Question: procedure function(x: integer, a1, a2, an: distinct integers) n and x # ai) i:-i+1 while (i if i n then location : else location :=

procedure function(x: integer, a1, a2, an: distinct integers) n and x

procedure function(x: integer, a1, a2, an: distinct integers) n and x # ai) i:-i+1 while (i if i n then location : else location := 0 return location{location is the subscript of the term that equals x, or is 0 if x is not found) What is the best-case scenario time complexity of this algorithm? O (logn) 0 (1) 0 (n)

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