Code Invicta Discussion

general
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Let’s make a general thread to discuss solutions and doubts of Code Invicta.

Contest Link : Contest Page | CodeChef

I was able to solve MOON and XORSEQ.

Those who have solved BCK2, can you share your approach?
Thank you!

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For BCK2:
for every edge add 2 * weight_of_edge * min_of_number_of_nodes_on_both_side_of_edge

For fairy tales:
you just have to check if that edge is a bridge or not using timestamp based dfs

How to solve the last 2 ones??

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question - moon
please give test cases where my code fails
code - CodeChef: Practical coding for everyone

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link is not working

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changed

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approach for XORSEQ

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@chaudhary_19 is it working for 1 0 2 0 ? answer should be NO.

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yeah…it is showing NO

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for XORSEQ approach ?
@chaudhary_19

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There is a pattern in consecutive numbers xor.
It’s explained really well here.

And after that, you can use binary search(upper or lower bound) to find the starting and ending consecutive range of 0’s.
I used two set’s for it, one for positions of 1 and other for positions of houses which are still 0 and in neighbourhood of 1.

Solution

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will this round effect my codechef rating?
I mean it was a rated round or not rated?

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Not rated.

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test case : 2 6 5 1
answer should be YES but your solution is getting NO.

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not able to understand your solution

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Can you explain the logic behind the conclusion of
For BCK2:
for every edge add 2 * weight_of_edge * min_of_number_of_nodes_on_both_side_of_edge

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No, the round was unrated for everyone.

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The approach is greedy.
In order to maximize my answer, for an edge i want all persons standing on nodes of one side of that edge to go towards other side of edge and vice versa. However only the minimum of both the sides of an edge can go passing that edge so taking minimum.

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Great! I got it. Thanks for sharing.

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then how will you balance the sweetness in this testcase