notes

Scenarios we will using Binary Search:

1) Array are sorted => in this case Binary Search works like magic
{BINARY SEARCH CAN ALSO BE DONE ON UNSORTED ARRAY IF IT WILL SATISFY SECOND POINT,
NOT MANDATORY FOR THE ARRAY TO BE SORTED}

2) Eliminating or Reducing the search space by half {This eliminating half is done in
   every iteration}

So, if we are able to eliminate half of input in each and every iteration OR
we are able to increase the search space by twice, in both the scenarios, we can go
for binary search.

Eg:
64, 32, 16, 8, 4, 2, 1 => Reducing input size by half in each and every iteration

OR

1, 2, 4, 8, 16, 32, 64 => If I am able to increase the input size by twice to search
for something

then this is also a binary search

Time Complexity:
log₂(n) => log₂(64) => log₂(2)^6 => 6 iterations to identify the target
=========

Logical Operators

0 && 1 = 0 {We will not be checking after 0}
1 && 0 = 0 {check both sides}
1 && 1 = 1 {check both sides}
0 && 1 = 0 {We will not be checking after 0}

0 || 0 = 0 {check both sides}
1 || 0 = 1 {We will not be checking after 1}
0 || 1 = 1 {check both sides}
1 || 1 = 1 {We will not be checking after 1}