Leetcode #464: Can I Win
In this guide, we solve Leetcode #464 Can I Win in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
In the "100 game" two players take turns adding, to a running total, any integer from 1 to 10. The player who first causes the running total to reach or exceed 100 wins.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Bit Manipulation, Memoization, Math, Dynamic Programming, Bitmask, Game Theory
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: maxChoosableInteger = 10, desiredTotal = 11
Output: false
Explanation:
No matter which integer the first player choose, the first player will lose.
The first player can choose an integer from 1 up to 10.
If the first player choose 1, the second player can only choose integers from 2 up to 10.
The second player will win by choosing 10 and get a total = 11, which is >= desiredTotal.
Same with other integers chosen by the first player, the second player will always win.
Python Solution
class Solution:
def canIWin(self, maxChoosableInteger: int, desiredTotal: int) -> bool:
def dfs(mask: int, s: int) -> bool:
for i in range(1, maxChoosableInteger + 1):
if mask >> i & 1 ^ 1:
if s + i >= desiredTotal or not dfs(mask | 1 << i, s + i):
return True
return False
if (1 + maxChoosableInteger) * maxChoosableInteger // 2 < desiredTotal:
return False
return dfs(0, 0)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.