[Codility] Lesson3 - Time Complexity: Tape Equilibrium (C++)

Problem

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].

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How to solve

- 정보

N개의 정수를 가진 배열 A: 테이프의 수

정수 P, 0<P<N 은 테이프를 2개의 파트로 자른다

2개의 파트의 차는 (A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])

 

- 구해야 하는것?

배열 A를 반으로 잘랐을 때, 두 부분의 차가 가장 작은 값 return

 

- 풀이

각 idx에서의 배열의 sum 을 구해놓고 계산

left sum =sum[i-1]

right sum = sum[len-1] - sum[i-1]

 

**주의**

배열의 갯수가 2인 경우 조건 !

 

Solution(c++)

// you can use includes, for example:
#include <bits/stdc++.h>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)
    int len = A.size();
    int minVal = INT_MAX;
    vector<int> sum(len, 0);

    if(len == 2){
        return A[1] - A[0];
    }

    sum[0] = A[0];
    for(int i=1; i<len; i++){
        sum[i] = sum[i-1] + A[i];
    }
    for(int i=1; i<len; i++){
        minVal = min(abs(sum[i-1] - (sum[len-1]- sum[i-1])), minVal);
    }

    return minVal;
}

Test Result

app.codility.com/demo/results/training5GHYY4-W3M/

Test results - Codility