Maximize the Minimum Game Score

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ArraysBinary Search

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 index must always remain within the bounds of the array after the first move.

Return the maximum possible minimum value in gameScore after at most m moves.

Example 1:

Input: points = [2,4], m = 3

Output: 4

Explanation:

Initially, index i = -1 and gameScore = [0, 0].

Move	Index	gameScore
Increase `i`	0	`[2, 0]`
Increase `i`	1	`[2, 4]`
Decrease `i`	0	`[4, 4]`

The minimum value in gameScore is 4, and this is the maximum possible minimum among all configurations. Hence, 4 is the output.

Example 2:

Input: points = [1,2,3], m = 5

Output: 2

Explanation:

Initially, index i = -1 and gameScore = [0, 0, 0].

Move	Index	gameScore
Increase `i`	0	`[1, 0, 0]`
Increase `i`	1	`[1, 2, 0]`
Decrease `i`	0	`[2, 2, 0]`
Increase `i`	1	`[2, 4, 0]`
Increase `i`	2	`[2, 4, 3]`

The minimum value in gameScore is 2, and this is the maximum possible minimum among all configurations. Hence, 2 is the output.

Constraints:

2 <= n == points.length <= 5 * 10⁴
1 <= points[i] <= 10⁶
1 <= m <= 10⁹

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the values in the input array?
Can the input array be empty or null?
Can a single bag have a negative number of items initially?
If it is impossible to achieve a positive minimum score for any distribution, what should I return?
Is the number of bags always less than or equal to the number of items in the array?

Brute Force Solution

Approach

The brute force method for this game problem involves exploring every single possible way the game could be played. We imagine trying every possible sequence of choices a player could make during the game. By exploring all possibilities, we are guaranteed to find the best possible game outcome, even if it takes a long time.

Here's how the algorithm would work step-by-step:

Consider all possible numbers of rounds the game could last.
For each possible number of rounds, think about all the different choices the player could make in each round.
Simulate the game being played for each unique sequence of choices, tracking the game score at the end of each round.
For each simulated game, find the lowest score that was achieved during the game.
Compare the lowest scores across all the different possible games (sequences of choices).
Select the game that had the highest of all these lowest scores. That will be the solution, representing the maximum possible minimum score the player could guarantee.

Code Implementation

def maximize_minimum_game_score_brute_force(possible_moves):
    maximum_minimum_score = float('-inf')

    # Iterate through all possible game lengths.
    for number_of_rounds in range(1, len(possible_moves) + 1):
        all_possible_game_states = get_all_possible_game_states(
            possible_moves, number_of_rounds
        )

        # Iterate through each possible game state
        for game_state in all_possible_game_states:
            minimum_score_for_game = min(game_state)

            # We want to maximize the minimum score.
            maximum_minimum_score = max(
                maximum_minimum_score, minimum_score_for_game
            )

    return maximum_minimum_score

def get_all_possible_game_states(possible_moves, number_of_rounds):
    if number_of_rounds == 0:
        return [[]]

    previous_game_states = get_all_possible_game_states(
        possible_moves, number_of_rounds - 1
    )

    all_game_states = []
    for game_state in previous_game_states:
        for move in possible_moves:
            new_game_state = game_state + [move]

            all_game_states.append(new_game_state)

    return all_game_states

# Example Usage (Test Case 1):
possible_moves_test_one = [1, 2, 3]
result_test_one = maximize_minimum_game_score_brute_force(possible_moves_test_one)

# Example Usage (Test Case 2):
possible_moves_test_two = [5, 2, 9, 1]
result_test_two = maximize_minimum_game_score_brute_force(possible_moves_test_two)

# Example Usage (Test Case 3):
possible_moves_test_three = [8, 6, 7, 5, 3, 10, 9]
result_test_three = maximize_minimum_game_score_brute_force(possible_moves_test_three)

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores all possible gameplays. The dominant factor in the time complexity is the number of possible choices the player can make in each round. Assuming there are two choices at each step and the game can last up to n rounds (where n is the input size, likely the number of elements or rounds), the total number of possible gameplays is approximately 2 multiplied by itself n times, thus 2^n. Therefore, the algorithm must simulate all possible games and calculate the minimum game score for each, leading to an exponential time complexity.

Space Complexity

O(N^R) – The brute force approach simulates every possible game sequence. Step 3 describes simulating the game and tracking scores, implying a data structure (like a list or array) to store these scores. If we consider 'R' to be the maximum number of rounds, and each round has, on average, 'N' choices, then exploring all sequences requires storing states for potentially N^R different game paths. The number of simulated games grows exponentially with the number of rounds, R, relative to the input size N, because for each possible game state a copy of the state is created. Thus the space complexity is O(N^R), where N is the input size and R is the number of rounds.

Optimal Solution

Approach

The goal is to pick numbers such that the smallest number you pick is as large as possible. We'll use a clever technique called binary search to efficiently find this optimal value, instead of checking every single possibility. This avoids exploring obviously bad scores.

Here's how the algorithm would work step-by-step:

Think about a possible minimum score. Is it achievable or not? If we can pick all the numbers so that none of the chosen numbers are less than the potential minimum score then it's achievable.
Use binary search to guess potential minimum scores. Start by guessing a minimum score in the middle of the range of possible scores.
Check if we can achieve this guessed minimum score. This means making sure we can pick numbers from each group that are all greater than or equal to our guessed score.
If we can achieve the guessed minimum score, then try a higher minimum score (search in the upper half).
If we can't achieve the guessed minimum score, then try a lower minimum score (search in the lower half).
Continue this process of guessing and checking until we find the highest possible minimum score that we can achieve.

Code Implementation

def maximize_minimum_game_score(number_of_players, group_values):
    left = 1
    right = max(max(group) for group in group_values)
    answer = 0

    while left <= right:
        potential_minimum_score = (left + right) // 2

        # Check if the current minimum score is achievable.
        if is_achievable(number_of_players, group_values, potential_minimum_score):
            # If achievable, try to increase the min score.
            answer = potential_minimum_score
            left = potential_minimum_score + 1

        else:
            # If not achievable, decrease the min score.
            right = potential_minimum_score - 1

    return answer

def is_achievable(number_of_players, group_values, minimum_score):
    players_needed = 0

    # Count how many players are needed to achieve min score.
    for group in group_values:
        found_value = False
        for value in group:
            if value >= minimum_score:
                found_value = True
                break
        if not found_value:
            return False

    return True

Big(O) Analysis

Time Complexity

O(n*m*log(maxValue - minValue)) – The algorithm uses binary search to find the optimal minimum score within the range of possible values (maxValue - minValue). The binary search performs log(maxValue - minValue) iterations. Inside each iteration, we check if a given minimum score is achievable. This check involves iterating through the n groups of numbers, and for each group, we need to find a number greater than or equal to the guessed minimum score. Finding this number in each group can take up to m operations where m is the size of the largest group. Therefore the time complexity to check for achievability is O(n*m). Combining the binary search with the achievability check gives us a total time complexity of O(n*m*log(maxValue - minValue)).

Space Complexity

O(1) – The described binary search algorithm, focused on guessing and checking potential minimum scores, primarily uses a fixed number of variables to store the lower and upper bounds for the binary search, and potentially a variable to store the current guess. The space used to store these variables does not scale with the size of the input, which we can consider as the total number of elements across all groups. Therefore, the algorithm exhibits constant auxiliary space complexity.

Edge Cases

Case	How to Handle
Empty input array	Return 0 or throw an IllegalArgumentException, depending on the problem statement, as no score is possible.
Input array with a single element	Return the value of the single element as that will become the only game score.
All elements in the array are identical	The minimum score would be the identical element, so handle with standard min/max operations.
Array contains negative numbers	The solution must handle negative numbers correctly in the calculation of the minimum game score (min function).
Array contains very large positive numbers, potentially leading to integer overflow	Use a data type with a larger range, like 'long', or consider using modular arithmetic if appropriate.
The array is extremely large, exceeding available memory	If possible, process the array in chunks or utilize an out-of-core algorithm to minimize memory usage.
Maximum size input array with minimal values will trigger min operations very often	Verify complexity to avoid timeout - optimal algorithm may be necessary.
Input array contains zero values	Zeroes need to be handled carefully in calculations, particularly when finding the minimum game score.

Empty input array

How to Handle:

Return 0 or throw an IllegalArgumentException, depending on the problem statement, as no score is possible.

Input array with a single element

How to Handle:

Return the value of the single element as that will become the only game score.

All elements in the array are identical

How to Handle:

The minimum score would be the identical element, so handle with standard min/max operations.

Array contains negative numbers

How to Handle:

The solution must handle negative numbers correctly in the calculation of the minimum game score (min function).

Array contains very large positive numbers, potentially leading to integer overflow

How to Handle:

Use a data type with a larger range, like 'long', or consider using modular arithmetic if appropriate.

The array is extremely large, exceeding available memory

How to Handle:

If possible, process the array in chunks or utilize an out-of-core algorithm to minimize memory usage.

Maximum size input array with minimal values will trigger min operations very often

How to Handle:

Verify complexity to avoid timeout - optimal algorithm may be necessary.

Input array contains zero values

How to Handle:

Zeroes need to be handled carefully in calculations, particularly when finding the minimum game score.