Solution to Min-Perimeter-Rectangle by codility

26 Jan

January 26, 2014 Sheng 3

Question: http://codility.com/demo/take-sample-test/min_perimeter_rectangle

Question Name: MinPerimeterRectangle

Let the length of one side be len_1, and the length of one adjacent side be len_2. For a rectangle with a constant area, the perimeter is minimized when the difference between len_1 and len_2, abs(len_1- len_2), is minimized.

def solution(N):

from math import sqrt

for i in xrange(int(sqrt(N)), 0, -1):

if N % i == 0:

return 2*(i+N/i)

3 Replies to “Solution to Min-Perimeter-Rectangle by codility”

micropentium6 says:
December 16, 2015 at 7:58 pm
Great observation! min(abs(len1,len2))! I didn’t notice it at all!
I wonder what the time complexity of sqrt is? 🙂
Reply
- Sheng says:
  December 16, 2015 at 11:48 pm
  sqrt takes much time. But it is still O(1) or O(logN).
  Reference:
  http://stackoverflow.com/questions/6884359/c-practical-computational-complexity-of-cmath-sqrt
  http://stackoverflow.com/questions/28815339/time-complexity-of-math-sqrt-java
  Reply

100%, 100% java

public int solution(int N) {

Integer x = ((Double)Math.sqrt(N)).intValue();

int y = 0 ;

for(int i = x ; x>0 ; x--){

if(N%x == 0){

y = N/x;

break;

}

return 2*(x+y) ;

}

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