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Leetcode 292 Nim Game.

Leetcode 292 Nim Game

You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove the stones.

Both of you are very clever and have optimal strategies for the game. Write a function to determine whether you can win the game given the number of stones in the heap.

For example, if there are 4 stones in the heap, then you will never win the game: no matter 1, 2, or 3 stones you remove, the last stone will always be removed by your friend.

Hint:

If there are 5 stones in the heap, could you figure out a way to remove the stones such that you will always be the winner?


My First thought was use Recursive, the code is below, but the problem is exceeded the time limit.


public class Solution {
    public boolean canWinNim(int n) {
        if(n==1||n==2||n==3){
            return true;
        }
        if(n==4){
            return false;
        }
        if(n==5){
            return true;
        }
        return (canWinNim(n-1)||canWinNim(n-2)||canWinNim(n-3));
    }
}

So I find a very good solution with full prove from the discussion part of the leetcode.

Here is the Prove:

Proof:

the base case: when n = 4, as suggested by the hint from the problem, no matter which number that that first player, the second player would always be able to pick the remaining number.

For 1* 4 < n < 2 * 4, (n = 5, 6, 7), the first player can reduce the initial number into 4 accordingly, which will leave the death number 4 to the second player. i.e. The numbers 5, 6, 7 are winning numbers for any player who got it first.

Now to the beginning of the next cycle, n = 8, no matter which number that the first player picks, it would always leave the winning numbers (5, 6, 7) to the second player. Therefore, 8 % 4 == 0, again is a death number.

Following the second case, for numbers between (24 = 8) and (34=12), which are 9, 10, 11, are winning numbers for the first player again, because the first player can always reduce the number into the death number 8.

Then the code should be very simple:


public class Solution {
    public boolean canWinNim(int n) {
        return n%4!=0;
    }
}
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