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This approach uses Dynamic Programming to determine if Alice can win given n stones. We can maintain a boolean array dp where dp[i] represents whether Alice can win with i stones. A state is winning if there's any move that leaves the opponent in a losing state.
Time Complexity: O(n*sqrt(n)), Space Complexity: O(n)
1import math
2
3def winnerSquareGame(n):
4 dp = [False] * (n + 1)
5 for i in range(1, n + 1):
6 for k in range(1, int(math.sqrt(i)) + 1):
7 if not dp[i - k * k]:
8 dp[i] = True
9 break
10 return dp[n]We initialize a boolean list dp and iterate through each state, marking it as winning if there exists a move that forces the opponent into a losing position.
This approach involves a recursive solution with memoization to cache previously computed results. If n is already solved, we return the cached result. Otherwise, we recursively check if any move leaves the opponent in a losing position.
Time Complexity: O(n*sqrt(n)), Space Complexity: O(n)
1#include <vector>
2#include <cmath>
bool isWinning(int n, std::vector<int>& memo) {
if (n == 0) return false;
if (memo[n] != -1) return memo[n];
for (int k = 1; k * k <= n; ++k) {
if (!isWinning(n - k * k, memo)) {
return memo[n] = true;
}
}
return memo[n] = false;
}
bool winnerSquareGame(int n) {
std::vector<int> memo(n+1, -1);
return isWinning(n, memo);
}This C++ solution uses a vector to memoize results. For each position, we recursively evaluate if there are moves leading to a losing state for the opponent. If a losing position is found, it is marked as winning in the memo table.