If there are even number of towers (say 2), the second player can copy the optimal strategy of the first player. Eventually, the first player will run out of moves. P2 Wins.

If there are odd number of towers, Player 1 can reduce the last tower to 1 in a single move(last move), leaving P2 without a move (post the copying of optimal strategy by P2). P1 Wins

If there is only 1 block, no move is possible. P2 Wins

The logic of my solution

-If the number of towers is even, it is always the second player who wins, since they always play the best game

-Likewise, if the number of towers is odd, the first player will always win

-But in the case that the height of the towers is equal to 1 from the beginning, the first player will lose before he plays because he does not have any moves to play

Python Solution

def towerBreakers(n, m):
    if m == 1:
        return 2
    elif (n%2 == 0):
        return 2
    else:
        return 1

how to know who has how many moves ?

Assume both players always play optimally.

Java 7 solution below...

public static int towerBreakers(int n, int m) {

    int winner = 0;

    if(n%2 == 1){
        if(m > 1){
            winner = 1;
        }else{
            winner = 2;
        }
    }else{
        winner = 2;
    }

    return winner;



}

}

Tip for anyone stuck on this: don't just look at the first example, theres a second one as sample input at the bottom. Playing optimally doesn't necessarily mean using the biggest factor