Black&White Ring Game CodeChef Solution

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Black&White Ring Game CodeChef Solution

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Problem

Alice and Bob are playing a game. Alice goes first.

They have a binary circular array $A$ with $N$ elements. In one move, a player can:

Choose a circular continuous segment of the array and perform a left cyclic shift on the chosen segment.

We define the term $d i ff (A)$ as the number of pairs of adjacent elements with different values. Note that we also consider $A_N$ and $A_1$ as adjacent.

A player can make a move only if, for the updated array $A^{'}$ , $diff(A’)\gt diff(A)$ .

The player who is unable to make a move, loses. Find the winner of the game if both players play optimally.

Note:

For an array $A$ with $N$ elements, some circular continuous segments are $[A_1], [A_1, A_2, A_3], [A_{N-1}, A_N, A_1, A_2]$ , etc.
A left cyclic shift of a segment $[A_1, A_2, A_3]$ is $[A_2, A_3, A_1]$ . Similarly, left cyclic shift of segment $[A_{N-1}, A_N, A_1, A_2]$ is $[A_N, A_1, A_2, A_{N-1}]$ .

Input Format

The first line of input contains a single integer $T$ , the number of test cases. The description of the test cases follows.
Each test cases consists of two lines of input.
- The first line of each test case contains a single integer $N$ , the number of elements.
- The second line of each test case contains $N$ space-separated integers $A_1,A_2, \ldots, A_N$ .

Output Format

For each test case, if Alice will win print Alice, otherwise print Bob.

You may print each character in Uppercase or Lowercase. For example: BOB, bob, boB, and Bob are all identical.

Constraints

$\leq T \leq 100$
$\leq N \leq 10^6$
$\leq A_i \leq 1$
The sum of $N$ over all test cases won’t exceed $10^6$ .

Sample 1:

Input

Output

3
3
1 0 1
4
1 1 0 0
8
1 0 1 0 0 0 1 1

Bob
Alice
Bob

Explanation:

Test case $1$ : Alice can not perform any move. Thus, Bob wins.

Test case $2$ : Alice can perform the following move:

Choose the segment $[A_1,A_2,A_3]$ and perform left cyclic shift. Thus, $A$ becomes $[1, 0, 1, 0]$ .
Also, $d i ff ([1, 1, 0, 0]) = 2$ as pairs of adjacent elements with different values are $(A_2, A_3)$ and $(A_4, A_1)$ . After this move, $d i ff ([1, 0, 1, 0]) = 4$ as all pairs of adjacent elements have different values. Since $\gt diff([1, 1, 0, 0])$ , this move is valid.

After performing the move, Bob has no valid move left. Thus, Alice wins.

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