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ambitious

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 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 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) (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 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 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) (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 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 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) (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));

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.

- 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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