallest arrangement.

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