Question: In python I need a function that takes an integer and returns the absolute value of n minus the nearest [triangle number](http://en.wikipedia.org/wiki/Triangular_number) to n. The way I'm doing it now is by generating a list of all the triangle numbers up to n. 
Then using:

```
def closest(myList, myNumber):
    c = min(myList, key=lambda x:abs(x-myNumber))

```

To find the closest triangle number. 

The end result list should look like:

```
[0, 0, 1 , 0 , 1 ,1 , 0 , 1, 2 , 1, 0 , 1, 2, 2, 1, 0, etc.]

```

If you have another method of generating the same result that's faster go for it.

Answer: 
In my solution I have used the fact that 
let K = sqrt(2\*N)
 if N is a triangular number then it will be Kth triangular number
We know Kth triangular number will be K\*(K+1)/2

Here is my solution in Python

```
from math import sqrt
from math import floor
def close_triangle(n):
m = floor(sqrt(2*n))
t1 = m*(m+1)/2       #next triangular number to n
m = m-1
t2 = m*(m+1)/2       #previous triangular number to n
diff1 = t1-n
diff2 = n - t2
if diff1 < diff2 :
  return t1
else:
  return t2
print "Nearest trianguler number";
print close_triangle(13)

```
May not be the best way but should be faster than your method:

```
def triangle_around(n):
    k = int((2*n)**0.5)
    p = (k**2 + k) / 2 
    while p < n:
        k, p = k + 1, p + k + 1 
    return (p - k, p)

def close_triangle(n):
    t = triangle_around(n)
    return min(t[1] - n, n - t[0])

print close_triangle(8)    # gives 2

```