How do you solve Maximum Number of Removable Characters on LeetCode?

Maximum Number of Removable Characters (LeetCode #1898) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log m) time complexity and O(m) space complexity. Key insight: Binary search allows us to efficiently find the maximum k.

What is the brute force solution for Maximum Number of Removable Characters?

The brute force approach for Maximum Number of Removable Characters is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible value of k from 0 to the length of the removable array. For each k, we will remove the characters from s and check if p is still a subsequence of the modified s. The optimal Optimal Solution approach improves this to O(n log m).

What algorithm or data structure is used to solve Maximum Number of Removable Characters?

Maximum Number of Removable Characters uses the following concepts: Array, Two Pointers, String, Binary Search. The recommended patterns to study are: Binary Search, Two Pointers.

What are the interview tips for Maximum Number of Removable Characters?

Explain your thought process clearly when using binary search. Make sure to handle edge cases in your code. Practice explaining the difference between subsequence and substring.

What are common mistakes when solving Maximum Number of Removable Characters?

Not considering edge cases where k = 0 or k = removable.length. Confusing the subsequence check with substring check.

#1898

Maximum Number of Removable Characters

Medium

ArrayTwo PointersStringBinary SearchBinary SearchTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log m)Space O(m)

We can use binary search to find the maximum k efficiently. For each mid value in our binary search, we will check if p is a subsequence of s after removing the first mid indices in removable.

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Algorithm

4 steps

1Step 1: Initialize left to 0 and right to the length of removable.
2Step 2: While left is less than or equal to right, calculate mid as (left + right) // 2.
3Step 3: Check if p is a subsequence of s after removing the first mid indices in removable.
4Step 4: If it is, update max_k to mid and move left to mid + 1. Otherwise, move right to mid - 1.

solution.py21 lines

1def is_subsequence(s, p, removable, k):
2    removed = set(removable[:k])
3    j = 0
4    for i in range(len(s)):
5        if i in removed:
6            continue
7        if j < len(p) and s[i] == p[j]:
8            j += 1
9    return j == len(p)
10
11def max_removable_characters(s, p, removable):
12    left, right = 0, len(removable)
13    max_k = 0
14    while left <= right:
15        mid = (left + right) // 2
16        if is_subsequence(s, p, removable, mid):
17            max_k = mid
18            left = mid + 1
19        else:
20            right = mid - 1
21    return max_k

ℹ

Complexity note: The time complexity is O(n log m) because we perform a binary search on the range of removable indices (m) and for each mid, we check if p is a subsequence of s, which takes O(n) time.

1Binary search allows us to efficiently find the maximum k.
2Checking for subsequence can be done in linear time.

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