Stone Game IX

Medium

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ArraysGreedy Algorithms

Alice and Bob continue their games with stones. There is a row of n stones, and each stone has an associated value. You are given an integer array stones, where stones[i] is the value of the i^th stone.

Alice and Bob take turns, with Alice starting first. On each turn, the player may remove any stone from stones. The player who removes a stone loses if the sum of the values of all removed stones is divisible by 3. Bob will win automatically if there are no remaining stones (even if it is Alice's turn).

Assuming both players play optimally, return true if Alice wins and false if Bob wins.

Example 1:

Input: stones = [2,1]
Output: true
Explanation: The game will be played as follows:
- Turn 1: Alice can remove either stone.
- Turn 2: Bob removes the remaining stone. 
The sum of the removed stones is 1 + 2 = 3 and is divisible by 3. Therefore, Bob loses and Alice wins the game.

Example 2:

Input: stones = [2]
Output: false
Explanation: Alice will remove the only stone, and the sum of the values on the removed stones is 2. 
Since all the stones are removed and the sum of values is not divisible by 3, Bob wins the game.

Example 3:

Input: stones = [5,1,2,4,3]
Output: false
Explanation: Bob will always win. One possible way for Bob to win is shown below:
- Turn 1: Alice can remove the second stone with value 1. Sum of removed stones = 1.
- Turn 2: Bob removes the fifth stone with value 3. Sum of removed stones = 1 + 3 = 4.
- Turn 3: Alices removes the fourth stone with value 4. Sum of removed stones = 1 + 3 + 4 = 8.
- Turn 4: Bob removes the third stone with value 2. Sum of removed stones = 1 + 3 + 4 + 2 = 10.
- Turn 5: Alice removes the first stone with value 5. Sum of removed stones = 1 + 3 + 4 + 2 + 5 = 15.
Alice loses the game because the sum of the removed stones (15) is divisible by 3. Bob wins the game.

Constraints:

1 <= stones.length <= 10⁵
1 <= stones[i] <= 10⁴

Try coding it in LeetCode

Stone Game IX

Medium

Asked by:

14 views

Topics:

ArraysGreedy Algorithms

Alice and Bob continue their games with stones. There is a row of n stones, and each stone has an associated value. You are given an integer array stones, where stones[i] is the value of the i^th stone.

Alice and Bob take turns, with Alice starting first. On each turn, the player may remove any stone from stones. The player who removes a stone loses if the sum of the values of all removed stones is divisible by 3. Bob will win automatically if there are no remaining stones (even if it is Alice's turn).

Assuming both players play optimally, return true if Alice wins and false if Bob wins.

Example 1:

Input: stones = [2,1]
Output: true
Explanation: The game will be played as follows:
- Turn 1: Alice can remove either stone.
- Turn 2: Bob removes the remaining stone. 
The sum of the removed stones is 1 + 2 = 3 and is divisible by 3. Therefore, Bob loses and Alice wins the game.

Example 2:

Input: stones = [2]
Output: false
Explanation: Alice will remove the only stone, and the sum of the values on the removed stones is 2. 
Since all the stones are removed and the sum of values is not divisible by 3, Bob wins the game.

Example 3:

Input: stones = [5,1,2,4,3]
Output: false
Explanation: Bob will always win. One possible way for Bob to win is shown below:
- Turn 1: Alice can remove the second stone with value 1. Sum of removed stones = 1.
- Turn 2: Bob removes the fifth stone with value 3. Sum of removed stones = 1 + 3 = 4.
- Turn 3: Alices removes the fourth stone with value 4. Sum of removed stones = 1 + 3 + 4 = 8.
- Turn 4: Bob removes the third stone with value 2. Sum of removed stones = 1 + 3 + 4 + 2 = 10.
- Turn 5: Alice removes the first stone with value 5. Sum of removed stones = 1 + 3 + 4 + 2 + 5 = 15.
Alice loses the game because the sum of the removed stones (15) is divisible by 3. Bob wins the game.

Constraints:

1 <= stones.length <= 10⁵
1 <= stones[i] <= 10⁴

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 is the maximum size of the stones array?
Can the values in the stones array be zero?
Are all the values in the stones array guaranteed to be non-negative integers?
If both Alice and Bob play optimally, and Alice cannot win, should I return false?
Are there any constraints on the values of the individual stones (maximum value)?

Brute Force Solution

Approach

The brute force approach to this game is to explore every possible move in every possible game state. We simulate every possible game from the beginning until one player wins, and then we determine who wins from the starting position. This ensures that we consider all possible scenarios and outcomes.

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

Consider all possible moves for the first player.
For each of those moves, consider all possible moves for the second player.
Continue alternating moves between players, exploring every possibility at each step.
Keep track of the running sum of the stones chosen.
At each turn, check if the running sum is divisible by three. If it is, the current player loses.
If one player cannot make a move because all stones have been taken, determine the winner based on whether the running sum is divisible by three.
Repeat this process for all possible sequences of moves to determine if the first player can always win, or if the second player can force a win, or if the first player can win in at least one possible game.
The first player wins if there exists a sequence of moves that lead to a winning state and the first player plays optimally.

Code Implementation

def stone_game_ix_brute_force(stones):
    def can_win(current_stones, current_sum):
        # If no stones are left, determine the winner based on the sum
        if not current_stones:
            return current_sum % 3 != 0

        for index, stone in enumerate(current_stones):
            # If the current sum is divisible by 3, the player loses
            if (current_sum + stone) % 3 == 0:
                continue

            remaining_stones = current_stones[:index] + current_stones[index+1:]

            # Recursively call the function for the next player's turn.
            if not can_win(remaining_stones, current_sum + stone):
                return True

        return False

    # This checks if Alice can win from the starting position
    return can_win(stones, 0)

Big(O) Analysis

Time Complexity

O(3^n) – The brute force approach explores every possible combination of moves. In the worst case, each player has multiple choices at each turn. Since each stone can potentially be selected, the number of possible game states grows exponentially with the number of stones, n. Each decision splits the game into multiple possibilities resembling a decision tree with a branching factor dependent on the remaining choices, leading to approximately 3^n possible game states that need to be evaluated. The total operations therefore approximate O(3^n).

Space Complexity

O(N) – The brute force approach explores every possible move through recursion. In the worst-case scenario, the recursion depth can go up to N, where N is the number of stones, because each move removes one stone. Each recursive call adds a new frame to the call stack. Therefore, the auxiliary space used by the recursion stack is proportional to N, resulting in O(N) space complexity.

Optimal Solution

Approach

The core idea is to analyze the remainders of the stones when divided by 3. The player who can force a draw state for their opponent will win. We want to count how many stones have remainder 0, 1, and 2 when divided by 3, and based on those counts, determine the winner.

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

First, count how many stones have a remainder of 0, 1, and 2 when divided by 3.
If there are no stones with remainder 1 or no stones with remainder 2, the first player loses if there are an even number of stones with remainder 0, and wins if there are an odd number of stones with remainder 0 and at least one stone with remainder 0.
Otherwise, analyze the difference between the counts of stones with remainder 1 and stones with remainder 2.
If the absolute difference between these counts is greater than 2, the first player wins.
If the absolute difference is less than or equal to 2, and there are any stones with remainder 0, the first player wins.
Otherwise, if the absolute difference is less than or equal to 2, and there are no stones with remainder 0, the second player wins.

Code Implementation

def stone_game_ix(stones):    remainder_0_count = 0
    remainder_1_count = 0
    remainder_2_count = 0
    for stone in stones:
        if stone % 3 == 0:
            remainder_0_count += 1
        elif stone % 3 == 1:
            remainder_1_count += 1
        else:
            remainder_2_count += 1

    # Handle base cases where either remainder 1 or 2 stones are missing.
    if remainder_1_count == 0 or remainder_2_count == 0:
        if remainder_0_count % 2 == 0:
            return False

        return remainder_0_count > 0

    # Analyze the difference between remainder 1 and 2 counts.
    difference = abs(remainder_1_count - remainder_2_count)

    #If diff is greater than 2, Alice always wins.
    if difference > 2:
        return True

    #Check if remainder 0 stones lead to Alice winning.
    if remainder_0_count > 0:
        return True

    #Otherwise, Bob wins.
    return False

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of n stones once to count the number of stones with remainder 0, 1, and 2 when divided by 3. The subsequent conditional checks and arithmetic operations are performed in constant time, independent of the input size. Therefore, the dominant factor in determining the time complexity is the initial counting step, which takes O(n) time. Thus the overall time complexity is O(n).

Space Complexity

O(1) – The algorithm uses a fixed number of integer variables to store the counts of stones with remainders 0, 1, and 2 when divided by 3. The number of these variables does not depend on the number of stones (N). Therefore, the auxiliary space required is constant, regardless of the input size, and is denoted as O(1).

Edge Cases

Case	How to Handle
Empty input array	Return False immediately as no moves are possible.
Array with only one element	Return False immediately as the first player will always lose.
All numbers in the array are divisible by 3	If the number of elements is odd, Alice wins; otherwise, Bob wins.
Array with all numbers leaving a remainder of 1 when divided by 3	Compare the counts of numbers with remainder 1 and 2, Alice wins if count(1) > count(2) + 1 or count(2) > count(1).
Array with all numbers leaving a remainder of 2 when divided by 3	Compare the counts of numbers with remainder 1 and 2, Alice wins if count(1) > count(2) + 1 or count(2) > count(1).
A large array with uneven distribution of remainders (1 and 2)	The solution's O(n) complexity should scale adequately for large arrays; analyze remainder distribution to predict winner.
Integer overflow if summing very large numbers	The solution doesn't involve summing the numbers themselves, so overflow isn't a concern; only the counts are compared.
Input array contains negative numbers	The problem description does not specify that the numbers should be non-negative; if so, the modulus operation must be handled carefully to always produce a number between 0 and 2.