TieRopes

🔗 문제 링크

https://app.codility.com/programmers/lessons/16-greedy_algorithms/tie_ropes/start/

❔ 문제 설명

There are N ropes numbered from 0 to N − 1, whose lengths are given in an array A, lying on the floor in a line. For each I (0 ≤ I < N), the length of rope I on the line is A[I].

We say that two ropes I and I + 1 are adjacent. Two adjacent ropes can be tied together with a knot, and the length of the tied rope is the sum of lengths of both ropes. The resulting new rope can then be tied again.

For a given integer K, the goal is to tie the ropes in such a way that the number of ropes whose length is greater than or equal to K is maximal.

For example, consider K = 4 and array A such that:

A[0] = 1
A[1] = 2
A[2] = 3
A[3] = 4
A[4] = 1
A[5] = 1
A[6] = 3

The ropes are shown in the figure below.

We can tie:

rope 1 with rope 2 to produce a rope of length A[1] + A[2] = 5;
rope 4 with rope 5 with rope 6 to produce a rope of length A[4] + A[5] + A[6] = 5.
After that, there will be three ropes whose lengths are greater than or equal to K = 4. It is not possible to produce four such ropes.

Write a function:

def solution(K, A)

that, given an integer K and a non-empty array A of N integers, returns the maximum number of ropes of length greater than or equal to K that can be created.

For example, given K = 4 and array A such that:

A[0] = 1
A[1] = 2
A[2] = 3
A[3] = 4
A[4] = 1
A[5] = 1
A[6] = 3

the function should return 3, as explained above

⚠️ 제한사항

• N is an integer within the range [1..100,000];

• K is an integer within the range [1..1,000,000,000];

• each element of array A is an integer within the range [1..1,000,000,000].

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💡 풀이 (언어 : Python)

쉬운 문제였지만, 어렵게 생각하다 버벅였다.

def solution(K, A):
    count = 0
    length = 0
    
    for rope in A:
        length += rope 
        if length >= K:     
            count += 1
            length = 0
            
    return count