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J, 9 bytes 1#.]=&><\ This is the 7 byte APL solve but makes use of #. in place of +/"1. I came up with 1#.]=[\ first, but bubbler pointed out it breaks when non zeros are present. At...
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#1: Initial revision
by
south
·
2022-07-19T04:37:38Z (over 2 years ago)
# [J](https://jsoftware.com/#/), 9 bytes ```J 1#.]=&><\ ``` This is the 7 byte APL solve but makes use of `#.` in place of `+/"1`. I came up with `1#.]=[\` first, but [bubbler](https://codegolf.codidact.com/users/55472) pointed out it breaks when non zeros are present. [Attempt it online!](https://ato.pxeger.com/run?1=m70wa8GCpaUlaboWNx3SbK0MlfVibdXsbGK4UpMz8hXSlAwVohUMgdAIDA3B0FjBWMfKWMFcwVTBRMESKGwMxGYKZhBzoMbBjAUA)