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Unicode-Find algorithm solution in Creative category for Count Chains by StefanPochmann
def count_chains(circles):
chains = ''.join(map(chr, range(len(circles))))
for i, (x, y, r) in enumerate(circles):
for j, (X, Y, R) in enumerate(circles):
d2 = (X-x)**2 + (Y-y)**2
if (R-r)**2 < d2 < (R+r)**2:
chains = chains.replace(chains[i], chains[j])
return len(set(chains))
March 28, 2024
