Problem 42290. GJam 2015 Rd1B: Noisy Neighbors

Created by Richard Zapor

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This Challenge is derived from <a href = "https://code.google.com/codejam/contest/8224486/dashboard#s=p1">GJam 2015 Rd 1B: Noisy Neighbors</a>. Fastest completion - 8 minutes.Determine minimum number of adjacencies for N placed people in an RxC hotel matrixInput: m, an RxC zeros array; N number of rooms to be filledOutput: NN, minimum number of common wallsExamples: Small Case 1&lt;=R*C&lt;=16, 0&lt;=N&lt;=R*C<pre class="language-matlab">[1 1 1;1 0 1;1 1 1] has minimum 8 common walls vs [1 1 1;1 1 1;1 1 0] has 10 common
[1;0;0;1] has 0 common walls
ones(2,3) has 7 common walls
</pre>Theory: The small case can be solved with brute force using vector set with nchoosek followed by processing of convolutions. The large case has 10000 rooms making brute force hopeless.Additional GJam solutions can be found at <a href = "http://www.go-hero.net/jam/15">Example GJam Matlab solutions</a>. Select Find Solutions, change Language to Matlab. The Test Suite, at the bottom, contains a full GJam Matlab solution.

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Last Solution submitted on Sep 18, 2020

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Rafael S.T. Vieira on 18 Sep 2020

There is no need to test all cases. The ideal case is a swiss cheese in which no guest share a wall. Just start filling rooms from there if needed (the only problems is that room filling must follow an order, which makes the code quite large (but repetitive and boring), and there are two base cases for the swiss cheese structure). Maybe it's possible to create a function and decrease code size, but unless you see it at the moment that you are coding it, it would mean loosing too much time during competition when a Ctrl+C, Ctrl+V would be faster (and better than testing all cases).

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Problem 42290. GJam 2015 Rd1B: Noisy Neighbors

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Problem 42290. GJam 2015 Rd1B: Noisy Neighbors

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