What is the time complexity of Bag of Tokens?

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

What data structures are used to solve Bag of Tokens?

This problem uses Array, Two Pointers, Greedy, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Bag of Tokens?

Insight 1: Sorting the tokens allows us to efficiently use a greedy approach, as we can prioritize playing smaller tokens face-up when we have enough power and larger tokens face-down when our score allows. Insight 2: A two-pointer approach helps optimize the process of deciding which tokens to play face-up or face-down based on current power and score.

Which companies ask LeetCode 948. Bag of Tokens in interviews?

This problem is frequently asked at Flexport.

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Bag of Tokens - LeetCode 948 Solution

Bag of Tokens - Complete Solution Guide

Bag of Tokens is LeetCode problem 948, 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 start with an initial power of power , an initial score of 0 , and a bag of tokens given as an integer array tokens , where each tokens[i] denotes the value of token i . Your goal is to maximize the total score by strategically playing these tokens. In one move, you can play an unplayed token in one of the two ways (but not both for the same token): Face-up : If your current power is at least tokens[i] , you may play token i , losing tokens[i] power and gaining 1 score. Face-down : If your c

Detailed Explanation

The problem asks us to maximize the score we can obtain by playing tokens, given an initial power and a list of token values. We can play a token in two ways: face-up (spend power to gain score) or face-down (spend score to gain power). The goal is to find the maximum score we can achieve through strategic token plays.

Solution Approach

The solution uses a greedy approach combined with a two-pointer technique. First, the tokens are sorted in ascending order. Then, two pointers, `left` and `right`, are initialized to the start and end of the sorted token list, respectively. We iterate while `left <= right`. In each iteration, we check if we can play the token at `left` face-up (power >= tokens[left]). If so, we increment the score and reduce the power. Otherwise, if the score is greater than 0 and there are at least two tokens left (left < right), we play the token at `right` face-down, increasing power and decreasing score. If neither of these conditions is met, it means we cannot play any more tokens, and we break the loop. We maintain the maximum score achieved during the process.

Step-by-Step Algorithm

Step 1: Sort the `tokens` array in ascending order.
Step 2: Initialize `score` to 0, `max_score` to 0, `left` to 0, and `right` to `len(tokens) - 1`.
Step 3: While `left <= right`, do the following:
Step 4: If `power >= tokens[left]`, play the token face-up: decrement `power` by `tokens[left]`, increment `score` by 1, increment `left` by 1, and update `max_score` to the maximum of its current value and `score`.
Step 5: Else if `score > 0` and `left < right`, play the token face-down: increment `power` by `tokens[right]`, decrement `score` by 1, and decrement `right` by 1.
Step 6: Else, break the loop because no more moves are possible.
Step 7: Return `max_score`.

Key Insights

Insight 1: Sorting the tokens allows us to efficiently use a greedy approach, as we can prioritize playing smaller tokens face-up when we have enough power and larger tokens face-down when our score allows.
Insight 2: A two-pointer approach helps optimize the process of deciding which tokens to play face-up or face-down based on current power and score.
Insight 3: The problem requires careful handling of edge cases such as not having enough power or score to play any tokens, or when there are no tokens at all.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers, Greedy, Sorting.

Companies

Asked at: Flexport.

Bag of Tokens | LeetCode 948

Medium

Array Two Pointers Greedy Sorting

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int bagOfTokensScore(int[] tokens, int power) {
        Arrays.sort(tokens);
        
        int score = 0;
        int max_score = 0;
        int left = 0;
        int right = tokens.length - 1;
        
        while (left <= right) {
            if (power >= tokens[left]) {
                power -= tokens[left];
                score += 1;
                left += 1;
                max_score = Math.max(max_score, score);
            } else if (score > 0 && left < right) {
                power += tokens[right];
                score -= 1;
                right -= 1;
            } else {
                break;
            }
        }
        
        return max_score;
    }
}

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Problem Info

Description

Problem 948: Bag of Tokens - Detailed explanation and solution approaches.

Topics

Array Two Pointers Greedy Sorting

Companies

Flexport

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:948

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: Sort the `tokens` array in ascending order.
2
Step 2: Initialize `score` to 0, `max_score` to 0, `left` to 0, and `right` to `len(tokens) - 1`.
3
Step 3: While `left <= right`, do the following:
4
Step 4: If `power >= tokens[left]`, play the token face-up: decrement `power` by `tokens[left]`, increment `score` by 1, increment `left` by 1, and update `max_score` to the maximum of its current value and `score`.
5
Step 5: Else if `score > 0` and `left < right`, play the token face-down: increment `power` by `tokens[right]`, decrement `score` by 1, and decrement `right` by 1.
6
Step 6: Else, break the loop because no more moves are possible.
7
Step 7: Return `max_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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is driven by the sorting step. After sorting, the two-pointer approach iterates through the array at most once, contributing O(n) to the overall time. However, O(n log n) dominates O(n). The space complexity is constant because we only use a few variables to store scores, pointers, and the maximum score, regardless of the input size.

📖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 combinations of playing tokens face-up or face-down. However, this would likely result in a higher time complexity.

Time: O(2^n), where n is the number of tokens, as each token has two choices (face-up or face-down).Space: O(n), to store the DP table or recursion stack.

Trade-offs: The dynamic programming approach is less efficient than the greedy approach due to the exponential time complexity. It may be useful if the constraints were different and the greedy approach didn't work, or if you needed all possible scores for some reason, rather than just the maximum.

✨

Key Insights

1Insight 1: Sorting the tokens allows us to efficiently use a greedy approach, as we can prioritize playing smaller tokens face-up when we have enough power and larger tokens face-down when our score allows.
2Insight 2: A two-pointer approach helps optimize the process of deciding which tokens to play face-up or face-down based on current power and score.
3Insight 3: The problem requires careful handling of edge cases such as not having enough power or score to play any tokens, or when there are no tokens at all.

📚

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.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `tokens` is empty. The loop won't execute, and the initial `max_score` (0) will be returned, which is correct.
Edge case 2: `power` is initially less than the smallest token, and `score` is 0. The loop will terminate immediately, returning 0, which is also correct.
Edge case 3: All tokens can be played face-up consecutively. The algorithm will correctly accumulate the score until the power runs out.
Edge case 4: A mix of tokens requires playing some face-up and some face-down. The greedy strategy ensures optimal score maximization.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the tokens. Without sorting, the greedy approach won't work correctly, and the optimal score might not be achieved.

✗Common mistake 2: Not handling the edge case where the score is 0 and there's not enough power to play any tokens face-up. This can lead to incorrect results or infinite loops.

✗Common mistake 3: Not checking `left < right` before attempting to play a token face down. This prevents out-of-bounds access and ensures the solution terminates correctly.

✅

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: Resource allocation - Imagine managing a budget (power) to acquire resources (tokens) that provide some utility (score). You might need to strategically spend resources to gain more budget or prioritize acquiring high-utility resources when available.

🏦Real-world scenario 2: Game strategy - Many games involve making choices between spending resources (like in-game currency or action points) to gain benefits (like score or in-game advantages) or conserving resources. This problem mirrors the strategic decision-making process in such games.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: July 19, 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 14

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

@TechLeadJul 7

❓

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 "Bag of Tokens"?

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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