Challenges

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# Note:

This challenge was underspecified and bad, and as such I would not encourage answering it in the future.

# 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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# 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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