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stone-game-iv.py
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stone-game-iv.py
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from functools import cache
import math
class Solution:
# Time complexity: O(sqrt(n) * n)
# Space complexity: O(n)
# Iterative bottom up
# Note: Running the inner loop in reverse produces consistent shorter times
# This must be due to the fact that False value tend to be more at larger numbers
# I still don't have the intuition for why that is though, in terms of the game play
def winnerSquareGame(self, n: int) -> bool:
dp = [False] * (n+1)
for i in range(1, n+1):
dp[i] = not all(dp[i - j**2] for j in range(math.floor(i**0.5), 0, -1))
return dp[-1]
# Time complexity: O(sqrt(n) * n)
# Space complexity: O(n)
# Iterative bottom up solution
def winnerSquareGame2(self, n: int) -> bool:
dp = [False]
for i in range(1, n+1):
dp.append(not all(dp[i - j**2] for j in range(1, math.floor(i**(1/2))+1)))
return dp[-1]
# Time complexity: O(sqrt(n) * n)
# Space complexity: O(n)
# Recursive top down memoized approach
# None: Optimized compared to the further below algorithm
def winnerSquareGame3(self, n: int) -> bool:
@cache
def solve(n):
if n == 0:
return False
for i in range(1, math.floor(n**(1/2))+1):
if not solve(n-i**2):
return True
return False
return solve(n)
# Time complexity: O(sqrt(n) * n)
# Space complexity: O(n)
# Recursive approach with memoization
def winnerSquareGame4(self, n: int) -> bool:
@cache
def solve(turn, n):
if n == 0:
if turn:
return False
else:
return True
if n == 1:
if turn:
return True
else:
return False
if n == 2:
if turn:
return False
else:
return True
if turn == True: # Alice turn
for i in range(1, math.floor(n**(1/2))+1):
if solve(False, n - i*i):
return True
return False
else: # Bob turn
for i in range(1, math.floor(n**(1/2))+1):
if solve(True, n - i*i) == False:
return False
return True
return solve(True, n)