25. Stone Game IV.cpp

/*
Stone Game IV
=============

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

Hint #1  
Use dynamic programming to keep track of winning and losing states. Given some number of stones, Alice can win if she can force Bob onto a losing state.
*/

class Solution
{
  bool helper(int n, vector<int> &dp)
  {
    if (n == 0)
    {
      return false;
    }

    if (dp[n] != -1)
    {
      return dp[n];
    }

    for (int i = 1; i <= sqrt(n); i++)
    {
      bool x = helper(n - i * i, dp);
      if (!x)
      {
        dp[n] = 1;
        return true;
      }
    }

    dp[n] = 0;

    return false;
  }

public:
  bool winnerSquareGame(int n)
  {
    vector<int> dp(n + 1, -1);
    return helper(n, dp);
  }
};