[Codility] Lesson3 - Time Complexity | TapeEquilibrium

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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[틀린풀이]

import sys

def solution(A):
    # 그룹으로 나눴을 때 그 차이가 최소가 되는 경계를 찾아라
    # RETURN: 그 차이값
    sum_all = sum(A)

    min_diff = sys.maxsize
    for i in range(len(A)):
        sum_1 = sum(A[:i+1])
        sum_2 = sum_all - sum_1
        diff = abs(sum_1-sum_2)
        if diff < min_diff: min_diff = diff
    
    return min_diff

sum 자체를 중복해서 계산해야 하는 시간을 고려 안함 

 

[맞는 풀이]

import sys

def solution(A):
    # 그룹으로 나눴을 때 그 차이가 최소가 되는 경계를 찾아라
    # RETURN: 그 차이값
    sum_all = sum(A)

    min_diff = sys.maxsize
    sum_tmp = 0
    for i in range(len(A)-1):
        sum_1 = sum_tmp + A[i]
        sum_2 = sum_all - sum_1
        diff = abs(sum_1-sum_2)

        if diff < min_diff: min_diff = diff

        sum_tmp = sum_1
    
    return min_diff