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2-liner: numpy rot90 * 3 & enumerate the indexing solution in 3rd party category for The Highest Building by Stensen
import numpy as np, operator as o
highest_building = lambda B: max([[i+1, np.sum(b)] for i, b in enumerate(np.rot90(B, 3))], key=o.itemgetter(1))
Oct. 9, 2020
