Editorial for Scary Big Chessboard

You can check if the chessboard is consistent by checking if all the black cells have same parity and all the white cells have different parity.

Let's a cell's position is (x, y), then let's define it's parity as (x+y)%2. You can easily see that, by our definition, every adjacent pair of cells will have different parity.

So the complexity for a test case is: O(Q)


    45% Solution Ratio

    NirjhorEarliest, May '17

    MarzukFastest, 0.0s

    NirjhorLightest, 131 kB

    SIR.24Shortest, 506B

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