Program for Square Root of Integer

Given a positive integer n, find its square root. If n is not a perfect square, then return floor of √n.

Examples :

Input: n = 4
Output: 2
Explanation: The square root of 4 is 2.

Input: n = 11
Output: 3
Explanation: The square root of 11 lies in between 3 and 4 so floor of the square root is 3.

[Naive Approach] Using a loop - O(sqrt(n)) Time and O(1) Space

Start from 1 and square each number until the square exceeds the given number. The last number whose square is less than or equal to n is the answer.

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[Expected Approach] Using Binary Search - O(log(n)) Time and O(1) Space

The square of a number increases as the number increases, so the square root of n must lie in a sorted (monotonic) range.
If a number's square is more than n, the square root must be smaller.
If it's less than or equal to n, the square root could be that number or greater.
Because of this pattern, we can apply binary search in the range 1 to n to efficiently find the square root.

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[Alternate Approach] Using Built In functions - O(log(n)) Time and O(1) Space

 We can directly use built in functions to find square root of an integer.

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[Alternate Approach] Using Formula Used by Pocket Calculators - O(1) Time and O(1) Space

The idea is to use mathematical formula √n = e^{1/2 × log(n)} to compute the square root of an integer n.

Let's say square root of n is x:
=> x = √n
Squaring both the sides:
=> x² =n
Taking log on both the sides:
=> log(x²) = log(n)
=> 2 × log(x) = log(n)
=> log(x) = 1/2 × log(n)
To isolate x, exponentiate both sides with base e:
=> x = e^{1/2 * log(n)}
x is the square root of n:
So, √n = e^{1/2 × log(n)}

Because of the way computations are done in computers in case of decimals, the result from the expression may be slightly less than the actual square root. Therefore, we will also consider the next integer after the calculated result as a potential answer.

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