1 line 0 ms C++ solution with explanation

  • 30


    At first this problem might seems a bit tough but it is easy and has a pattern which is as follow.

    I have applied the bottom up dynamic programming approach to fill the array and noticed that only number divisible by 4 are the positions where player1(playing first chance) is losing.

    class Solution {
        bool canWinNim(int n) {
            return n%4 ;
    1. Base case :

      If the numbers of stones are 1,2 or 3, then player 1 will win.

      If the numbers of stones are 4, then player 1 will lose irrespective of the number of stones he/she remove

      So lookup table will look like this : W[1]->W[2]->W[3]->L[4].
    2. For num_stones=5, the player can either remove 1,2 or 3 stones i.e. the other player (player 2) will win if the number of stones left are 1,2 or 3 and will lose only when the number of stones left are 4 ( see the lookup table in step 1) .
      So, if Player1 remove 1 stone, the number of stones left will be 4, which will defeat player2. So, now the lookup entry for num_stones=5 will be W.
      Lookup now will look like this : W->W->W->L->W (for player 1-> who is taking the first chance).
    3. Likewise, we can fill the complete lookup table by looking at the values at last three index. If anyone of them is L => Player 1 will win the game as he will remove only that many number of stones which will bring player 2 to the L position
    4. In the end, you will notice that only positions 4->8->12->16 will contain L for player 1 thus answer is simple n%4.

  • 5

    Great explanation! Thanks a lot!

Log in to reply

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.