Tech It Yourself
tháng 1 17, 2020
0
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:
int solution(vector<int> &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].
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:
int solution(vector<int> &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].
#include <limits> int solution(vector<int> &A) { // write your code in C++11 (g++ 4.8.2) vector<int> pre_sum; int s = 0; for (auto i: A) { s += i; pre_sum.push_back(s); } int min_abs_diff = numeric_limits<int>::max(); for (size_t i = 1; i < pre_sum.size(); i++) { int abs_diff = abs(pre_sum[i - 1] * 2 - pre_sum[pre_sum.size() - 1]); min_abs_diff = min(min_abs_diff, abs_diff); } return min_abs_diff; }
