What is the time complexity of Maximum Number of Removable Characters?

The optimal solution has a time complexity of O(log(R) * (S + P)).

What data structures are used to solve Maximum Number of Removable Characters?

This problem uses Array, Two Pointers, String, Binary Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Number of Removable Characters?

Insight 1: The problem can be efficiently solved using Binary Search. We can binary search for the maximum possible `k` value. Insight 2: For a given `k`, we can check if `p` is a subsequence of `s` after removing the first `k` characters from `s` specified by `removable`. This involves simulating the removal and then checking the subsequence condition.

Which companies ask LeetCode 1898. Maximum Number of Removable Characters in interviews?

This problem is frequently asked at Moveworks, Snowflake.

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Maximum Number of Removable Characters - LeetCode 1898 Solution

Q: How to solve Maximum Number of Removable Characters?

The solution uses a binary search approach to find the maximum value of `k`. For each `k` value being tested in the binary search, we simulate removing the characters at the indices specified by the first `k` elements of the `removable` array. Then, we check if `p` is still a subsequence of the modified string `s`. The `is_subsequence` function determines this. If `p` is a subsequence, we try a larger `k` by updating the left boundary of the binary search. Otherwise, we try a smaller `k` by updating the right boundary.

Maximum Number of Removable Characters - Complete Solution Guide

Maximum Number of Removable Characters is LeetCode problem 1898, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given two strings s and p where p is a subsequence of s . You are also given a distinct 0-indexed integer array removable containing a subset of indices of s ( s is also 0-indexed ). You want to choose an integer k ( 0 <= k <= removable.length ) such that, after removing k characters from s using the first k indices in removable , p is still a subsequence of s . More formally, you will mark the character at s[removable[i]] for each 0 <= i < k , then remove all marked characters and check

Detailed Explanation

The problem requires us to find the maximum number of characters we can remove from a given string `s` such that another string `p` remains a subsequence of `s`. We are given a list of indices `removable` which specifies the order in which characters can be removed from `s`. We need to find the largest `k` (from 0 to the length of `removable`) such that removing the first `k` characters from `s` (as specified by the `removable` array) still leaves `p` as a subsequence of the modified `s`. A subsequence is formed by deleting some (or none) characters from the original string without changing the order of the remaining characters.

Solution Approach

The solution uses a binary search approach to find the maximum value of `k`. For each `k` value being tested in the binary search, we simulate removing the characters at the indices specified by the first `k` elements of the `removable` array. Then, we check if `p` is still a subsequence of the modified string `s`. The `is_subsequence` function determines this. If `p` is a subsequence, we try a larger `k` by updating the left boundary of the binary search. Otherwise, we try a smaller `k` by updating the right boundary.

Step-by-Step Algorithm

Step 1: Initialize `left` to 0 and `right` to the length of `removable`. Initialize `result` to store the maximum `k` found so far.
Step 2: While `left` is less than or equal to `right`, calculate `mid` as the average of `left` and `right`.
Step 3: Call the `is_subsequence` function with `mid` as the number of characters to remove.
Step 4: If `is_subsequence(mid)` returns `true` (meaning `p` is still a subsequence after removing `mid` characters), update `result` to `mid` and set `left` to `mid + 1` to search for a larger value of `k`.
Step 5: If `is_subsequence(mid)` returns `false`, set `right` to `mid - 1` to search for a smaller value of `k`.
Step 6: After the binary search completes, return the `result`.
Step 7: The `is_subsequence` function creates a set of removed indices. It then uses two pointers, one for `s` and one for `p`, to iterate through the strings. If a character in `s` at the current index is not removed and matches the current character in `p`, the `p` pointer is incremented. The `s` pointer is incremented regardless. If the `p` pointer reaches the end of `p`, it means `p` is a subsequence of `s`.

Key Insights

Insight 1: The problem can be efficiently solved using Binary Search. We can binary search for the maximum possible `k` value.
Insight 2: For a given `k`, we can check if `p` is a subsequence of `s` after removing the first `k` characters from `s` specified by `removable`. This involves simulating the removal and then checking the subsequence condition.
Insight 3: Using a `set` or `boolean array` to keep track of removed indices is efficient for checking if a character at a given index should be considered during subsequence checking.

Complexity Analysis

Time Complexity: O(log(R) * (S + P))

Space Complexity: O(R)

Topics

This problem involves: Array, Two Pointers, String, Binary Search.

Companies

Asked at: Moveworks, Snowflake.

Maximum Number of Removable Characters | LeetCode 1898

Medium

Array Two Pointers String Binary Search

Time: O(log(R) * (S + P))Space: O(R)

Solution Code

Available in:

class Solution {
    public int maximumRemovals(String s, String p, int[] removable) {
        
        
        
        
        
        
        int result = 0;
        int left = 0;
        int right = removable.length;

        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (isSubsequence(s, p, removable, mid)) {
                result = mid;
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }

        return result;
    }

    private boolean isSubsequence(String s, String p, int[] removable, int k) {
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        java.util.Set<Integer> removedIndices = new java.util.HashSet<>();
        for (int i = 0; i < k; i++) {
            removedIndices.add(removable[i]);
        }

        int pPtr = 0;
        int sPtr = 0;

        while (sPtr < s.length() && pPtr < p.length()) {
            if (!removedIndices.contains(sPtr) && s.charAt(sPtr) == p.charAt(pPtr)) {
                pPtr++;
            }
            sPtr++;
        }

        return pPtr == p.length();
    }
}

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Problem Info

Description

Problem 1898: Maximum Number of Removable Characters - Detailed explanation and solution approaches.

Topics

Array Two Pointers String Binary Search

Companies

MoveworksSnowflake

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1898

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize `left` to 0 and `right` to the length of `removable`. Initialize `result` to store the maximum `k` found so far.
2
Step 2: While `left` is less than or equal to `right`, calculate `mid` as the average of `left` and `right`.
3
Step 3: Call the `is_subsequence` function with `mid` as the number of characters to remove.
4
Step 4: If `is_subsequence(mid)` returns `true` (meaning `p` is still a subsequence after removing `mid` characters), update `result` to `mid` and set `left` to `mid + 1` to search for a larger value of `k`.
5
Step 5: If `is_subsequence(mid)` returns `false`, set `right` to `mid - 1` to search for a smaller value of `k`.
6
Step 6: After the binary search completes, return the `result`.
7
Step 7: The `is_subsequence` function creates a set of removed indices. It then uses two pointers, one for `s` and one for `p`, to iterate through the strings. If a character in `s` at the current index is not removed and matches the current character in `p`, the `p` pointer is incremented. The `s` pointer is incremented regardless. If the `p` pointer reaches the end of `p`, it means `p` is a subsequence of `s`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log(R) * (S + P))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(R)

📝

Why?

The binary search has a logarithmic time complexity because it halves the search space in each iteration. The `is_subsequence` function iterates through the strings `s` and `p` using pointers, resulting in a linear time complexity with respect to their lengths. The space complexity is dominated by the space used to store the removed indices in `is_subsequence` function.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Linear Search

Iterate through each possible value of `k` from 0 to the length of `removable` and check if `p` is a subsequence of `s` after removing the characters. Return the largest `k` for which `p` is a subsequence.

Time: O(R * (S + P)), where R is the length of the `removable` array, S is the length of the string `s`, and P is the length of the string `p`.Space: O(min(S, R))

Trade-offs: The linear search approach is simpler to implement but has a higher time complexity than the binary search approach. It is suitable for smaller input sizes where the difference in performance is not significant. For larger input sizes, the binary search approach is more efficient.

✨

Key Insights

1Insight 1: The problem can be efficiently solved using Binary Search. We can binary search for the maximum possible `k` value.
2Insight 2: For a given `k`, we can check if `p` is a subsequence of `s` after removing the first `k` characters from `s` specified by `removable`. This involves simulating the removal and then checking the subsequence condition.
3Insight 3: Using a `set` or `boolean array` to keep track of removed indices is efficient for checking if a character at a given index should be considered during subsequence checking.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: `removable` array is empty. In this case, `left` and `right` are initialized to 0. The binary search will still execute once with mid = 0. is_subsequence(0) will be called, which correctly returns whether p is a subsequence of the original s. The result will be 0.
Edge case 2: `p` is an empty string. In this case, `p_ptr` will always be equal to `len(p)` (which is 0), so `is_subsequence` will always return `true`. The binary search will find the largest possible `k` which is the length of the `removable` array.
Edge case 3: `s` is an empty string but `p` is not. The `is_subsequence` will always return false. The result will be 0.
Edge case 4: `p` is not a subsequence of `s` even before removing any characters. `is_subsequence(0)` will return false, and the result will be 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge case where `removable` is empty. The binary search must still function correctly in this case and return the correct result.

✗Common mistake 2: Not correctly implementing the `is_subsequence` function. It is important to ensure that the subsequence check is accurate after removing the characters.

✗Common mistake 3: Incorrectly updating the `left` and `right` boundaries in the binary search. Ensure that `left = mid + 1` and `right = mid - 1` are used correctly to avoid infinite loops or incorrect results.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Text editing software. Imagine a user undoing a series of edits (represented by `removable` indices). The software needs to quickly determine the latest state where a certain keyword (`p`) is still present in the text (`s`).

🏦Real-world scenario 2: Data streaming validation. A stream of data arrives (`s`), and certain data points might be flagged for removal (`removable`). Before accepting a stream, we want to ensure a required pattern (`p`) is still present. This algorithm helps determine how many flagged points can be removed while still maintaining the necessary pattern.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 7, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Number of Removable Characters"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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