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] = 3We 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] = 3the 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].
# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
# -*- coding：utf-8 -*-
# &Author AnFany
# Lesson 3：Time Complexity
# P 3.3 TapeEquilibrium
def solution_list(A):
"""
返回数组2部分差的最小值，时间复杂度O(N * N)
:param A: N个整数组成的数组
:return: 差的最小值
"""
sum_list = sum(A)
sub_abs_value = []
for i in range(1, len(A)):
sub_abs_value.append(abs(sum_list - 2 * sum(A[:i])))
return min(sub_abs_value)
def solution(A):
"""
返回数组2部分差的最小值，时间复杂度O(N)
:param A: N个整数组成的数组
:return: 差的最小值
"""
sum_value = sum(A)
if len(A) == 2:
first_sub_abs = abs(A[1] - A[0])
return first_sub_abs
else:
sub_abs_list = []
first_sum = 0
for i in A[:-1]:
first_sum += i
sub_abs = abs(sum_value - 2 * first_sum)
if sub_abs == 0:
return 0
sub_abs_list.append(sub_abs)
return min(sub_abs_list)# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
# -*- coding：utf-8 -*-
# &Author AnFany
# Lesson 3：Time Complexity
# P 3.3 TapeEquilibrium
def solution(A):
"""
返回数组2部分差的最小值，时间复杂度O(N)
:param A: N个整数组成的数组
:return: 差的最小值
"""
sum_value = sum(A)
if len(A) == 2:
first_sub_abs = abs(A[1] - A[0])
return first_sub_abs
else:
sub_abs_list = []
first_sum = 0
for i in A[:-1]:
first_sum += i
sub_abs = abs(sum_value - 2 * first_sum)
if sub_abs == 0:
return 0
sub_abs_list.append(sub_abs)
return min(sub_abs_list)# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")
# -*- coding：utf-8 -*-
# &Author AnFany
# Lesson 3：Time Complexity
# P 3.3 TapeEquilibrium
def solution(A):
"""
返回数组2部分差的最小值，时间复杂度O(N)
:param A: N个整数组成的数组
:return: 差的最小值
"""
sum_value = sum(A)
if len(A) == 2:
first_sub_abs = abs(A[1] - A[0])
return first_sub_abs
else:
sub_abs_list = []
first_sum = 0
for i in A[:-1]:
first_sum += i
sub_abs = abs(sum_value - 2 * first_sum)
if sub_abs == 0:
return 0
sub_abs_list.append(sub_abs)
return min(sub_abs_list)The solution obtained perfect score.