How do you solve Find Maximum Removals From Source String on LeetCode?

Find Maximum Removals From Source String (LeetCode #3316) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding subsequences is crucial for this problem.

What is the time complexity of Find Maximum Removals From Source String?

The optimal solution for Find Maximum Removals From Source String runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only iterate through the source string once for each removal attempt, making it efficient.

What is the brute force solution for Find Maximum Removals From Source String?

The brute force approach for Find Maximum Removals From Source String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying every possible combination of removals from the source string based on the target indices. For each combination, we check if the pattern remains a subsequence of the modified source. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Maximum Removals From Source String?

Find Maximum Removals From Source String uses the following concepts: Array, Hash Table, Two Pointers, String, Dynamic Programming. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Maximum Removals From Source String?

Always clarify the problem and constraints before diving into coding. Discuss your thought process and approach before writing code. Test your solution with edge cases to ensure robustness.

What are common mistakes when solving Find Maximum Removals From Source String?

Not checking all combinations in brute force. Failing to correctly implement the subsequence check.

#3316

Find Maximum Removals From Source String

Medium

ArrayHash TableTwo PointersStringDynamic ProgrammingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a two-pointer technique to efficiently check how many characters can be removed while ensuring the pattern remains a subsequence. We iterate through the target indices and check if removing characters still allows the pattern to be formed.

⚙️

Algorithm

4 steps

1Step 1: Initialize two pointers, one for the source string and one for the pattern.
2Step 2: Iterate through the targetIndices and try to remove characters from the source string.
3Step 3: For each character removal, check if the pattern can still be formed using the two-pointer technique.
4Step 4: Count the maximum number of valid removals.

solution.py21 lines

1# Full working Python code
2
3def max_removals_optimal(source, pattern, targetIndices):
4    left = 0
5    right = 0
6    max_removals = 0
7    while right < len(targetIndices):
8        # Try to remove the character at targetIndices[right]
9        removed = targetIndices[right]
10        left_ptr, pattern_ptr = 0, 0
11        while left_ptr < len(source):
12            if left_ptr == removed:
13                left_ptr += 1
14                continue
15            if pattern_ptr < len(pattern) and source[left_ptr] == pattern[pattern_ptr]:
16                pattern_ptr += 1
17            left_ptr += 1
18        if pattern_ptr == len(pattern):
19            max_removals += 1
20        right += 1
21    return max_removals

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the source string once for each removal attempt, making it efficient.

1Understanding subsequences is crucial for this problem.
2Using two pointers can significantly reduce time complexity.

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