Sharing an answer code of mine about TieRopes problem of Codility lesson 16.

# Lesson 16: TieRopes

There are N ropes numbered from 0 to N − 1, whose lengths are given in a zero-indexed 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
```

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 zero-indexed 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.

Assume that:

- 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].

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).

# Answer Code in Python 3

- Time complexity:
- Space complexity:

```
# Python 3.6
# Time complexity: O(N)
# Space complexity: O(N)
def solution(K, A):
tied = []
i=0
for _ in range(len(A)):
temp = 0
while temp < K and i<len(A):
temp += A[i]
i += 1
if temp < K:
break
else: tied.append(temp)
return len(tied)
```