2 is less than 2 31.
Floor sqrt 1 0 8 x 1 2.
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Well it has to be an integer.
There are lots of integers less than 2 31.
The floor function also known as the greatest integer function.
The binary search can be further optimized to start with start 0 and end x 2.
So which one do we choose.
Choose the greatest one which is 2 in.
Floor of square root of x cannot be more than x 2 when x 1.
The outer one is not redundant because the square root of a number x only results in an integer if x is a square number.
How do we define the floor of 2 31.
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Thanks to vinit for suggesting above optimization.
And it has to be less than or maybe equal to 2 31 right.