[Codility] Lesson 5 - MinAvgTwoSlice

Task Discription

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 &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].

Source Code

#include <algorithm>

int solution(vector<int> &A) {
    float minAvg = (A[0] + A[1]) / 2;
    int minIdx = 0;

    for (int i = 2; i < A.size(); i++) {
        float avg = (float)(A[i-2] + A[i-1] + A[i]) / (float)3;
        if (minAvg > avg) {
            minAvg = avg;
            minIdx = i - 2;
        }

        avg = (float)(A[i-1] + A[i]) / (float)2;
        if (minAvg > avg) {
            minAvg = avg;
            minIdx = i - 1;
        }
    }

    return minIdx;
}

Review

수학적 지식이 있어야 100% 정답을 볼 수 있는 문제였다.. ㅜ
4개 이상의 원소는 고려할 필요 없는 문제이다.
2개와 3개일 때만 고려하면 된다.