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AVAILABLE LESSONS:

Lesson 1

Iterations

Lesson 2

Arrays

Lesson 3

Time Complexity

Lesson 4

Counting Elements

Lesson 5

Prefix Sums

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Lesson 7

Stacks and Queues

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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

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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

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Compute number of inversion in an array.

Programming language:
Spoken language:

An array A consisting of N integers is given. An *inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

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

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

An array A consisting of N integers is given. An *inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

int solution(vector<int> &A);

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

An array A consisting of N integers is given. An *inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

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

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

*inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

func Solution(A []int) int

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

*inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

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

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

*inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

function solution(A);

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

*inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

function solution(A)

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

- expected worst-case time complexity is O(N*log(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.

*inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

int solution(NSMutableArray *A);

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

*inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

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

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

*inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

function solution($A);

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

*inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

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

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

*inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

def solution(A)

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

*inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

def solution(a)

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

*inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

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

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

*inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

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

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

*inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

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

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

*inversion* is a pair of indexes (P, Q) such that P < Q and A[Q] < A[P].

Write a function:

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

that computes the number of inversions in A, or returns −1 if it exceeds 1,000,000,000.

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, in the following array:

there are four inversions:

so the function should return 4.

Complexity:

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

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