Binary search steps

#1

why binary search needs log n steps

#2

Hey @manish.bharadwajsbh
In Binary Search we first find mid and then after comparison we either discard first half or second half of array
So lets assume we start from array of size n
then
n -> n/2 -> n/4 -> … ->1
Now assume it took k steps to reach 1 i.e worst case
then
n/2^k == 1
=> n== 2^k
=> logn=k
hence Binary search take logn time in worst case

#3

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