#### Task:

Your job here is to implement a method, `approx_root` in Ruby/Python/Crystal and `approxRoot` in JavaScript/CoffeeScript, that takes one argument, `n`, and returns the approximate square root of that number, rounded to the nearest hundredth and computed in the following manner.

1.      Start with `n = 213` (as an example). 
2.      To approximate the square root of n, we will first find the greatest perfect square that is below or equal to `n`. (In this example, that would be 196, or 14 squared.) We will call the square root of this number (which means sqrt 196, or 14) `base`.
3.      Then, we will take the lowest perfect square that is greater than or equal to `n`. (In this example, that would be 225, or 15 squared.)
4.      Next, subtract 196 (greatest perfect square less than or equal to `n`) from `n`. (213 - 196 = **17**) We will call this value `diff_gn`.
5.      Find the difference between the lowest perfect square greater than or equal to `n` and the greatest perfect square less than or equal to `n`. (225 – 196 = **29**) We will call this value `diff_lg`.
6.      Your final answer is `base` + (`diff_gn` / `diff_lg`). In this example: 14 + (17 / 29) which is 14.59, rounded to the nearest hundredth.

Just to clarify, if the input is a perfect square itself, you should return the exact square of the input.

In case you are curious, the approximation (computed like above) for 213 rounded to four decimal places is 14.5862. The actual square root of 213 is 14.5945. 

Inputs will always be positive whole numbers. If you are having trouble understanding it, let me know with a comment, or take a look at the second group of the example cases.

#### Some examples:

```python
approx_root(400) #=> 20
approx_root(401) #=> 
  # smallest perfect square above 401 is 441 or 21 squared
  # greatest perfect square below 401 is 400 or 20 squared
  # difference between 441 and 400 is 41
  # difference between 401 and 400 is 1
  # answer is 20 + (1 / 41) which becomes 20.02, rounded to the nearest hundredth
  # final answer = 20.02.
approx_root(2) #=>
  # smallest perfect square above 2 is 4 or 2 squared
  # greatest perfect square below 2 is 1 or 1 squared
  # difference between 4 and 1 is 3
  # difference between 2 and 1 is 1
  # answer is 1 + (1 / 3), which becomes 1.33, rounded to the nearest hundredth
  # final answer = 1.33.

# math.sqrt() isn't disabled.
```

Also check out my other creations — [Square Roots: Simplifying/Desimplifying](https://www.codewars.com/kata/square-roots-simplify-slash-desimplify/), [Square and Cubic Factors](https://www.codewars.com/kata/square-and-cubic-factors), [Keep the Order](https://www.codewars.com/kata/keep-the-order), [Naming Files](https://www.codewars.com/kata/naming-files), [Elections: Weighted Average](https://www.codewars.com/kata/elections-weighted-average), [Identify Case](https://www.codewars.com/kata/identify-case), [Split Without Loss](https://www.codewars.com/kata/split-without-loss), [Adding Fractions](https://www.codewars.com/kata/adding-fractions),
[Random Integers](https://www.codewars.com/kata/random-integers), [Implement String#transpose](https://www.codewars.com/kata/implement-string-number-transpose), [Implement Array#transpose!](https://www.codewars.com/kata/implement-array-number-transpose), [Arrays and Procs #1](https://www.codewars.com/kata/arrays-and-procs-number-1), and [Arrays and Procs #2](https://www.codewars.com/kata/arrays-and-procs-number-2).

If you notice any issues or have any suggestions/comments whatsoever, please don't hesitate to mark an issue or just comment. Thanks!

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def approx_root(n):
	

Solutions:
def approx_root(n):
	base = int(n ** 0.5)
	return round(base + (n - base ** 2) / ((base + 1) ** 2 - base ** 2), 2)


def approx_root(n):
	m = n ** 0.5
	(a, b) = [int(x) for x in [m, m + 1]]
	c = a ** 2
	return round(a + (n - c) / (b ** 2 - c), 2)


def approx_root(n):
	b = int(n ** 0.5)
	diff_gn = n - b ** 2
	diff_lg = (b + 1) ** 2 - b ** 2
	return round(b + diff_gn / diff_lg, 2)


def approx_root(n):
	import math
	s = int(math.sqrt(n))
	diff_gn = n - s ** 2
	diff_lg = (s + 1) ** 2 - s ** 2
	return round(s + diff_gn / diff_lg, 2)


approx_root = r = lambda n, i=0: n > (i + 1) ** 2 and r(n, i + 1) or round(i + (n - i * i) / (2 * i + 1), 2)


approx_root = lambda n: round(next((i + (n - i * i) / (2 * i + 1) for i in range(n) if (i + 1) ** 2 >= n)), 2)


def approx_root(n):
	for i in range(1, int(n ** 0.5) + 2):
		if i ** 2 > n:
			base = i - 1
			break
	base2 = base + 1
	diff_gn = n - base ** 2
	diff_lg = base2 ** 2 - base ** 2
	return round(base + diff_gn / diff_lg, 2)


import math

def approx_root(n):
	a = n ** 0.5
	(b, c) = (math.floor(a) ** 2, (math.floor(a) + 1) ** 2)
	return a if type(a) == int else round(b ** 0.5 + (n - b) / (c - b), 2)


def approx_root(n):
	b = int(n ** 0.5)
	return round(b + (n - b * b) / ((b + 1) ** 2 - b * b), 2)