[Codility] Lesson-05.1: CountDiv
This post deals with the problem of counting the number of elements whose remainder is 0 when divided by a specific value in a given array.
Task Description
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.
Write an efficient algorithm for the following assumptions:
- 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.
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Key Point
- When linear searching for a long-range array, time consumption will be longer.
- After dividing 0 with any value, the result is 0.
Solution (using Python)
def solution(A, B, K):
bound_1, bound_2 = (A / K), (B / K) // 1
if(bound_1 != 0 and bound_1 % 1 == 0): bound_1 -= 1
bound_1 = bound_1 // 1
answer = max(int(bound_2 - bound_1), 0)
if(A == 0): answer += 1
return answerPlease use the above solution for reference.
I recommend you to write your own source code.
