What is the brute force solution for Lexicographically Minimum String After Removing Stars?

The brute force approach for Lexicographically Minimum String After Removing Stars is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves iterating through the string and removing '*' characters one by one. For each '*', we find the leftmost non-'*' character, remove it, and continue until all '*' are processed. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Lexicographically Minimum String After Removing Stars?

Lexicographically Minimum String After Removing Stars uses the following concepts: Hash Table, String, Stack, Greedy, Heap (Priority Queue). The recommended patterns to study are: Stack, Greedy Algorithms.

What are the interview tips for Lexicographically Minimum String After Removing Stars?

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases, such as strings with no '*' or strings that are entirely '*'. Practice explaining your thought process as you code; this helps interviewers understand your reasoning.

What are common mistakes when solving Lexicographically Minimum String After Removing Stars?

Not handling the case where multiple '*' characters are present correctly, leading to incorrect removals. Failing to understand that popping from a stack represents removing the most recent character, which is key to achieving the lexicographically smallest result.

#3170

Lexicographically Minimum String After Removing Stars

Medium

Hash TableStringStackGreedyHeap (Priority Queue)StackGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach uses a stack to efficiently manage the characters. We push non-'*' characters onto the stack, and when we encounter a '*', we pop the top character from the stack, effectively removing the smallest non-'*' character.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty stack.
2Step 2: Iterate through each character in the string.
3Step 3: If the character is '*', pop the top character from the stack (if not empty).
4Step 4: If the character is not '*', push it onto the stack.
5Step 5: After processing all characters, join the stack to form the result string.

solution.py9 lines

1def removeStars(s):
2    stack = []
3    for char in s:
4        if char == '*':
5            if stack:
6                stack.pop()
7        else:
8            stack.append(char)
9    return ''.join(stack)

ℹ

Complexity note: The time complexity is O(n) since we traverse the string once and each character is pushed or popped from the stack at most once. The space complexity is O(n) due to the stack storing characters.

1Using a stack allows us to efficiently manage the removal of characters, ensuring we always remove the most recent character when a '*' is encountered.
2The order of operations is crucial; processing characters in a single pass ensures we maintain the correct lexicographical order.

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