### PROBLEM LINKS

### DIFFICULTY

EASY

### EXPLANATION

This game is a modification to the game of Kayles. Like other impartial games, this problem can be solved using Sprague-Grundy theorem.

Let’s define a game as a substring of S. For convenience, we refer to a substring by its left and right index S(i, j). Note that after a player erases a substring S(a, b), in S(i, j), the game is split into two subgames, S(i, a-1) and S(b+1, j). Because a player can choose any game in his/her turn, the Grundy number of the game is simply the XOR of the Grundy numbers of the two subgames.

Let G(i, j) be the Grundy number of the substring S(i, j). Then G(i, j) = mex({G(i, a-1) XOR G(b+1, j) : S(a, b) exists in the dictionary and is a substring of S(i, j)}

A game is losing iff its Grundy number is zero. So, Teddy, the first player, will win if G(1, |S|) is not zero. On the other hand, Tracy will win if G(1, |S|) is zero.

### SETTER’S SOLUTION

Can be found here.

### TESTER’S SOLUTION

Can be found here.