How do you solve Process String with Special Operations II on LeetCode?

Process String with Special Operations II (LeetCode #3614) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Operations can drastically change the string length.

What is the brute force solution for Process String with Special Operations II?

The brute force approach for Process String with Special Operations II is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach simulates the operations directly on the string, building the result step by step. It is straightforward but inefficient for larger strings due to repeated duplications and reversals. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Process String with Special Operations II?

Process String with Special Operations II uses the following concepts: String, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Process String with Special Operations II?

Clarify the problem constraints. Discuss potential edge cases upfront. Explain your thought process clearly.

What are common mistakes when solving Process String with Special Operations II?

Forgetting to handle edge cases with k. Incorrectly applying operations in reverse.

#3614

Process String with Special Operations II

Hard

StringSimulationHash MapArray

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 building the final string, we track the effective length and operations to determine the k-th character directly. This avoids unnecessary memory usage and time-consuming operations.

⚙️

Algorithm

3 steps

1Step 1: Traverse the string in reverse, maintaining a list of effective lengths after each operation.
2Step 2: For each operation, adjust the index k based on the operation type (undo duplication or reversal).
3Step 3: After processing, return the character at the adjusted k-th index or '.' if out of bounds.

solution.py20 lines

1def process_string(s, k):
2    lengths = [0]
3    for char in s:
4        if char.islower():
5            lengths.append(lengths[-1] + 1)
6        elif char == '*':
7            lengths.append(lengths[-1] - 1)
8        elif char == '#':
9            lengths.append(lengths[-1] * 2)
10        elif char == '%':
11            lengths.append(lengths[-1])
12    if k >= lengths[-1]: return '.'
13    for i in range(len(s) - 1, -1, -1):
14        if lengths[i] == k:
15            return s[i]
16        if s[i] == '#':
17            k %= lengths[i]
18        elif s[i] == '%':
19            k = lengths[i] - 1 - k
20    return '.'

ℹ

Complexity note: The optimal solution processes the string in linear time, maintaining a list of lengths, which allows us to efficiently determine the k-th character without building the entire result.

1Operations can drastically change the string length.
2Tracking lengths allows us to avoid full string construction.

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