You have to climb up a ladder. The ladder has exactly N rungs, numbered from 1 to N. With each step, you can ascend by one or two rungs. More precisely:
- with your first step you can stand on rung 1 or 2,
- if you are on rung K, you can move to rungs K + 1 or K + 2,
- finally you have to stand on rung N.
Your task is to count the number of different ways of climbing to the top of the ladder.
For example, given N = 4, you have five different ways of climbing, ascending by:
- 1, 1, 1 and 1 rung,
- 1, 1 and 2 rungs,
- 1, 2 and 1 rung,
- 2, 1 and 1 rungs, and
- 2 and 2 rungs.
Given N = 5, you have eight different ways of climbing, ascending by:
- 1, 1, 1, 1 and 1 rung,
- 1, 1, 1 and 2 rungs,
- 1, 1, 2 and 1 rung,
- 1, 2, 1 and 1 rung,
- 1, 2 and 2 rungs,
- 2, 1, 1 and 1 rungs,
- 2, 1 and 2 rungs, and
- 2, 2 and 1 rung.
The number of different ways can be very large, so it is sufficient to return the result modulo 2P, for a given integer P.
Write a function:
def solution(A, B)
that, given two non-empty arrays A and B of L integers, returns an array consisting of L integers specifying the consecutive answers; position I should contain the number of different ways of climbing the ladder with A[I] rungs modulo 2B[I].
For example, given L = 5 and:
A[0] = 4 B[0] = 3 A[1] = 4 B[1] = 2 A[2] = 5 B[2] = 4 A[3] = 5 B[3] = 3 A[4] = 1 B[4] = 1the function should return the sequence [5, 1, 8, 0, 1], as explained above.
Write an efficient algorithm for the following assumptions:
- L is an integer within the range [1..50,000];
- each element of array A is an integer within the range [1..L];
- each element of array B is an integer within the range [1..30].