Formatted question description: https://leetcode.ca/all/1510.html

# 1510. Stone Game IV (Hard)

Alice and Bob take turns playing a game, with Alice starting first.

Initially, there are n stones in a pile.  On each player's turn, that player makes a move consisting of removing any non-zero square number of stones in the pile.

Also, if a player cannot make a move, he/she loses the game.

Given a positive integer n. Return True if and only if Alice wins the game otherwise return False, assuming both players play optimally.

Example 1:

Input: n = 1
Output: true
Explanation: Alice can remove 1 stone winning the game because Bob doesn't have any moves.

Example 2:

Input: n = 2
Output: false
Explanation: Alice can only remove 1 stone, after that Bob removes the last one winning the game (2 -> 1 -> 0).

Example 3:

Input: n = 4
Output: true
Explanation: n is already a perfect square, Alice can win with one move, removing 4 stones (4 -> 0).


Example 4:

Input: n = 7
Output: false
Explanation: Alice can't win the game if Bob plays optimally.
If Alice starts removing 4 stones, Bob will remove 1 stone then Alice should remove only 1 stone and finally Bob removes the last one (7 -> 3 -> 2 -> 1 -> 0).
If Alice starts removing 1 stone, Bob will remove 4 stones then Alice only can remove 1 stone and finally Bob removes the last one (7 -> 6 -> 2 -> 1 -> 0).

Example 5:

Input: n = 17
Output: false
Explanation: Alice can't win the game if Bob plays optimally.


Constraints:

• 1 <= n <= 10^5

Related Topics:
Dynamic Programming

## Solution 1. Bottom-up DP

Let dp[i] be whether Alice and win starting with i stones.

dp[i] = true    // If any dp[i - j * j] is false
= false   // otherwise
where 1 <= j * j <= i  -- the stone taken by Bob.

// OJ: https://leetcode.com/problems/stone-game-iv/

// Time: O(N * sqrt(N))
// Space: O(N)
class Solution {
public:
bool winnerSquareGame(int n) {
vector<bool> dp(n + 1);
for (int i = 1; i <= n; ++i) {
for (int j = 1; j * j <= i && !dp[i]; ++j) dp[i] = !dp[i - j * j];
}
return dp[n];
}
};


Or use static variable to save computation

// OJ: https://leetcode.com/problems/stone-game-iv/

// Time: O(N * sqrt(N))
// Space: O(N)
class Solution {
public:
bool winnerSquareGame(int n) {
static vector<bool> dp(1);
for (int i = dp.size(); i <= n; ++i) {
dp.push_back(false);
for (int j = 1; j * j <= i && !dp[i]; ++j) dp[i] = !dp[i - j * j];
}
return dp[n];
}
};


## Solution 2. Top-down DP

// OJ: https://leetcode.com/problems/stone-game-iv/

// Time: O(N * sqrt(N))
// Space: O(N)
class Solution {
vector<int> dp; // -1 unvisited, 0 lose, 1 win
bool dfs(int n) {
if (n == 0) return false;
if (dp[n] != -1) return dp[n];
for (int i = 1; i * i <= n && dp[n] != 1; ++i) dp[n] = !dfs(n - i * i);
return dp[n];
}
public:
bool winnerSquareGame(int n) {
dp.assign(n + 1, -1);
return dfs(n);
}
};


Java

class Solution {
public boolean winnerSquareGame(int n) {
boolean[] dp = new boolean[n + 1];
dp = false;
int sqrt = (int) Math.sqrt(n);
for (int i = 1; i <= sqrt; i++)
dp[i * i] = true;
for (int i = 2; i <= n; i++) {
if (!dp[i]) {
int curSqrt = (int) Math.sqrt(i);
boolean flag = false;
for (int j = 1; j <= curSqrt; j++) {
if (!dp[i - j * j]) {
flag = true;
break;
}
}
if (flag)
dp[i] = true;
else
dp[i] = false;
}
}
return dp[n];
}
}