[Codility] Lesson-03.3: TapeEquilibrium
In this post, the third solution for lesson 3 with reducing time complexity is handled. For accessing the problem, click the following link.
Task Description
A non-empty array A consisting of N integers is given. Array A represents numbers on a tape.
Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], …, A[P − 1] and A[P], A[P + 1], …, A[N − 1].
The difference between the two parts is the value of: |(A[0] + A[1] + … + A[P − 1]) − (A[P] + A[P + 1] + … + A[N − 1])|
In other words, it is the absolute difference between the sum of the first part and the sum of the second part.
For example, consider array A such that:
A[0] = 3
A[1] = 1
A[2] = 2
A[3] = 4
A[4] = 3
We can split this tape in four places:
- P = 1, difference = |3 − 10| = 7
- P = 2, difference = |4 − 9| = 5
- P = 3, difference = |6 − 7| = 1
- P = 4, difference = |10 − 3| = 7
Write a function:
def solution(A)
that, given a non-empty array A of N integers, returns the minimal difference that can be achieved.
For example, given:
A[0] = 3
A[1] = 1
A[2] = 2
A[3] = 4
A[4] = 3
the function should return 1, as explained above.
Write an efficient algorithm for the following assumptions:
- N is an integer within the range [2..100,000];
- each element of array A is an integer within the range [−1,000..1,000].
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Key Point
- Efficient! Efficient! Efficient!
- You need to avoid calculating the sum of each two tapes for every loop.
- Before entering the loop, prepare the sum of the original tape before splitting.
Solution (using Python)
def solution(A):
sum_front, sum_back = 0, sum(A)
diff_list = []
for elm in A:
sum_front += elm
sum_back -= elm
diff_list.append(abs(sum_front - sum_back)) diff_list = diff_list[:-1]
return min(diff_list)
Please use the above solution for reference.
I recommend you to write your own source code.
