What is the time complexity of Minimum Levels to Gain More Points?

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

What data structures are used to solve Minimum Levels to Gain More Points?

This problem uses Array, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Levels to Gain More Points?

Insight 1: Calculate the total possible score if all levels are played, which represents the combined potential score for Alice and Bob. Insight 2: Iteratively calculate Alice's score and determine Bob's score by subtracting Alice's score from the total score. This avoids iterating through Bob's section separately.

Which companies ask LeetCode 3096. Minimum Levels to Gain More Points in interviews?

This problem is frequently asked at IBM.

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Minimum Levels to Gain More Points - LeetCode 3096 Solution

Minimum Levels to Gain More Points - Complete Solution Guide

Minimum Levels to Gain More Points is LeetCode problem 3096, 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 binary array possible of length n . Alice and Bob are playing a game that consists of n levels. Some of the levels in the game are impossible to clear while others can always be cleared. In particular, if possible[i] == 0 , then the i th level is impossible to clear for both the players. A player gains 1 point on clearing a level and loses 1 point if the player fails to clear it. At the start of the game, Alice will play some levels in the given order starting from the 0 th level

Detailed Explanation

The problem asks us to find the minimum number of levels Alice needs to play in a game such that her score is greater than Bob's. The game consists of 'n' levels, where each level can either be cleared (possible[i] = 1) or impossible to clear (possible[i] = 0). Clearing a level gives a player 1 point, and failing loses 1 point. Alice plays some initial levels and Bob plays the remaining levels. Both players play optimally, aiming to maximize their scores. We need to return the minimum number of levels Alice should play to have a strictly higher score than Bob. If it is impossible for Alice to have a higher score, return -1.

Solution Approach

The solution iterates through all possible numbers of levels Alice can play, from 1 to n-1. For each number of levels, it calculates Alice's score and implicitly calculates Bob's score as the total score minus Alice's score. It then checks if Alice's score is greater than Bob's score. If it finds a valid number of levels where Alice's score is greater, it returns the number of levels. If no such number of levels is found, it returns -1.

Step-by-Step Algorithm

Step 1: Calculate the total possible score by iterating through the 'possible' array. Add 1 to the total_score if possible[i] is 1, subtract 1 if possible[i] is 0.
Step 2: Initialize alice_score to 0.
Step 3: Iterate from i = 0 to n-2 (inclusive), representing the number of levels Alice will play up to i+1.
Step 4: In each iteration, update alice_score by adding 1 if possible[i] is 1 and subtracting 1 if possible[i] is 0.
Step 5: Check if 2 * alice_score > total_score. If this condition is met, it means Alice's score is greater than Bob's score. Return i + 1, which is the minimum number of levels Alice needs to play.
Step 6: If the loop completes without finding a valid number of levels, return -1.

Key Insights

Insight 1: Calculate the total possible score if all levels are played, which represents the combined potential score for Alice and Bob.
Insight 2: Iteratively calculate Alice's score and determine Bob's score by subtracting Alice's score from the total score. This avoids iterating through Bob's section separately.
Insight 3: The condition Alice needs to satisfy is that her score is strictly greater than Bob's score: alice_score > total_score - alice_score, simplified as 2 * alice_score > total_score.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Prefix Sum.

Companies

Asked at: IBM.

Minimum Levels to Gain More Points | LeetCode 3096

Medium

Array Prefix Sum

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

Solution Code

Available in:

class Solution {
    public int minimumLevels(int[] possible) {
        int n = possible.length;
        
        int total_score = 0;
        for (int p : possible) {
            if (p == 1) {
                total_score += 1;
            } else {
                total_score -= 1;
            }
        }
        
        int alice_score = 0;
        for (int i = 0; i < n - 1; i++) {
            if (possible[i] == 1) {
                alice_score += 1;
            } else {
                alice_score -= 1;
            }
            
            // Condition is Alice's score > Bob's score
            // alice_score > total_score - alice_score
            // 2 * alice_score > total_score
            if (2 * alice_score > total_score) {
                return i + 1;
            }
        }
        
        return -1;
    }
}

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Problem Info

Description

Problem 3096: Minimum Levels to Gain More Points - Detailed explanation and solution approaches.

Topics

Array Prefix Sum

Companies

IBM

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3096

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

6 steps

1
Step 1: Calculate the total possible score by iterating through the 'possible' array. Add 1 to the total_score if possible[i] is 1, subtract 1 if possible[i] is 0.
2
Step 2: Initialize alice_score to 0.
3
Step 3: Iterate from i = 0 to n-2 (inclusive), representing the number of levels Alice will play up to i+1.
4
Step 4: In each iteration, update alice_score by adding 1 if possible[i] is 1 and subtracting 1 if possible[i] is 0.
5
Step 5: Check if 2 * alice_score > total_score. If this condition is met, it means Alice's score is greater than Bob's score. Return i + 1, which is the minimum number of levels Alice needs to play.
6
Step 6: If the loop completes without finding a valid number of levels, return -1.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because the algorithm iterates through the array 'possible' a fixed number of times (twice). The space complexity is O(1) since the algorithm only uses a few integer variables, and their memory usage doesn't depend on the size of the input array.

📖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

Prefix Sum with Binary Search

Calculate prefix sums for the scores. Then, use binary search to find the minimum number of levels for Alice to play such that her score exceeds Bob's. Requires more setup than the current linear solution.

Time: O(n + log n), for prefix sum calculation and binary search respectively.Space: O(n) for the prefix sum array.

Trade-offs: This approach is more complex to implement and doesn't offer significant performance advantages over the linear solution due to the linear setup cost. The overhead of binary search is only beneficial for extremely large inputs where the log n part starts mattering but the problem constraints don't make it particularly relevant in the contest context.

✨

Key Insights

1Insight 1: Calculate the total possible score if all levels are played, which represents the combined potential score for Alice and Bob.
2Insight 2: Iteratively calculate Alice's score and determine Bob's score by subtracting Alice's score from the total score. This avoids iterating through Bob's section separately.
3Insight 3: The condition Alice needs to satisfy is that her score is strictly greater than Bob's score: alice_score > total_score - alice_score, simplified as 2 * alice_score > total_score.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: An array with only 0s. In this case, the total score will be negative, and Alice will never be able to achieve a higher score than Bob. The algorithm correctly returns -1.
Edge case 2: An array where Alice needs to play all but the last level. This is explicitly handled in the for loop that goes until n-2.
Edge case 3: An array with a mix of 0s and 1s where Alice can win after playing just one level. The loop correctly identifies this scenario.
Edge case 4: An array where it is not possible for Alice to have a higher score than Bob. The algorithm correctly returns -1 after checking all possibilities.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the case where it's impossible for Alice to have a higher score and forgetting to return -1 in such scenarios.

✗Common mistake 2: Incorrectly calculating Bob's score. Using the total score and Alice's score avoids having to iterate through Bob's portion.

✗Common mistake 3: Not checking if 2 * alice_score > total_score but instead checking alice_score > total_score directly, which is incorrect.

✗Common mistake 4: Forgetting that each player has to play at least one level.

✅

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: Game theory and decision-making problems, where resources or actions are allocated between two entities to achieve a specific outcome.

🏦Real-world scenario 2: Resource allocation in a project, where the goal is to find the minimum resources required for one team to outperform another based on project completion probabilities.

💡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 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

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

@InterviewProJul 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.

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 "Minimum Levels to Gain More Points"?

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 prefix sum techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?