Lesson 5. Perfix Sums - MinAvgTwoSlice #1

1. 문제

A non-empty array A consisting of N integers is given. A pair of integers (P, Q), such that 0 ≤ P < Q < N, is called a slice of array A (notice that the slice contains at least two elements). The average of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).
For example, array A such that:
A[0] = 4
A[1] = 2
A[2] = 2
A[3] = 5
A[4] = 1
A[5] = 5
A[6] = 8

contains the following example slices:
slice (1, 2), whose average is (2 + 2) / 2 = 2;
slice (3, 4), whose average is (5 + 1) / 2 = 3;
slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.
The goal is to find the starting position of a slice whose average is minimal.

Write a function:
  int solution(vector<int> &A);

that, given a non-empty array A consisting of N integers, returns the starting position of the slice with the minimal average. If there is more than one slice with a minimal average, you should return the smallest starting position of such a slice.
For example, given array A such that:
A[0] = 4
A[1] = 2
A[2] = 2
A[3] = 5
A[4] = 1
A[5] = 5
A[6] = 8
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 [−10,000..10,000].

 

주어진 배열에서 최저 평균 값을 가지는 구간의 시작 index를 구하는 문제.

예) return 1

A[0]	A[1]	A[2]	A[3]	A[4]	A[5]	A[6]
4	2	2	5	1	5	8

Slice	Average
(1, 2)	(2 + 2) / 2 = 4
(3, 4)	(5 + 1) / 2 =3
(1, 4)	(2 + 2 + 5 + 1) / 4 = 2.5

 

 

2. 매개변수 조건

• vector A 크기: 2 ~ 1,000,000
• vector A 요소 값: -10,000 ~ 10,000

 

 

3. 코드 (C++)

모든 평균 값을 구한다.

int solution(vector<int> &A) {

    double min = 10001;
    int pos = 0;

    for(size_t i = 0 ; i < A.size(); i++)
    {
        double avg = 0;
        int cnt = 1;
        int sum = A[i];
        
        //모든 경우의 평균값을 구한다
        for(size_t j = i + 1; j < A.size(); j++)
        {
            sum += A[j];
            cnt++;
            avg = static_cast<double>(sum) / cnt;
            
            //현재 구한 평균 값이, 가장 작은 값이라면
            if(avg < min)
            {
                min = avg;
                pos = i;
            }
        }
    }

    return pos;
}

 

 

4. 결과