Play with friend Complete Solution

This article is part of the Scaler Topics Fortnightly Contest - 19

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Solution Approach

We'll use the concepts of Grundy numbers and Sprague-Grundy's Theorem in this solution.
The idea is that every game state can be assigned an integer number, and if there are many elements of a game, then the value assigned to that total game state is the xor of the values of each element individually.
The Grundy number of a state is the minimum number that is not achieved among any state that the state can move to.
we have to separate the problem into 2 cases, B even and odd.
Let f(n) denote the Grundy number of a element of value n. By definition f(0)=0.
If B is even, then when you split the element of value 2n into B elements of value n, the resulting Grundy number of that state is (f(n) xor'ed n times = 0).
as B is even. Given this, it is easy to compute that f(0)=0,f(1)=1,f(2)=2,f(3)=0, f(4)=1. Now I will show by induction that for n>=2,f(2n-1)=0,f(2n)=1.
The case where B is odd is similar but requires more work. Let's look at the splitting operation first. This time, from a element of value 2n we can move to B element of value n, with Grundy number (f(n) xor'ed n times = f(n)) as B is odd.
So from 2n we can achieve the Grundy numbers f(2n-1) and f(n).

Time Complexity - O(|A|*log(max(A[i])) Space Complexity - O(1)