Winner of the game - algorithm

Question

I came across this question in recent hiring challenge :
Given N strings, each of '0'-'9' digits and an integer K. Two players A and B play a game as follows :
A plays first and picks any string of first K strings.
Next player, B is allowed to pick string starting with digit equal to last digit in string A picked in his last turn and among next K strings to last picked string by A. e.g. if A picked "045679" indexed i, then B can only pick string starting with digit '9' from strings from indices i+1 to i+K.
The one who can't pick a string loses
Both players play optimally.
We need to tell the winner and the least index of string picked up by winner.
I have seen such problems before and need a suggestion to how to think of such problems. Thanks in advance.

Answer 1

This is similar to the game of nim. You have to go from a winning position into a losing position in order to still win.
We can go backwards: if the last player took the last string, then the player to move now is losing, if the 2nd last string was picked, then the player to move is losing, unless the last digit of 2nd last string and starting digit of last string match, and so on.
With correct implementation you can get O(n) time complexity and O(k) space complexity, because you only need to keep track of a window of size k. Something like this:
// return a winning first move or -1 if there is none
public int solve(String[] strings, int k) {
Queue<Boolean> q = new Deque<>();
int[] lossCount = new int[10];
for (int i = strings.length - 1; i >= 0; i--) {
int start = strings[i].charAt(0) - '0';
int end = strings[i].charAt(strings[i].length() - 1) - '0';
if (i + k + 1 < strings.length && !q.remove())
lossCount[strings[i + k + 1].charAt(0) - '0']--;
boolean win = lossCount[end] == 0;
if (!win)
lossCount[start]++;
else if (i < k)
return i;
q.add(win);
}
return -1;
}