Another solution:
is_close_to_n = lambda x: n-10 <= x <= n+10
result = filter(is_close_to_n, N)
Generalization bits:
def is_close_to(n):
f = lambda x: n-10 <= x <= n+10
return f
result12 = filter(is_close_to(12), N)
result123 = filter(is_close_to(123), N)
Do not sort. Sorting is, generally speaking, O (n log n); brute force search is O (n).
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