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Was literally waiting for this!

<3

waiting...

@shayan please sir can you explain this : PROBLEM D: we are using f[] array to make the node dead and lets say we get node z and node y' for operation x then we make node y' as dead .How are we ensuring that down the line for any other x ,node y' will be not be critical node(only node forming pair with some other node divisible by operation number )??? or in simple words please prove choosing any node among two would be fine .

So if you look closely, actually the Pegion-hole principle ensures that. At every step down the line, let's say we have (x) alive nodes and at that point we will be requiring (x-1) edges. So there will surely be a pair that will be having the same modulo value (as per Pegion-hole principle). Hence it does not matter which edge we had removed previously as we are sure to find a pair at every step!

Thanks for the answering!!! But I got this answer after thinking like this see for any operation x ,I will have x+1 numbers and their remainder would be in range from 0 to x-1 and that means there will always be at least two numbers with same remainder and that's the pigeon hole principle.

waited for this , thanks brother.

Chinese video editorial by me