What is the time complexity of Lexicographically Smallest String After Adjacent Removals?

The optimal solution for Lexicographically Smallest String After Adjacent Removals runs in O(n) time and uses O(n) space. The stack efficiently manages removals, allowing us to process each character in linear time.

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

The brute force approach for Lexicographically Smallest String After Adjacent Removals is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible combination of removing adjacent characters and check the resulting strings. This method is exhaustive but inefficient. The optimal Optimal Solution approach improves this to O(n).

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

Lexicographically Smallest String After Adjacent Removals uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Stack, Greedy.

What are the interview tips for Lexicographically Smallest String After Adjacent Removals?

Explain your thought process clearly. Discuss edge cases upfront. Practice similar string manipulation problems.

What are common mistakes when solving Lexicographically Smallest String After Adjacent Removals?

Not considering circular adjacency (e.g., 'a' and 'z'). Failing to handle empty strings correctly.

#3563

Lexicographically Smallest String After Adjacent Removals

Hard

StringDynamic ProgrammingStackGreedy

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)

Using a stack-like approach allows us to efficiently manage removals and build the smallest string without generating all combinations.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty stack to build the result string.
2Step 2: Iterate through each character in the string. If the current character and the top of the stack are consecutive, pop the stack.
3Step 3: Push the current character onto the stack. Finally, join the stack to form the result.

solution.py8 lines

1def lexicographically_smallest(s):
2    stack = []
3    for char in s:
4        while stack and (abs(ord(stack[-1]) - ord(char)) == 1 or (stack[-1] == 'a' and char == 'z') or (stack[-1] == 'z' and char == 'a')):
5            stack.pop()
6        stack.append(char)
7    return ''.join(stack)
8

ℹ

Complexity note: The stack efficiently manages removals, allowing us to process each character in linear time.

1Adjacent characters can be removed if they are consecutive in the alphabet.
2Using a stack helps efficiently manage removals.

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