What is the time complexity of Lexicographically Smallest String After Reverse?

The optimal solution for Lexicographically Smallest String After Reverse runs in O(n) time and uses O(n) space. We only iterate through the string once for each k, leading to O(n) time complexity.

What is the brute force solution for Lexicographically Smallest String After Reverse?

The brute force approach for Lexicographically Smallest String After Reverse is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible strings by reversing the first k or last k characters for each k from 1 to n. Then, we simply find the lexicographically smallest string from these results. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Lexicographically Smallest String After Reverse?

Lexicographically Smallest String After Reverse uses the following concepts: Two Pointers, Binary Search, Enumeration. The recommended patterns to study are: String Manipulation, Two Pointers.

What are the interview tips for Lexicographically Smallest String After Reverse?

Think about edge cases like single-character strings. Practice string manipulation techniques. Understand the properties of lexicographical order.

What are common mistakes when solving Lexicographically Smallest String After Reverse?

Not considering all values of k. Forgetting to compare the results correctly.

#3722

Lexicographically Smallest String After Reverse

Medium

Two PointersBinary SearchEnumerationString ManipulationTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Instead of generating all combinations, we can directly compare the results of reversing the first k and last k characters, keeping track of the smallest string as we iterate.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to store the smallest string as the original string.
2Step 2: Loop through k from 1 to n.
3Step 3: For each k, compute the two potential strings and update the smallest string accordingly.

solution.py7 lines

1def smallestString(s):
2    smallest = s
3    for k in range(1, len(s) + 1):
4        rev_first = s[:k][::-1] + s[k:]
5        rev_last = s[:-k] + s[-k:][::-1]
6        smallest = min(smallest, rev_first, rev_last)
7    return smallest

ℹ

Complexity note: We only iterate through the string once for each k, leading to O(n) time complexity.

1Reversing affects the order of characters significantly.
2Lexicographical order can be determined by string comparisons.

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