LeetCode 464: Can I Win Solution in Python – A Step-by-Step Guide

What Is LeetCode 464: Can I Win?

In LeetCode 464: Can I Win, you’re given two integers: maxChoosableInteger (the range of numbers 1 to maxChoosableInteger) and desiredTotal (the target sum). Two players take turns picking a number from the pool without replacement, adding it to their running total. The first to reach or exceed desiredTotal wins. Your task is to determine if the first player can force a win. For example, with maxChoosableInteger = 10 and desiredTotal = 11, the first player can pick 10, forcing a win no matter what. It’s like a number-picking duel where strategy decides the victor.

Problem Statement

• Input: maxChoosableInteger (int)—range 1 to max; desiredTotal (int)—target sum.
• Output: bool—True if first player can win, False otherwise.
• Rules:
• Players alternate, pick from 1 to maxChoosableInteger, no reuse.
• First to reach or exceed desiredTotal wins.
• First player’s move.

Constraints

• 1 <= maxChoosableInteger <= 20.
• 0 <= desiredTotal <= 300.

Examples to Get Us Started

• Input: maxChoosableInteger = 10, desiredTotal = 11
• Output: True (Pick 10, win immediately).
• Input: maxChoosableInteger = 10, desiredTotal = 0
• Output: True (No moves needed).
• Input: maxChoosableInteger = 10, desiredTotal = 40
• Output: False (Max sum = 55, but strategy fails).

Understanding the Problem: Winning the Game

To solve LeetCode 464: Can I Win in Python, we need to determine if the first player can force a win by choosing numbers strategically, assuming the second player plays optimally. A naive approach—trying all sequences—is O(2^n) or worse, infeasible for maxChoosableInteger = 20. The key? Use game theory with memoization to optimize recursive choices. We’ll explore:

• Best Solution (DP with Memoization): O(2^n) time with memo, O(2^n) space—fast and optimal.
• Alternative Solution (Brute Force Recursion): O(n!) time, O(n) space—simple but slow.

Let’s dive into the DP solution—it’s the strategist’s winning playbook we need.

Best Solution: Dynamic Programming with Memoization

Why This Is the Best Solution

The DP with memoization approach is the top pick because it’s O(2^n) time (n = maxChoosableInteger) and O(2^n) space, efficiently solving the game by caching subproblem results. It uses a recursive function with a state (remaining numbers, current total) and memoizes outcomes, avoiding redundant calculations. It’s like mapping every possible move in a game tree, then reusing your notes—clever and swift!

How It Works

Here’s the strategy:

• Step 1: Check base cases:
• If desiredTotal ≤ 0, first player wins (no moves needed).
• If max sum < desiredTotal, no win possible.
• Step 2: Define recursive function can_win(choices, total, memo):
• choices: tuple of remaining numbers.
• total: current sum needed to win.
• memo: cache of (choices, total) → bool.
• Step 3: For each choice:
• Pick number, recurse with remaining choices and reduced total.
• If opponent can’t win after this move, current player wins.
• Step 4: Return True if first player can force a win.
• Why It Works:
• Memoization prunes exponential tree.
• Optimal play assumption ensures correct outcome.

Step-by-Step Example

Example: maxChoosableInteger = 4, desiredTotal = 6

• Choices: [1, 2, 3, 4], max sum = 10 ≥ 6.
• Start (total=6):
• Pick 1: total=5, opponent: [2,3,4]:
• 2: 3, [3,4] → 3: 0 (win), False.
• Pick 3: total=3, opponent: [1,2,4]:
• 1: 2, [2,4] → 2: 0 (win), False.
• Pick 4: total=2, opponent: [1,2,3]:
• 1: 1, [2,3] → no win, True (first wins).
• Result: True.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, broken down clearly:

Copy codeclass Solution:
    def canIWin(self, maxChoosableInteger: int, desiredTotal: int) -> bool:
        # Step 1: Base cases
        if desiredTotal <= 0:
            return True
        max_sum = (maxChoosableInteger * (maxChoosableInteger + 1)) // 2
        if max_sum < desiredTotal:
            return False

        # Initial choices
        choices = tuple(range(1, maxChoosableInteger + 1))
        memo = {}

        # Step 2: Recursive function
        def can_win(choices, total, memo):
            state = (choices, total)
            if state in memo:
                return memo[state]

            # Try each number
            for i, num in enumerate(choices):
                if num >= total:  # Win immediately
                    memo[state] = True
                    return True
                # Opponent's turn
                next_choices = choices[:i] + choices[i+1:]
                if not can_win(next_choices, total - num, memo):
                    memo[state] = True
                    return True

            memo[state] = False
            return False

        return can_win(choices, desiredTotal, memo)

• Line 4-8: Base cases: win if total ≤ 0, lose if max sum < target.
• Line 11-12: Set up choices and memo.
• Line 15-30: Recursive can_win:
• Check memo for cached result.
• For each choice:
• If num ≥ total, win.
• Recurse for opponent; if they lose, we win.
• Cache and return False if no win found.
• Line 32: Start recursion.
• Time Complexity: O(2^n)—subsets with memoization.
• Space Complexity: O(2^n)—memo size.

It’s like a game-winning crystal ball!

Alternative Solution: Brute Force Recursion

Why an Alternative Approach?

The brute force recursion tries all possible sequences—O(n!) time, O(n) space—without caching, exploring every game path. It’s slow but intuitive, like playing out every move in your head. Good for small n or learning!

How It Works

• Step 1: Recurse with choices and total.
• Step 2: For each move, check if opponent loses.
• Step 3: Return True if win possible.

Step-by-Step Example

Example: maxChoosableInteger = 2, desiredTotal = 3

• Choices: [1,2]:
• 1: total=2, opp: [2] → 2: 0 (win), False.
• 2: total=1, opp: [1] → 1: 0 (win), False.
• Result: True (pick 2 wins).

Code for Brute Force

Copy codeclass Solution:
    def canIWin(self, maxChoosableInteger: int, desiredTotal: int) -> bool:
        if desiredTotal <= 0:
            return True
        max_sum = (maxChoosableInteger * (maxChoosableInteger + 1)) // 2
        if max_sum < desiredTotal:
            return False

        choices = list(range(1, maxChoosableInteger + 1))

        def can_win(choices, total):
            for i, num in enumerate(choices):
                if num >= total:
                    return True
                next_choices = choices[:i] + choices[i+1:]
                if not can_win(next_choices, total - num):
                    return True
            return False

        return can_win(choices, desiredTotal)

• Time Complexity: O(n!)—factorial paths.
• Space Complexity: O(n)—recursion stack.

It’s a slow game simulator!

Comparing the Two Solutions

• DP with Memoization (Best):
• Pros: O(2^n), fast, scalable.
• Cons: O(2^n) space.
• Brute Force (Alternative):
• Pros: O(n!), simple.
• Cons: Impractical for n > 10.

DP wins for efficiency.

Edge Cases and Examples

• Input: 10, 0 → True.
• Input: 1, 2 → False.
• Input: 5, 50 → False.

DP handles all well.

Complexity Recap

• DP: Time O(2^n), Space O(2^n).
• Brute Force: Time O(n!), Space O(n).

DP’s the champ.

Key Takeaways

• DP: Memoize for speed.
• Brute Force: Try all plays.
• Python Tip: Recursion wins games—see [Python Basics](/python/basics).

Final Thoughts: Claim That Victory

LeetCode 464: Can I Win in Python is a strategic game showdown. DP with memoization is your winning move, while brute force is a slow gambit. Want more game theory? Try LeetCode 292: Nim Game or LeetCode 877: Stone Game. Ready to play? Head to Solve LeetCode 464 on LeetCode and strategize today—your coding skills are game-ready!