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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

painless

Find a maximal set of non-overlapping segments.

Programming language:
Spoken language:

Located on a line are N segments, numbered from 0 to N − 1, whose positions are given in arrays A and B. For each I (0 ≤ I < N) the position of segment I is from A[I] to B[I] (inclusive). The segments are sorted by their ends, which means that B[K] ≤ B[K + 1] for K such that 0 ≤ K < N − 1.

Two segments I and J, such that I ≠ J, are *overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

We say that the set of segments is *non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

The size of a non-overlapping set containing a maximal number of segments is 3. For example, possible sets are {0, 2, 3}, {0, 2, 4}, {1, 2, 3} or {1, 2, 4}. There is no non-overlapping set with four segments.

Write a function:

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

that, given two arrays A and B consisting of N integers, returns the size of a non-overlapping set containing a maximal number of segments.

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

Copyright 2009–2019 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

Located on a line are N segments, numbered from 0 to N − 1, whose positions are given in arrays A and B. For each I (0 ≤ I < N) the position of segment I is from A[I] to B[I] (inclusive). The segments are sorted by their ends, which means that B[K] ≤ B[K + 1] for K such that 0 ≤ K < N − 1.

Two segments I and J, such that I ≠ J, are *overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

We say that the set of segments is *non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

The size of a non-overlapping set containing a maximal number of segments is 3. For example, possible sets are {0, 2, 3}, {0, 2, 4}, {1, 2, 3} or {1, 2, 4}. There is no non-overlapping set with four segments.

Write a function:

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

that, given two arrays A and B consisting of N integers, returns the size of a non-overlapping set containing a maximal number of segments.

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

Copyright 2009–2019 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

Located on a line are N segments, numbered from 0 to N − 1, whose positions are given in arrays A and B. For each I (0 ≤ I < N) the position of segment I is from A[I] to B[I] (inclusive). The segments are sorted by their ends, which means that B[K] ≤ B[K + 1] for K such that 0 ≤ K < N − 1.

Two segments I and J, such that I ≠ J, are *overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

We say that the set of segments is *non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

The size of a non-overlapping set containing a maximal number of segments is 3. For example, possible sets are {0, 2, 3}, {0, 2, 4}, {1, 2, 3} or {1, 2, 4}. There is no non-overlapping set with four segments.

Write a function:

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

that, given two arrays A and B consisting of N integers, returns the size of a non-overlapping set containing a maximal number of segments.

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

Copyright 2009–2019 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

*overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

*non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

Write a function:

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

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

*overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

*non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

Write a function:

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

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

*overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

*non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

Write a function:

function solution(A, B);

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

*overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

*non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

Write a function:

fun solution(A: IntArray, B: IntArray): Int

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

*overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

*non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

Write a function:

function solution(A, B)

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

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.

*overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

*non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

Write a function:

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

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

*overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

*non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

Write a function:

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

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

*overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

*non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

Write a function:

function solution($A, $B);

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

*overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

*non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

Write a function:

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

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

*overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

*non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

Write a function:

def solution(A, B)

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

*overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

*non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

Write a function:

def solution(a, b)

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

*overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

*non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

Write a function:

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

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

*overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

*non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

Write a function:

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

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

*overlapping* if they share at least one common point. In other words, A[I] ≤ A[J] ≤ B[I] or A[J] ≤ A[I] ≤ B[J].

*non-overlapping* if it contains no two overlapping segments. The goal is to find the size of a non-overlapping set containing the maximal number of segments.

For example, consider arrays A, B such that:

The segments are shown in the figure below.

Write a function:

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

For example, given arrays A, B shown above, the function should return 3, as explained above.

Write an ** efficient** algorithm for the following assumptions:

- N is an integer within the range [0..30,000];
- each element of arrays A, B is an integer within the range [0..1,000,000,000];
- A[I] ≤ B[I], for each I (0 ≤ I < N);
- B[K] ≤ B[K + 1], for each K (0 ≤ K < N − 1).

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