Problem Description

This problem is about a two-player game where players take turns choosing numbers to add to a running total, with the goal of being the first to reach or exceed a target value.

Game Rules:

• Two players take turns choosing integers from 1 to maxChoosableInteger
• Each integer can only be used once throughout the entire game (no replacement)
• Players add their chosen number to a running total that starts at 0
• The first player to make the total reach or exceed desiredTotal wins
• Both players play optimally (always make the best possible move)

Your Task: Given two integers:

• maxChoosableInteger: The maximum number that can be chosen (players can pick from 1 to this number)
• desiredTotal: The target sum to reach or exceed to win

Return true if the first player (the one who starts) can force a win, otherwise return false.

Example Scenario: If maxChoosableInteger = 10 and desiredTotal = 11:

• Player 1 could choose 10, bringing the total to 10
• Player 2 would need to choose from the remaining numbers (1-9), and any choice would make the total ≥ 11
• Therefore, Player 2 would win, so the function should return false

Key Points:

• Numbers cannot be reused once chosen
• "Optimal play" means each player always makes the move that gives them the best chance to win
• The first player wins if they have a guaranteed winning strategy regardless of how the second player plays

Intuition

The key insight is that this is a classic game theory problem where we need to determine if the current player has a winning strategy. Since both players play optimally, we need to think recursively: a player can win if they can make a move that puts their opponent in a losing position.

Let's think about when the first player can guarantee a win:

1. If they can choose a number that immediately reaches or exceeds the target, they win
2. If they can choose a number that leaves the opponent in a position where the opponent cannot win no matter what they choose

This naturally leads to a recursive approach where we try all possible moves and check if any move leads to the opponent losing.

However, there's an important edge case to consider first: What if the sum of all available numbers (1 + 2 + ... + maxChoosableInteger) is less than desiredTotal? In this case, even if both players use all numbers, they can't reach the target, so nobody can win. We can calculate this sum using the formula (1 + maxChoosableInteger) * maxChoosableInteger / 2.

For the main game logic, we need to track:

• Which numbers have been used (since each number can only be used once)
• The current running total

We can represent the used numbers as a bitmask where bit i indicates whether number i has been used. For example, if numbers 1 and 3 are used, the mask would have bits 1 and 3 set.

The recursive strategy works as follows:

• For each available number i (where bit i in mask is 0)
• If choosing i makes the total ≥ desiredTotal, current player wins
• Otherwise, simulate choosing i and check if the opponent loses in the resulting state
• If the opponent loses, then the current player wins by choosing i
• If no choice leads to a win, the current player loses

Since the same game state (same used numbers and same total) can occur through different sequences of moves, we use memoization to avoid recalculating the same states.

Solution Approach

The solution uses state compression with bitmask and memoization to efficiently solve this game theory problem.

Step 1: Check if winning is possible

First, we calculate if the sum of all available numbers is sufficient to reach the target:

if (1 + maxChoosableInteger) * maxChoosableInteger // 2 < desiredTotal:
    return False

If the total sum is less than desiredTotal, nobody can win, so we return False.

Step 2: Define the recursive function with memoization

We define dfs(mask, s) where:

• mask: A bitmask representing which numbers have been used (bit i is 1 if number i is used)
• s: The current cumulative sum
• Returns: True if the current player can win from this state

The @cache decorator provides automatic memoization to avoid recalculating the same states.

Step 3: Try all possible moves

Inside dfs, we iterate through each number from 1 to maxChoosableInteger:

for i in range(1, maxChoosableInteger + 1):
    if mask >> i & 1 ^ 1:  # Check if number i is not used

The bit operation mask >> i & 1 ^ 1 checks if bit i is 0 (number i hasn't been used):

• mask >> i shifts the mask right by i positions
• & 1 extracts the least significant bit
• ^ 1 flips the bit (so we get True if the bit was 0)

Step 4: Check winning conditions

For each available number i, we check two winning conditions:

if s + i >= desiredTotal or not dfs(mask | 1 << i, s + i):
    return True

1. Immediate win: If s + i >= desiredTotal, choosing i immediately wins the game
2. Opponent loses: If not dfs(mask | 1 << i, s + i) returns True, it means after we choose i, the opponent will be in a losing position

The expression mask | 1 << i creates a new mask with bit i set to 1, marking number i as used.

Step 5: Return the result

• If any move leads to a win, we return True
• If no move leads to a win after trying all possibilities, we return False

The initial call dfs(0, 0) starts with no numbers used (mask = 0) and sum = 0.

Time Complexity: O(2^n * n) where n = maxChoosableInteger, as there are 2^n possible states and each state requires checking up to n numbers.

Space Complexity: O(2^n) for the memoization cache.

Example Walkthrough

Let's walk through a small example with maxChoosableInteger = 4 and desiredTotal = 6.

Initial Setup:

• Available numbers: {1, 2, 3, 4}
• Sum of all numbers: 1 + 2 + 3 + 4 = 10 ≥ 6 (game is winnable)
• Starting state: mask = 0000 (binary), sum = 0

Player 1's Turn (First Move): Player 1 evaluates each possible choice:

1. Choose 1: New sum = 1, mask = 0001

   • Player 2 could then choose 4 → sum becomes 5
   • Player 1 would need to win from (mask=1001, sum=5)
   • Any remaining choice (2 or 3) would make sum ≥ 6, so Player 1 wins
   • This path eventually leads to Player 1 winning ✓

2. Choose 2: New sum = 2, mask = 0010

   • Player 2 could choose 4 → sum becomes 6 (Player 2 wins immediately) ✗

3. Choose 3: New sum = 3, mask = 0100

   • Player 2 could choose 4 → sum becomes 7 (Player 2 wins immediately) ✗

4. Choose 4: New sum = 4, mask = 1000

   • Player 2 could choose 2 → sum becomes 6 (Player 2 wins immediately) ✗

Analysis: Player 1 can force a win by choosing 1. Here's why:

• After Player 1 chooses 1 (sum = 1), Player 2 has three options:
  • Choose 2 (sum = 3): Player 1 chooses 3 or 4 and wins
  • Choose 3 (sum = 4): Player 1 chooses 2 or 4 and wins
  • Choose 4 (sum = 5): Player 1 chooses 2 or 3 and wins

Since Player 1 has at least one move (choosing 1) that guarantees a win regardless of Player 2's response, the function returns true.

How the Algorithm Works: The recursive function dfs(mask, s) explores this game tree:

• At each state, it tries all unused numbers
• If any choice leads to immediate victory (sum ≥ 6) or puts the opponent in a losing position, it returns True
• The memoization prevents re-evaluating identical game states (same used numbers and sum)
• The initial call dfs(0, 0) returns True because Player 1 can force a win by choosing 1

Solution Implementation

Time and Space Complexity

The time complexity is O(2^n × n), where n is maxChoosableInteger.

The function uses memoization with @cache decorator to store results for different game states. Each state is represented by a bitmask that indicates which numbers have been used. There are 2^n possible states (each number from 1 to n can be either used or unused). For each state, the function iterates through all n numbers to check if they can be chosen, resulting in O(n) work per state. Therefore, the total time complexity is O(2^n × n).

The space complexity is O(2^n), where n is maxChoosableInteger.

The space is dominated by the memoization cache, which stores results for up to 2^n different states (all possible combinations of used/unused numbers). Additionally, the recursion depth can be at most n (when all numbers are chosen one by one), contributing O(n) to the call stack. Since O(2^n) + O(n) = O(2^n), the overall space complexity is O(2^n).

Common Pitfalls

1. Forgetting the Early Termination Check

Pitfall: Not checking if it's mathematically possible to reach the desiredTotal with all available numbers combined. Without this check, the algorithm will unnecessarily explore all game states even when nobody can win.

Problem Example:

# If maxChoosableInteger = 5, desiredTotal = 100
# Sum of 1+2+3+4+5 = 15, which is less than 100
# The game will never end, but the algorithm might still try to compute states

Solution: Always include the early termination check before starting the recursive computation:

total_available_sum = (1 + maxChoosableInteger) * maxChoosableInteger // 2
if total_available_sum < desiredTotal:
    return False

2. Incorrect Bit Manipulation for Checking Used Numbers

Pitfall: Using wrong bit operations to check if a number has been used. A common mistake is confusing the bit positions or using incorrect operators.

Incorrect implementations:

# Wrong: Using bit position 0 for number 1 (off-by-one error)
if (used_numbers_mask >> (number - 1)) & 1 == 0:  # WRONG!

# Wrong: Checking if bit is set instead of unset
if (used_numbers_mask >> number) & 1 == 1:  # WRONG! This checks if number IS used

# Wrong: Not properly isolating the bit
if used_numbers_mask >> number == 0:  # WRONG! This checks all bits from position number onwards

Solution: Use the correct bit manipulation pattern:

# Correct: Check if bit at position 'number' is 0 (number hasn't been used)
if (used_numbers_mask >> number) & 1 == 0:
    # Number is available to use

3. Not Using Memoization or Using It Incorrectly

Pitfall: Implementing the solution without memoization leads to exponential time complexity with massive redundant calculations. The same game state (same used numbers and sum) can be reached through different move sequences.

Problem Example:

# Without memoization - this will time out for larger inputs
def can_win_from_state(used_numbers_mask, current_sum):
    # ... recursive logic without caching ...
  
# Or incorrect memoization key (forgetting current_sum)
memo = {}
def can_win_from_state(used_numbers_mask, current_sum):
    if used_numbers_mask in memo:  # WRONG! Ignores current_sum
        return memo[used_numbers_mask]

Solution: Use proper memoization with both state parameters:

from functools import cache

@cache
def can_win_from_state(used_numbers_mask: int, current_sum: int) -> bool:
    # The @cache decorator automatically handles memoization with both parameters

Or manually with a dictionary:

memo = {}
def can_win_from_state(used_numbers_mask, current_sum):
    key = (used_numbers_mask, current_sum)
    if key in memo:
        return memo[key]
    # ... compute result ...
    memo[key] = result
    return result

4. Misunderstanding the Win Condition Logic

Pitfall: Incorrectly implementing the logic for determining when the current player wins. A common mistake is only checking for immediate wins and forgetting about forcing the opponent into a losing position.

Incorrect implementation:

# Wrong: Only checking for immediate win
if new_sum >= desiredTotal:
    return True
# Forgot to check if opponent loses!

# Wrong: Incorrect opponent check
if can_win_from_state(new_mask, new_sum):  # WRONG! Should be NOT
    return True

Solution: Check both winning conditions:

if new_sum >= desiredTotal:  # Immediate win
    return True

if not can_win_from_state(new_mask, new_sum):  # Opponent loses
    return True

5. Edge Case: desiredTotal ≤ 0

Pitfall: Not handling the edge case where desiredTotal is 0 or negative. The first player would win immediately without choosing any number.

Solution: Add a check at the beginning:

if desiredTotal <= 0:
    return True  # First player wins immediately