# Codility – ladder

The problem:

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:

func Solution(A []int, B []int) []int

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] = 1

the function should return the sequence [5, 1, 8, 0, 1], as explained above.Assume that:

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].

Complexity:expected worst-case time complexity is O(L);

expected worst-case space complexity is O(L) (not counting the storage required for input arguments).

Solution: Screen cap of sublime text, below. Has all the code. It uses fibonacci series. It is so because no. of unique ways of climbing upto a step N, is sum of no. of unique ways of upto its two previous rungs. The code comment (in the screen grab, explains the reason.

Also, the above solution, worked (100%, Yay!) in the first try. Results below.