#3561

Medium

auto_awesomeStack-based state management

LeetCode Problem Workspace

Resulting String After Adjacent Removals

This problem focuses on removing adjacent characters in a string using a stack-based approach until no more operations are possible.

string stack simulation

Problem Statement

You are given a string s consisting of lowercase English letters. Repeatedly perform the following operation: remove any two consecutive characters in the string that are the same.

Return the resulting string after no more such operations can be performed. The operation stops when no more adjacent matching characters exist.

Examples

Example 1

Input: s = "abc"

Output: "c"

Example 2

Input: s = "adcb"

Output: ""

Example 3

Input: s = "zadb"

Output: "db"

Constraints

1 <= s.length <= 105
s consists only of lowercase English letters.

Solution Approach

Use a Stack for Character Removal

Traverse the string from left to right, pushing each character onto a stack. If the current character matches the one at the top of the stack, pop the stack to simulate removal. If not, push the character onto the stack. At the end of the traversal, the stack contains the resulting string.

Optimize for Performance

Since the string length can be large (up to 10^5), this approach ensures that each character is processed once, resulting in a time complexity of O(n). The space complexity is O(n) as well, due to the stack storing the characters.

Edge Cases Consideration

Handle edge cases such as strings with no adjacent characters to remove, strings with alternating characters, and strings that completely reduce to an empty string. Ensure the solution works for strings of varying lengths within the input constraints.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this solution is O(n), where n is the length of the string. Each character is pushed to and popped from the stack at most once. The space complexity is O(n) as the stack may hold up to n characters in the worst case.

What Interviewers Usually Probe

Efficient handling of large strings with minimal overhead.
Understanding of stack-based algorithms.
Ability to identify and address edge cases in string manipulation.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle cases where no characters can be removed.
Misunderstanding the stack's behavior when removing characters.
Failing to optimize for performance with large input sizes.

Follow-up variants

Consider implementing this approach using a different data structure, such as a queue, to explore alternative solutions.
Implement a variant that handles strings with special characters or case sensitivity.
Modify the problem to allow multiple removals of different types of adjacent characters.

FAQ

What is the time complexity of this approach?

The time complexity is O(n), where n is the length of the string, as each character is processed once.

How does the stack help in solving the problem?

The stack allows for efficient management of the string state, by pushing characters and removing them when adjacent matching ones are found.

What happens if the string has no adjacent matching characters?

If no adjacent matching characters exist, the stack will simply contain the original string as is, with no changes made.

What are the edge cases for this problem?

Edge cases include strings that are already reduced, strings with alternating characters, and strings that result in an empty string after removals.

How can this solution be optimized for large inputs?

The solution is already optimized with O(n) time and space complexity, ensuring efficient processing even for strings with lengths up to 10^5.

terminal

Solution

Solution 1: Stack

We can use a stack to simulate the process of removing adjacent characters. Iterate through each character in the string. If the character at the top of the stack and the current character are consecutive (i.e., their ASCII values differ by 1 or 25), pop the top character from the stack; otherwise, push the current character onto the stack. Finally, the characters remaining in the stack are those that can no longer be removed. Join the characters in the stack into a string and return it.

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class Solution:
    def resultingString(self, s: str) -> str:
        stk = []
        for c in s:
            if stk and abs(ord(c) - ord(stk[-1])) in (1, 25):
                stk.pop()
            else:
                stk.append(c)
        return "".join(stk)

Continue Practicing

#3412 find mirror score of a string #3174 clear digits #2696 minimum string length after removing substrings