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Challenge

You will be given a single 2D boolean array $M$. You may take its dimensions as a separate argument if needed.

Find the longest parallel lines in the matrix. Parallel lines can be vertical, horizontal, or diagonal, and there can't be a 1 anywhere between the two lines. (Imagine 0s are air, and the 1s are solid, the two lines have to be able to 'see' each other) For example, in the matrix

0 1 1 1 0 0
0 0 0 0 0 0
0 1 1 1 1 0
0 0 1 1 0 0
0 0 0 0 0 0


the longest parallel line is length 3. If there are no parallel lines, output 0.

Test Cases

I:
0 0 0 1
0 0 0 0
0 0 0 0
0 1 0 0
O: 0
I:
0 0 1 0 1
0 1 0 1 0
1 0 1 0 0
O: 3
I:
0 0 1 0 1
0 0 1 0 1
0 0 1 1 1
0 0 1 0 1
0 0 1 0 1
O: 2
(The 1 in the middle is obstructing the line of sight)

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1 comment

You may want to add some testcases with broken lines i.e. 5 5 ⍴ 1 0 1 1 0 0 Razetime‭ 3 months ago

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JavaScript (Node.js), 541 540 bytes

for(_='=>)==!(+==0h,.map(M,R(=1,,e,b	&&Nl=[-1,))s(n,=(a	a=u,v(i=g(z,u,f=(a,w,cs=Math.max,r?[...r(--a	)	(a)]:[],R>&b>&a<w&b<g=(f	,c,d(v(i(fc	d++f	)?a[b][f]:e)),v=fr(w+s).some(f),na(c,uc(f,z!f||(M1])olNO(oO)|(o--O(Po,l(Qx=z,yxN,yOx,y)a[y][x]&g(x,y,o,l)PQ++,n=s(n,Q,m1,0||(d0,1),n0,k=1]p[dp),1]q(jt=ze++,tpt)a[e][t]&g(t,q,!q)(q?m:d)++j,k=s(k,jkcf(r(w,gr(fa[f][g],w,0';G=/[-]/.exec(_);)with(_.split(G))_=join(shift());eval(_)


Try it online!

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3 comments

Wow! Did not expect this one to be answered so quickly. Unfortunately it seems to fail for [[0, 1, 0], [1, 0, 0], [0, 0, 0]], two lines of length 1 that are parallel diagonally. rak1507‭ 3 months ago

@rak1507 I updated the answer to cover the [[0, 1, 0], [1, 0, 0], [0, 0, 0]] case, but I think you need to provide a better definition of what you mean by "diagonal". From your examples I thought that that particular test case that you show should return 0 and not 1. You probably should provide more test cases and/or a reference implementation. Hakerh400‭ 3 months ago

Yeah, my bad, I realise now it wasn't fully thought through. rak1507‭ 3 months ago

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