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CheezItMan
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May 12, 2019
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This definitely works and with the space & time complexity you listed. There is a counterintuitive way to solve the problem without the sets you used, instead using the 1st row and 1st column to store the row and columns to zero out. However this is a good solution. Nice work.
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| row_set.each_with_index do |k| | ||
| for l in (0...matrix[k].length) |
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Two questions:
1). Why use .each_with_index for the outer loop if you don't end up using the index variable.
2). Why use a for-in loop here if you do an each with index for the outer loop?
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