Your browser (Unknown 0) is no longer supported. Some parts of the website may not work correctly. Please update your browser.

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:

There are N ropes numbered from 0 to N − 1, whose lengths are given in an array A, lying on the floor in a line. For each I (0 ≤ I < N), the length of rope I on the line is A[I].

We say that two ropes I and I + 1 are *adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For a given integer K, the goal is to tie the ropes in such a way that the number of ropes whose length is greater than or equal to K is maximal.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

After that, there will be three ropes whose lengths are greater than or equal to K = 4. It is not possible to produce four such ropes.

Write a function:

int solution(int K, int A[], int N);

that, given an integer K and a non-empty array A of N integers, returns the maximum number of ropes of length greater than or equal to K that can be created.

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

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

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

There are N ropes numbered from 0 to N − 1, whose lengths are given in an array A, lying on the floor in a line. For each I (0 ≤ I < N), the length of rope I on the line is A[I].

We say that two ropes I and I + 1 are *adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For a given integer K, the goal is to tie the ropes in such a way that the number of ropes whose length is greater than or equal to K is maximal.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

After that, there will be three ropes whose lengths are greater than or equal to K = 4. It is not possible to produce four such ropes.

Write a function:

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

that, given an integer K and a non-empty array A of N integers, returns the maximum number of ropes of length greater than or equal to K that can be created.

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

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

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

There are N ropes numbered from 0 to N − 1, whose lengths are given in an array A, lying on the floor in a line. For each I (0 ≤ I < N), the length of rope I on the line is A[I].

We say that two ropes I and I + 1 are *adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For a given integer K, the goal is to tie the ropes in such a way that the number of ropes whose length is greater than or equal to K is maximal.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

After that, there will be three ropes whose lengths are greater than or equal to K = 4. It is not possible to produce four such ropes.

Write a function:

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

that, given an integer K and a non-empty array A of N integers, returns the maximum number of ropes of length greater than or equal to K that can be created.

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

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

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

*adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

Write a function:

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

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

- expected worst-case time complexity is O(N);

*adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

Write a function:

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

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

- expected worst-case time complexity is O(N);

*adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

Write a function:

function solution(K, A);

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

- expected worst-case time complexity is O(N);

*adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

Write a function:

function solution(K, A)

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

- expected worst-case time complexity is O(N);

Note: All arrays in this task are zero-indexed, unlike the common Lua convention. You can use `#A` to get the length of the array A.

*adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

Write a function:

int solution(int K, NSMutableArray *A);

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

- expected worst-case time complexity is O(N);

*adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

Write a function:

function solution(K: longint; A: array of longint; N: longint): longint;

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

- expected worst-case time complexity is O(N);

*adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

Write a function:

function solution($K, $A);

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

- expected worst-case time complexity is O(N);

*adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

Write a function:

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

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

- expected worst-case time complexity is O(N);

*adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

Write a function:

def solution(K, A)

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

- expected worst-case time complexity is O(N);

*adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

Write a function:

def solution(k, a)

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

- expected worst-case time complexity is O(N);

*adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

Write a function:

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

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

- expected worst-case time complexity is O(N);

*adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

Write a function:

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

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

- expected worst-case time complexity is O(N);

*adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

Write a function:

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

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

- expected worst-case time complexity is O(N);

*adjacent*. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For example, consider K = 4 and array A such that:

The ropes are shown in the figure below.

We can tie:

- rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
- rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.

Write a function:

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

For example, given K = 4 and array A such that:

the function should return 3, as explained above.

Assume that:

- N is an integer within the range [1..100,000];
- K is an integer within the range [1..1,000,000,000];
- each element of array A is an integer within the range [1..1,000,000,000].

Complexity:

- expected worst-case time complexity is O(N);

Information about upcoming challenges, solutions and lessons directly in your inbox.

© 2009–2018 Codility Ltd., registered in England and Wales (No. 7048726). VAT ID GB981191408. Registered office: 107 Cheapside, London EC2V 6DN