#3039

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Apply Operations to Make String Empty

Name: Interview AiBox
Availability: InStock
Rating: 4.9 (10000 reviews)

Learn how to systematically apply operations on a string using array scanning and hash lookups to reduce it efficiently.

array hash table sorting counting

Problem Statement

Given a string s, repeatedly perform operations that remove characters until s is empty. Each operation targets specific characters identified using their counts and positions, requiring careful tracking.

For example, with s initially set to "aabcbbca", you scan the string and remove selected characters step by step. The last characters remaining after all operations demonstrate the process outcome. Implement an approach that maintains correct order and efficiently updates counts.

Examples

Example 1

Input: s = "aabcbbca"

Output: "ba"

Explained in the statement.

Example 2

Input: s = "abcd"

Output: "abcd"

We do the following operation:

Remove the underlined characters s = "abcd". The resulting string is s = "". The string just before the last operation is "abcd".

Constraints

1 <= s.length <= 5 * 105
s consists only of lowercase English letters.

Solution Approach

Count character frequencies with a hash map

Create a hash table to store the frequency of each character in s. This allows quick identification of characters that are candidates for removal in each operation and helps track which characters will remain in the last steps.

Scan the array and remove characters

Iterate over the string array, applying the operation rules to remove characters while updating the hash map. Use in-place marking or a stack to maintain the remaining characters efficiently without repeatedly shifting the array.

Construct the resulting string

After processing all operations, collect the remaining characters according to the original order of first appearance. This step uses the updated frequency table to ensure only the characters that survive the removal operations are included.

Complexity Analysis

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

Time complexity depends on the number of operations and frequency updates, roughly O(n) if each character is scanned once. Space complexity is O(1) for character counts plus O(n) for the output string storage.

What Interviewers Usually Probe

Tracking character counts efficiently hints at using a hash table.
Maintaining the correct order while removing characters is key for correct output.
Optimizing repeated scans of the string can be done with array marking or stack-based approaches.

Common Pitfalls or Variants

Common pitfalls

Failing to update character counts after each operation can lead to incorrect results.
Rescanning the entire string unnecessarily increases time complexity.
Ignoring the order of remaining characters can produce a wrong final string.

Follow-up variants

Allowing removal operations based on custom character priority instead of frequency.
Restricting operations to contiguous substrings rather than individual characters.
Tracking multiple strings simultaneously and applying operations in parallel.

FAQ

What is the main pattern used in Apply Operations to Make String Empty?

The primary pattern is array scanning combined with hash table lookups to track character frequencies and guide removal operations.

Can I use sorting to solve this problem?

Sorting alone is insufficient because it may disrupt original character order; combine it with frequency tracking to preserve sequence.

What should I do if multiple characters have the same frequency?

Maintain their original relative positions during removal, as the last remaining characters depend on order as well as frequency.

What is the space complexity of this approach?

Space complexity is O(n) for storing the resulting string plus O(1) for the hash map tracking character counts.

How does GhostInterview handle Apply Operations to Make String Empty?

It guides you step by step, highlighting which characters to remove, how to update counts, and ensures the correct final string is produced.

terminal

Solution

Solution 1: Hash Table or Array

We use a hash table or array $cnt$ to record the occurrence times of each character in string $s$, and use another hash table or array $last$ to record the last occurrence position of each character in string $s$. The maximum occurrence times of characters in string $s$ is denoted as $mx$.

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class Solution:
    def lastNonEmptyString(self, s: str) -> str:
        cnt = Counter(s)
        mx = cnt.most_common(1)[0][1]
        last = {c: i for i, c in enumerate(s)}
        return "".join(c for i, c in enumerate(s) if cnt[c] == mx and last[c] == i)

Continue Practicing

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