What is the time complexity of Maximize the Minimum Game Score?

The optimal solution has a time complexity of O(N*log(M*P)).

What data structures are used to solve Maximize the Minimum Game Score?

This problem uses Array, Binary Search, Greedy. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximize the Minimum Game Score?

Insight 1: The core idea is to use binary search to find the maximum possible minimum score. We can check if a given minimum score `x` is achievable within `m` moves. Insight 2: The `is_possible` function is critical. It calculates the minimum number of moves needed to make all game scores at least `x`. This involves figuring out how many times each index `i` needs to be visited, which can be calculated as `(x + points[i] - 1) // points[i]`.

Which companies ask LeetCode 3449. Maximize the Minimum Game Score in interviews?

This problem is frequently asked at Infosys.

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Maximize the Minimum Game Score - LeetCode 3449 Solution

Maximize the Minimum Game Score - Complete Solution Guide

Maximize the Minimum Game Score is LeetCode problem 3449, a Hard 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 an array points of size n and an integer m . There is another array gameScore of size n , where gameScore[i] represents the score achieved at the i th game. Initially, gameScore[i] == 0 for all i . You start at index -1, which is outside the array (before the first position at index 0). You can make at most m moves. In each move, you can either: Increase the index by 1 and add points[i] to gameScore[i] . Decrease the index by 1 and add points[i] to gameScore[i] . Note that the inde

Detailed Explanation

The problem asks us to maximize the minimum score across all games, given an array `points` representing the points achievable at each game and a maximum number of moves `m`. We start at index -1 and can move left or right, adding `points[i]` to `gameScore[i]` each time. The goal is to find the largest possible value `x` such that we can achieve a `gameScore` where every element is at least `x`, using at most `m` moves.

Solution Approach

The solution uses binary search to find the maximum possible minimum score. The search space is from 0 to `m * max(points)`, which represents the maximum possible score any game could have. For each potential minimum score `x` in the search space, the `is_possible` function checks if it's achievable within the given `m` moves. The `is_possible` function calculates the minimum moves required to achieve `x` by considering two cases: visiting all indices from 0 to n-1 and returning back to index 0 from some index. The lower cost amongst the two cases is compared against `m` to ascertain the possibility.

Step-by-Step Algorithm

Step 1: Initialize `low` to 0 and `high` to `m * max(points) + 1`. This defines the search space for the binary search.
Step 2: Perform binary search within the range `[low, high]` to find the maximum achievable minimum score. In each iteration calculate the mid point `mid = (low + high) // 2`.
Step 3: Call the `is_possible` function to check if a minimum score of `mid` can be achieved within `m` moves.
Step 4: Implement the `is_possible` function: Calculate the minimum number of visits required for each index using `c[i] = (x + points[i] - 1) // points[i]`. Consider two movement patterns.
Step 5: Case A: Calculate the minimum moves to visit indices 0 to n-1. Calculate `b[i] = c[i] - 1`. Compute prefix and suffix L values to decide how to achieve minimal left-to-right movement.
Step 6: Case B: Calculate the minimum moves to return back to 0. Calculate `b_prime[i]`, compute prefix and suffix L values, and find the minimum number of moves.
Step 7: Calculate cost_A and cost_B, return `min(cost_A, cost_B) <= m`.
Step 8: If `is_possible(mid)` returns `True`, it means `mid` is achievable, so update `ans = mid` and `low = mid + 1` to search for a higher possible score.
Step 9: If `is_possible(mid)` returns `False`, it means `mid` is not achievable, so update `high = mid - 1` to search for a lower possible score.
Step 10: After the binary search completes, return the final `ans`.

Key Insights

Insight 1: The core idea is to use binary search to find the maximum possible minimum score. We can check if a given minimum score `x` is achievable within `m` moves.
Insight 2: The `is_possible` function is critical. It calculates the minimum number of moves needed to make all game scores at least `x`. This involves figuring out how many times each index `i` needs to be visited, which can be calculated as `(x + points[i] - 1) // points[i]`.
Insight 3: There are two main movement patterns to consider: one where we visit all indices from 0 to n-1, and another where we go back to index 0 after visiting some indices. We choose the movement pattern that requires the least number of moves.

Complexity Analysis

Time Complexity: O(N*log(M*P))

Space Complexity: O(N)

Topics

This problem involves: Array, Binary Search, Greedy.

Companies

Asked at: Infosys.

Maximize the Minimum Game Score | LeetCode 3449

Hard

Array Binary Search Greedy

Time: O(N*log(M*P))Space: O(N)

Solution Code

Available in:

class Solution {
    public long maxScore(int[] points, int m) {
        int n = points.length;

        class Helper {
            boolean isPossible(long x) {
                if (x == 0) {
                    return true;
                }

                long[] c = new long[n];
                for (int i = 0; i < n; i++) {
                    c[i] = (x + points[i] - 1) / points[i];
                }

                // Case 1: Path starts at 0 and ends at n-1
                // Relations: R_i = L_i + 1 for i=0..n-2
                // Visits: v_0=1+L_0, v_i=L_{i-1}+L_i+1, v_{n-1}=L_{n-2}+1
                // Constraints on L_i:
                // L_0 >= c_0-1
                // L_{i-1}+L_i >= c_i-1 for i=1..n-1 (with L_{n-1}=0)
                long[] b = new long[n];
                b[0] = c[0] - 1;
                for (int i = 1; i < n; i++) {
                    b[i] = c[i] - 1;
                }

                long[] pref_L = new long[n - 1];
                pref_L[0] = Math.max(0, b[0]);
                for (int i = 1; i < n - 1; i++) {
                    pref_L[i] = Math.max(0, b[i] - pref_L[i - 1]);
                }

                long[] suff_L = new long[n - 1];
                suff_L[n - 2] = Math.max(0, b[n - 1]);
                for (int i = n - 3; i >= 0; i--) {
                    suff_L[i] = Math.max(0, b[i + 1] - suff_L[i + 1]);
                }

                long sum_L = 0;
                for (int i = 0; i < n - 1; i++) {
                    sum_L += Math.max(pref_L[i], suff_L[i]);
                }

                // Total moves = 1 (initial) + sum(R_i) + sum(L_i) = 1 + sum(L_i+1) + sum(L_i) = n + 2*sum_L
                long cost_A = n + 2 * sum_L;

                // Case 2: Path starts at 0 and ends at 0
                // Relations: R_i = L_i for i=0..n-2
                // Visits: v_0=1+L_0, v_i=L_{i-1}+L_i, v_{n-1}=L_{n-2}
                // Constraints on L_i:
                // L_0 >= c_0-1
                // L_{i-1}+L_i >= c_i for i=1..n-1 (with L_{n-1}=0)
                long[] b_prime = new long[n];
                b_prime[0] = c[0] - 1;
                for (int i = 1; i < n; i++) {
                    b_prime[i] = c[i];
                }

                long[] pref_L_prime = new long[n - 1];
                pref_L_prime[0] = Math.max(0, b_prime[0]);
                for (int i = 1; i < n - 1; i++) {
                    pref_L_prime[i] = Math.max(0, b_prime[i] - pref_L_prime[i - 1]);
                }

                long[] suff_L_prime = new long[n - 1];
                suff_L_prime[n - 2] = Math.max(0, b_prime[n - 1]);
                for (int i = n - 3; i >= 0; i--) {
                    suff_L_prime[i] = Math.max(0, b_prime[i + 1] - suff_L_prime[i + 1]);
                }

                long sum_L_prime = 0;
                for (int i = 0; i < n - 1; i++) {
                    sum_L_prime += Math.max(pref_L_prime[i], suff_L_prime[i]);
                }

                // Total moves = 1 (initial) + sum(R_i) + sum(L_i) = 1 + 2*sum_L
                long cost_B = 1 + 2 * sum_L_prime;

                return Math.min(cost_A, cost_B) <= m;
            }
        }

        Helper helper = new Helper();

        long low = 0;
        long high = (long)m * (long)java.util.Arrays.stream(points).max().getAsInt() + 1;
        long ans = 0;

        while (low <= high) {
            long mid = low + (high - low) / 2;
            if (mid == 0) {
                low = mid + 1;
                continue;
            }

            if (helper.isPossible(mid)) {
                ans = mid;
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }

        return ans;
    }
}

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Problem Info

Description

Problem 3449: Maximize the Minimum Game Score - Detailed explanation and solution approaches.

Topics

Array Binary Search Greedy

Companies

Infosys

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3449

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

10 steps

1
Step 1: Initialize `low` to 0 and `high` to `m * max(points) + 1`. This defines the search space for the binary search.
2
Step 2: Perform binary search within the range `[low, high]` to find the maximum achievable minimum score. In each iteration calculate the mid point `mid = (low + high) // 2`.
3
Step 3: Call the `is_possible` function to check if a minimum score of `mid` can be achieved within `m` moves.
4
Step 4: Implement the `is_possible` function: Calculate the minimum number of visits required for each index using `c[i] = (x + points[i] - 1) // points[i]`. Consider two movement patterns.
5
Step 5: Case A: Calculate the minimum moves to visit indices 0 to n-1. Calculate `b[i] = c[i] - 1`. Compute prefix and suffix L values to decide how to achieve minimal left-to-right movement.
6
Step 6: Case B: Calculate the minimum moves to return back to 0. Calculate `b_prime[i]`, compute prefix and suffix L values, and find the minimum number of moves.
7
Step 7: Calculate cost_A and cost_B, return `min(cost_A, cost_B) <= m`.
8
Step 8: If `is_possible(mid)` returns `True`, it means `mid` is achievable, so update `ans = mid` and `low = mid + 1` to search for a higher possible score.
9
Step 9: If `is_possible(mid)` returns `False`, it means `mid` is not achievable, so update `high = mid - 1` to search for a lower possible score.
10
Step 10: After the binary search completes, return the final `ans`.

💡

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(M*P))

💾

Space Complexity

O(N)

📝

Why?

The binary search dominates the time complexity. The `is_possible` function performs calculations that are linear in the size of the input array `points`. Since the binary search calls `is_possible` in each iteration, the overall time complexity is O(N * log(M * P)). The `is_possible` function creates several arrays of size `n`, contributing to O(n) space complexity.

📖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

A dynamic programming approach could be explored to determine the achievable minimum score for different numbers of moves. However, due to the large constraint on `m` (up to 10^9), it would be memory and computationally expensive.

Time: O(N * M), where M is the maximum movesSpace: O(N * M)

Trade-offs: DP is not efficient because of the large constraint on M. It is only useful if the constraint of the number of moves 'm' is low. Binary search is much more optimal in this context because of the large possible m.

✨

Key Insights

1Insight 1: The core idea is to use binary search to find the maximum possible minimum score. We can check if a given minimum score `x` is achievable within `m` moves.
2Insight 2: The `is_possible` function is critical. It calculates the minimum number of moves needed to make all game scores at least `x`. This involves figuring out how many times each index `i` needs to be visited, which can be calculated as `(x + points[i] - 1) // points[i]`.
3Insight 3: There are two main movement patterns to consider: one where we visit all indices from 0 to n-1, and another where we go back to index 0 after visiting some indices. We choose the movement pattern that requires the least number of moves.

📚

Key Terms Explained

For Students

Array

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

Binary Search

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: `x = 0`. If the target minimum score `x` is 0, the `is_possible` function should immediately return `True` as it requires no moves, handled by an `if x == 0: return True` condition.
Edge case 2: Zero points at an index. Since the problem states points[i] >=1, this case is not possible and needs no specific handling.
Edge case 3: `m = 1`. The code correctly handles the case by considering different movement strategies, and finding the lowest-cost movement possible. In the case of m=1, moving to the very first element will result in the first game score being increased by the value of the first element in points, and checking if `points[0] >= x`.
Edge case 4: Large values for points and m. The solution uses `long long` to avoid potential integer overflows when calculating visit counts and move counts.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow. Multiplying `m` with a value from `points` can lead to integer overflow, hence using `long long` is important.

✗Common mistake 2: Not considering both possible movement patterns (visiting all indices or returning to 0). Failing to consider both scenarios may yield a sub-optimal solution and failing the tests.

✗Common mistake 3: Inaccurate calculation of the number of visits to each game. The integer division `(x + points[i] - 1) // points[i]` needs to be handled carefully to get the correct number of visits.

✅

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 allocating a fixed budget (moves) to improve the performance (score) of different teams (games) such that all teams meet a minimum performance threshold.

🏦Real-world scenario 2: Network optimization. Optimizing network bandwidth (moves) to ensure a minimum quality of service (score) for all users (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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: June 30, 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

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

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Maximize the Minimum Game Score"?

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

🤖 Need help with this problem?