Write a function:

def solution(A, B, K)

that, given three integers A, B and K, returns the number of integers within the range [A..B] that are divisible by K, i.e.:

{ i : A ≤ i ≤ B, i mod K = 0 }

For example, for A = 6, B = 11 and K = 2, your function should return 3, because there are three numbers divisible by 2 within the range [6..11], namely 6, 8 and 10.

Assume that:

  • A and B are integers within the range [0..2,000,000,000];
  • K is an integer within the range [1..2,000,000,000];
  • A ≤ B.


  • expected worst-case time complexity is O(1);
  • expected worst-case space complexity is O(1).


def solution(A, B, K):
    a = A//K
    b = B//K
    if A%K == 0: a -= 1
    return b-a

Ok este problema esta fácil, lo único es que tiene que ser 0(1) y que estaba corriendo el script en python 3 y estaba dividiendo con “/” y no con “//”.

  • Primero sacamos la cantidad de números divisibles entre K de 0 a A
  • Y luego de 0 a B
  • En la tercera linea chacemos si A es divisible si es así, le restamos 1 a “a”.
  • Y por ultimo sacamos la diferencia de b entre a y listo

Aquí esta el repo con todas mis soluciones de codility:

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