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Find the maximum number of ropes that can be attached in order, without breaking any of the ropes.

Programming language:
Spoken language:

In a room there are N ropes and N weights. Each rope is connected to exactly one weight (at just one end), and each rope has a particular durability − the maximum weight that it can suspend.

There is also a hook, attached to the ceiling. The ropes can be attached to the hook by tying the end without the weight. The ropes can also be attached to other weights; that is, the ropes and weights can be attached to one another in a chain. A rope will break if the sum of weights connected to it, directly or indirectly, is greater than its durability.

We know the order in which we want to attach N ropes. More precisely, we know the parameters of the rope (durability and weight) and the position of each attachment. Durabilities, weights and positions are given in three zero-indexed arrays A, B, C of lengths N. For each I (0 ≤ I < N):

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,
- C[I] (such that C[I] < I) is the position to which we attach the I-th rope; if C[I] equals −1 we attach to the hook, otherwise we attach to the weight connected to the C[I]-th rope.

The goal is to find the maximum number of ropes that can be attached in the specified order without breaking any of the ropes.

Write a function:

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

that, given three zero-indexed arrays A, B, C of N integers, returns the maximum number of ropes that can be attached in a given order.

For example, given the following arrays:

the function should return 3, as if we attach a fourth rope then one rope will break, because the sum of weights is greater than its durability (2 + 3 + 1 = 6 and 6 > 5).

Given the following arrays:

the function should return 2, as if we attach a third rope then one rope will break, because the sum of weights is greater than its durability (2 + 2 + 1 = 5 and 5 > 4).

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

In a room there are N ropes and N weights. Each rope is connected to exactly one weight (at just one end), and each rope has a particular durability − the maximum weight that it can suspend.

There is also a hook, attached to the ceiling. The ropes can be attached to the hook by tying the end without the weight. The ropes can also be attached to other weights; that is, the ropes and weights can be attached to one another in a chain. A rope will break if the sum of weights connected to it, directly or indirectly, is greater than its durability.

We know the order in which we want to attach N ropes. More precisely, we know the parameters of the rope (durability and weight) and the position of each attachment. Durabilities, weights and positions are given in three zero-indexed arrays A, B, C of lengths N. For each I (0 ≤ I < N):

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,
- C[I] (such that C[I] < I) is the position to which we attach the I-th rope; if C[I] equals −1 we attach to the hook, otherwise we attach to the weight connected to the C[I]-th rope.

The goal is to find the maximum number of ropes that can be attached in the specified order without breaking any of the ropes.

Write a function:

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

that, given three zero-indexed arrays A, B, C of N integers, returns the maximum number of ropes that can be attached in a given order.

For example, given the following arrays:

the function should return 3, as if we attach a fourth rope then one rope will break, because the sum of weights is greater than its durability (2 + 3 + 1 = 6 and 6 > 5).

Given the following arrays:

the function should return 2, as if we attach a third rope then one rope will break, because the sum of weights is greater than its durability (2 + 2 + 1 = 5 and 5 > 4).

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

In a room there are N ropes and N weights. Each rope is connected to exactly one weight (at just one end), and each rope has a particular durability − the maximum weight that it can suspend.

There is also a hook, attached to the ceiling. The ropes can be attached to the hook by tying the end without the weight. The ropes can also be attached to other weights; that is, the ropes and weights can be attached to one another in a chain. A rope will break if the sum of weights connected to it, directly or indirectly, is greater than its durability.

We know the order in which we want to attach N ropes. More precisely, we know the parameters of the rope (durability and weight) and the position of each attachment. Durabilities, weights and positions are given in three zero-indexed arrays A, B, C of lengths N. For each I (0 ≤ I < N):

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,
- C[I] (such that C[I] < I) is the position to which we attach the I-th rope; if C[I] equals −1 we attach to the hook, otherwise we attach to the weight connected to the C[I]-th rope.

The goal is to find the maximum number of ropes that can be attached in the specified order without breaking any of the ropes.

Write a function:

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

that, given three zero-indexed arrays A, B, C of N integers, returns the maximum number of ropes that can be attached in a given order.

For example, given the following arrays:

the function should return 3, as if we attach a fourth rope then one rope will break, because the sum of weights is greater than its durability (2 + 3 + 1 = 6 and 6 > 5).

Given the following arrays:

the function should return 2, as if we attach a third rope then one rope will break, because the sum of weights is greater than its durability (2 + 2 + 1 = 5 and 5 > 4).

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,

Write a function:

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

For example, given the following arrays:

Given the following arrays:

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,

Write a function:

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

For example, given the following arrays:

Given the following arrays:

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,

Write a function:

function solution(A, B, C);

For example, given the following arrays:

Given the following arrays:

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,

Write a function:

function solution(A, B, C)

For example, given the following arrays:

Given the following arrays:

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,

Write a function:

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

For example, given the following arrays:

Given the following arrays:

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,

Write a function:

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

For example, given the following arrays:

Given the following arrays:

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,

Write a function:

function solution($A, $B, $C);

For example, given the following arrays:

Given the following arrays:

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,

Write a function:

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

For example, given the following arrays:

Given the following arrays:

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,

Write a function:

def solution(A, B, C)

For example, given the following arrays:

Given the following arrays:

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,

Write a function:

def solution(a, b, c)

For example, given the following arrays:

Given the following arrays:

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,

Write a function:

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

For example, given the following arrays:

Given the following arrays:

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,

Write a function:

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

For example, given the following arrays:

Given the following arrays:

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,

Write a function:

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

For example, given the following arrays:

Given the following arrays:

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

- A[I] is the durability of the I-th rope,
- B[I] is the weight connected to the I-th rope,

Write a function:

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

For example, given the following arrays:

Given the following arrays:

Assume that:

- N is an integer within the range [0..100,000];
- each element of array A is an integer within the range [1..1,000,000];
- each element of array B is an integer within the range [1..5,000];
- each element of array C is an integer such that −1 ≤ C[I] < I, for each I (0 ≤ I < N).

Complexity:

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

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