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

Programming language:
Spoken language:

Consider an 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 an array A of N integers, returns an 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) (not counting the storage required for input arguments).

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

Consider an 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 an array A of N integers, returns an 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) (not counting the storage required for input arguments).

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

Consider an 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 an array A of N integers, returns an 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) (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);

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.

*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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