Not able to understand the approach

#1

Actually i am not able to clearly understand the approach

#2

hello @priyansh781

Let’s take some examples and see how we can simplify the condition.

Original sorted array
[1, 2, 3, 4, 5, 6, 7]

After rotation, it might be something like
[3, 4, 5, 6, 7, 1, 2]
[6, 7, 1, 2, 3, 4, 5]
[1, 2, 3, 4, 5, 6, 7] <-- rotated and end up the same
and etc…

When you divide the rotated array into two halves, using mid index, at least one of subarray should remain sorted ALWAYS .

[3, 4, 5, 6, 7, 1, 2]
-> [3, 4, 5] [ 6, 7, 1, 2]
the left side remains sorted

[6, 7, 1, 2, 3, 4, 5]
-> [6, 7, 1] [2, 3, 4, 5]
the right side remains sorted

[1, 2, 3, 4, 5, 6, 7]
-> [1, 2, 3] [4, 5, 6, 7]
Both sides remain sorted.

If you know one side is sorted, the rest of logic becomes very simple.
If one side is sorted, check if the target is in the boundary, otherwise it’s on the other side.

IF smallest <= target <= biggest
  then target is here
ELSE
  then target is on the other side