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AVAILABLE LESSONS:

Lesson 1

Iterations

Lesson 2

Arrays

Lesson 3

Time Complexity

Lesson 4

Counting Elements

Lesson 5

Prefix Sums

Lesson 6

Sorting

Lesson 7

Stacks and Queues

Lesson 8

Leader

Lesson 9

Maximum slice problem

Lesson 10

Prime and composite numbers

Lesson 11

Sieve of Eratosthenes

Lesson 12

Euclidean algorithm

Lesson 13

Fibonacci numbers

Lesson 14

Binary search algorithm

Lesson 15

Caterpillar method

Lesson 16

Greedy algorithms

Lesson 17

Dynamic programming

Lesson 90

Tasks from Indeed Prime 2015 challenge

Lesson 91

Tasks from Indeed Prime 2016 challenge

Lesson 92

Tasks from Indeed Prime 2016 College Coders challenge

Lesson 99

Future training

Programming language:
Spoken language:

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 2^{P}, for a given integer P.

Assume that the following declarations are given:

struct Results { int * C; int L; // Length of the array };

Write a function:

struct Results solution(int A[], int B[], int L);

that, given two non-empty zero-indexed 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 2^{B[I]}.

For example, given L = 5 and:

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), beyond input storage (not counting the storage required for input arguments).

Copyright 2009–2018 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

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 2^{P}, for a given integer P.

Write a function:

vector<int> solution(vector<int> &A, vector<int> &B);

that, given two non-empty zero-indexed 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 2^{B[I]}.

For example, given L = 5 and:

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), beyond input storage (not counting the storage required for input arguments).

Copyright 2009–2018 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

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 2^{P}, for a given integer P.

Write a function:

class Solution { public int[] solution(int[] A, int[] B); }

that, given two non-empty zero-indexed 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 2^{B[I]}.

For example, given L = 5 and:

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), beyond input storage (not counting the storage required for input arguments).

Copyright 2009–2018 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

- 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.

^{P}, for a given integer P.

Write a function:

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

^{B[I]}.

For example, given L = 5 and:

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);

- 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.

^{P}, for a given integer P.

Write a function:

class Solution { public int[] solution(int[] A, int[] B); }

^{B[I]}.

For example, given L = 5 and:

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);

- 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.

^{P}, for a given integer P.

Write a function:

function solution(A, B);

^{B[I]}.

For example, given L = 5 and:

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);

- 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.

^{P}, for a given integer P.

Write a function:

function solution(A, B)

^{B[I]}.

For example, given L = 5 and:

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);

- 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.

^{P}, for a given integer P.

Write a function:

NSMutableArray * solution(NSMutableArray *A, NSMutableArray *B);

^{B[I]}.

For example, given L = 5 and:

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);

- 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.

^{P}, for a given integer P.

Assume that the following declarations are given:

Results = record C : array of longint; L : longint; {Length of the array} end;

Write a function:

function solution(A: array of longint; B: array of longint; L: longint): Results;

^{B[I]}.

For example, given L = 5 and:

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);

- 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.

^{P}, for a given integer P.

Write a function:

function solution($A, $B);

^{B[I]}.

For example, given L = 5 and:

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);

- 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.

^{P}, for a given integer P.

Write a function:

sub solution { my ($A, $B)=@_; my @A=@$A; my @B=@$B; ... }

^{B[I]}.

For example, given L = 5 and:

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);

- 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.

^{P}, for a given integer P.

Write a function:

def solution(A, B)

^{B[I]}.

For example, given L = 5 and:

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);

- 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.

^{P}, for a given integer P.

Write a function:

def solution(a, b)

^{B[I]}.

For example, given L = 5 and:

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);

- 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.

^{P}, for a given integer P.

Write a function:

object Solution { def solution(a: Array[Int], b: Array[Int]): Array[Int] }

^{B[I]}.

For example, given L = 5 and:

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);

- 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.

^{P}, for a given integer P.

Write a function:

public func solution(inout A : [Int], inout _ B : [Int]) -> [Int]

^{B[I]}.

For example, given L = 5 and:

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);

- 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.

^{P}, for a given integer P.

Write a function:

public func solution(_ A : inout [Int], _ B : inout [Int]) -> [Int]

^{B[I]}.

For example, given L = 5 and:

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);

- 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.

^{P}, for a given integer P.

Write a function:

Private Function solution(A As Integer(), B As Integer()) As Integer()

^{B[I]}.

For example, given L = 5 and:

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);

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