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Steps 2 through to 11 are essentially the same step repeated, just cleaning a different tooth every time. Iteration can be used to greatly simplify the algorithm. Look at this alternative: put ...

# longest_increasing_subsequence[prev] + array[i] if the length is longer.

# - are vectorized to work elementwise on arrays, # - implement non-trivial, iterative algorithms with a callback interface, and # - return rich objects with iteration count, termination status, etc. ...