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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 java.util.*;
2
3public class Solution {
4 public boolean winnerSquareGame(int n) {
5 boolean[] dp = new boolean[n + 1];
6 for (int i = 1; i <= n; i++) {
7 for (int k = 1; k * k <= i; k++) {
8 if (!dp[i - k * k]) {
9 dp[i] = true;
10 break;
11 }
12 }
13 }
14 return dp[n];
15 }
16}A boolean array is used to keep track of winning positions. For each number of stones, we check if removing any perfect square leaves a losing state for the opponent.
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.