What is the time complexity of Maximum Score From Removing Substrings?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Maximum Score From Removing Substrings?

This problem uses String, Stack, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Score From Removing Substrings?

Insight 1: The order of removals matters because removing one substring might expose other substrings that can then be removed. Insight 2: A greedy approach of prioritizing removals based on the higher score (x or y) can lead to the optimal solution.

Which companies ask LeetCode 1717. Maximum Score From Removing Substrings in interviews?

This problem is frequently asked at Swiggy.

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Maximum Score From Removing Substrings - LeetCode 1717 Solution

Maximum Score From Removing Substrings - Complete Solution Guide

Maximum Score From Removing Substrings is LeetCode problem 1717, 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 a string s and two integers x and y . You can perform two types of operations any number of times. Remove substring "ab" and gain x points. For example, when removing "ab" from "c ab xbae" it becomes "cxbae" . Remove substring "ba" and gain y points. For example, when removing "ba" from "cabx ba e" it becomes "cabxe" . Return the maximum points you can gain after applying the above operations on s . Example 1: Input: s = "cdbcbbaaabab", x = 4, y = 5 Output: 19 Explanation: - Remove

Detailed Explanation

The problem asks us to find the maximum score that can be obtained by repeatedly removing either the substring "ab" with score x or the substring "ba" with score y from a given string s. We can perform these removals any number of times and in any order. The goal is to maximize the total score from these removals.

Solution Approach

The solution employs a greedy strategy combined with a stack data structure. First, it determines whether x > y or y > x and prioritizes removal of the substring corresponding to the larger score. It iterates through the input string and uses a stack to keep track of characters. If the top of the stack combined with the current character forms either 'ab' or 'ba' (depending on which is being prioritized), the pair is removed (the top character is popped from the stack, and the current character is skipped), and the corresponding score is added. If not, the current character is pushed onto the stack. This process is repeated for both 'ab' and 'ba', regardless of the prioritization.

Step-by-Step Algorithm

Step 1: Define a helper function `remove_and_score(text, pair, points)` that takes a string, a substring pair ("ab" or "ba"), and the score for removing that pair as input.
Step 2: Inside `remove_and_score`, initialize an empty stack and a `current_score` to 0.
Step 3: Iterate through the input `text` character by character.
Step 4: If the stack is not empty and the top of the stack combined with the current character forms the target `pair`, pop the top of the stack and increment `current_score` by `points`.
Step 5: Otherwise, push the current character onto the stack.
Step 6: After processing the entire `text`, return the remaining string (represented by the characters left in the stack) and the `current_score`.
Step 7: In the `maximumGain` function, compare x and y to determine which substring to prioritize.
Step 8: Call `remove_and_score` with the prioritized substring and corresponding score, then call it again with the remaining string and the other substring and score.
Step 9: Sum the scores from both calls to `remove_and_score` and return the final score.

Key Insights

Insight 1: The order of removals matters because removing one substring might expose other substrings that can then be removed.
Insight 2: A greedy approach of prioritizing removals based on the higher score (x or y) can lead to the optimal solution.
Insight 3: Using a stack data structure can efficiently identify and remove the targeted substrings "ab" and "ba".

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Stack, Greedy.

Companies

Asked at: Swiggy.

Maximum Score From Removing Substrings | LeetCode 1717

Medium

String Stack Greedy

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int maximumGain(String s, int x, int y) {
        
        class Helper {
            public Pair<String, Integer> removeAndScore(String text, String pair, int points) {
                StringBuilder stack = new StringBuilder();
                int currentScore = 0;
                char p1 = pair.charAt(0);
                char p2 = pair.charAt(1);

                for (char character : text.toCharArray()) {
                    if (stack.length() > 0 && stack.charAt(stack.length() - 1) == p1 && character == p2) {
                        stack.deleteCharAt(stack.length() - 1);
                        currentScore += points;
                    } else {
                        stack.append(character);
                    }
                }

                return new Pair<>(stack.toString(), currentScore);
            }
        }

        Helper helper = new Helper();
        int score = 0;
        if (x > y) {
            // Prioritize removing "ab"
            Pair<String, Integer> result1 = helper.removeAndScore(s, "ab", x);
            String s1 = result1.getKey();
            int score1 = result1.getValue();
            
            Pair<String, Integer> result2 = helper.removeAndScore(s1, "ba", y);
            int score2 = result2.getValue();
            
            score = score1 + score2;
        } else {
            // Prioritize removing "ba"
            Pair<String, Integer> result1 = helper.removeAndScore(s, "ba", y);
            String s1 = result1.getKey();
            int score1 = result1.getValue();
            
            Pair<String, Integer> result2 = helper.removeAndScore(s1, "ab", x);
            int score2 = result2.getValue();
            
            score = score1 + score2;
        }
        
        return score;
    }

    static class Pair<K, V> {
        private final K key;
        private final V value;

        public Pair(K key, V value) {
            this.key = key;
            this.value = value;
        }

        public K getKey() {
            return key;
        }

        public V getValue() {
            return value;
        }
    }
}

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Problem Info

Description

Problem 1717: Maximum Score From Removing Substrings - Detailed explanation and solution approaches.

Topics

String Stack Greedy

Companies

Swiggy

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1717

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

9 steps

1
Step 1: Define a helper function `remove_and_score(text, pair, points)` that takes a string, a substring pair ("ab" or "ba"), and the score for removing that pair as input.
2
Step 2: Inside `remove_and_score`, initialize an empty stack and a `current_score` to 0.
3
Step 3: Iterate through the input `text` character by character.
4
Step 4: If the stack is not empty and the top of the stack combined with the current character forms the target `pair`, pop the top of the stack and increment `current_score` by `points`.
5
Step 5: Otherwise, push the current character onto the stack.
6
Step 6: After processing the entire `text`, return the remaining string (represented by the characters left in the stack) and the `current_score`.
7
Step 7: In the `maximumGain` function, compare x and y to determine which substring to prioritize.
8
Step 8: Call `remove_and_score` with the prioritized substring and corresponding score, then call it again with the remaining string and the other substring and score.
9
Step 9: Sum the scores from both calls to `remove_and_score` and return the final score.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because the algorithm processes each character of the input string a constant number of times. The space complexity is also linear because the stack used in the algorithm can potentially store all the characters of the input string in the worst case.

📖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

Dynamic Programming

Dynamic programming could potentially be used to explore all possible removal combinations. However, it's generally less efficient for this problem due to the potential for overlapping subproblems, which aren't easily memoizable because of the string transformations.

Time: Potentially O(n^3) or higher due to the string manipulations involved in DP.Space: O(n^2) or higher for storing DP table.

Trade-offs: DP is generally overkill for this problem due to the simpler and more efficient greedy approach. DP might be considered if the problem constraints were significantly different (e.g., limited number of removals, or more complex scoring rules), but the greedy approach is typically preferred here.

✨

Key Insights

1Insight 1: The order of removals matters because removing one substring might expose other substrings that can then be removed.
2Insight 2: A greedy approach of prioritizing removals based on the higher score (x or y) can lead to the optimal solution.
3Insight 3: Using a stack data structure can efficiently identify and remove the targeted substrings "ab" and "ba".

📚

Key Terms Explained

For Students

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

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

Edge Cases

Edge case 1: Empty string. The algorithm handles this implicitly because the loops won't execute, and the score will be 0.
Edge case 2: String with no occurrences of 'a' or 'b'. The algorithm correctly processes these strings without errors, resulting in a score of 0.
Edge case 3: String with only one type of pair ('ab' but no 'ba', or vice versa). The algorithm correctly removes all possible pairs of one type and then moves on to the second type.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not prioritizing the substring with the higher score can lead to a suboptimal solution.

✗Common mistake 2: Failing to handle cases where one substring cannot be removed after the other is removed can lead to incorrect results. The greedy algorithm covers this correctly.

✗Common mistake 3: Incorrectly implementing the stack operations can lead to errors in identifying and removing the substrings.

✅

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: This algorithm can be adapted for optimizing resource allocation in manufacturing, where resources need to be combined in specific pairs for maximum profit. The 'ab' and 'ba' could represent resources and x and y represent the profits for each combination.

🏦Real-world scenario 2: In text compression, similar techniques can be used to identify and remove frequently occurring patterns to reduce the size of the data, with the 'score' representing the amount of compression achieved.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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 6, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

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

@CodeNinjaJul 6

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

@TechLeadJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

What is a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Score From Removing Substrings"?

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

🤖 Need help with this problem?