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respectable

For each element of an array of integers, find the closest larger element.

Programming language:
Spoken language:

Consider a zero-indexed array A of N integers. Indices of this array are integers from 0 to N−1. Take an index K. Index J is called an *ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

Ascender J of K is called *the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

If K = 3 then K has two ascenders: 7 and 8. Its closest ascender is 7 and distance between K and 7 equals **abs**(K−7) = 4.

Assume that the following declarations are given:

struct Results { int * R; int N; // Length of the array };

Write a function:

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

that, given a zero-indexed array A of N integers, returns a zero-indexed array R of N integers, such that (for K = 0,..., N−1):

- if K has the closest ascender J, then R[K] =
abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

Consider a zero-indexed array A of N integers. Indices of this array are integers from 0 to N−1. Take an index K. Index J is called an *ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

Ascender J of K is called *the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

If K = 3 then K has two ascenders: 7 and 8. Its closest ascender is 7 and distance between K and 7 equals **abs**(K−7) = 4.

Write a function:

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

that, given a zero-indexed array A of N integers, returns a zero-indexed array R of N integers, such that (for K = 0,..., N−1):

- if K has the closest ascender J, then R[K] =
abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

Consider a zero-indexed array A of N integers. Indices of this array are integers from 0 to N−1. Take an index K. Index J is called an *ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

Ascender J of K is called *the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

If K = 3 then K has two ascenders: 7 and 8. Its closest ascender is 7 and distance between K and 7 equals **abs**(K−7) = 4.

Write a function:

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

that, given a zero-indexed array A of N integers, returns a zero-indexed array R of N integers, such that (for K = 0,..., N−1):

- if K has the closest ascender J, then R[K] =
abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

*ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

*the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

**abs**(K−7) = 4.

Write a function:

func Solution(A []int) []int

abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

*ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

*the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

**abs**(K−7) = 4.

Write a function:

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

abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

*ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

*the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

**abs**(K−7) = 4.

Write a function:

function solution(A);

abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

*ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

*the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

**abs**(K−7) = 4.

Write a function:

function solution(A)

abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

*ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

*the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

**abs**(K−7) = 4.

Write a function:

NSMutableArray * solution(NSMutableArray *A);

abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

*ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

*the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

**abs**(K−7) = 4.

Assume that the following declarations are given:

Results = record R : array of longint; N : longint; {Length of the array} end;

Write a function:

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

abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

*ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

*the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

**abs**(K−7) = 4.

Write a function:

function solution($A);

abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

*ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

*the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

**abs**(K−7) = 4.

Write a function:

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

abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

*ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

*the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

**abs**(K−7) = 4.

Write a function:

def solution(A)

abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

*ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

*the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

**abs**(K−7) = 4.

Write a function:

def solution(a)

abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

*ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

*the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

**abs**(K−7) = 4.

Write a function:

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

abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

*ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

*the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

**abs**(K−7) = 4.

Write a function:

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

abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

*ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

*the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

**abs**(K−7) = 4.

Write a function:

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

abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

*ascender* of K if A[J] > A[K]. Note that if A[K] is a maximal value in the array A, then K has no ascenders.

*the closest ascender* of K if **abs**(K−J) is the smallest possible value (that is, if the distance between J and K is minimal). Note that K can have at most two closest ascenders: one smaller and one larger than K.

For example, let us consider the following array A:

**abs**(K−7) = 4.

Write a function:

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

abs(K−J); that is, R[K] is equal to the distance between J and K,- if K has no ascenders then R[K] = 0.

For example, given the following array A:

the function should return the following array R:

Array R should be returned as:

- a structure
Results(in C), or- a vector of integers (in C++), or
- a record
Results(in Pascal), or- an array of integers (in any other programming language).

Assume that:

- N is an integer within the range [0..50,000];
- each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

Complexity:

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

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