Codility 13번 MinAvgTwoSlice

MinAvgTwoSlice

 

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:

class Solution { public int solution(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].

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import java.util.*;

class Solution {
    public int solution(int[] A) {
        float min = 10001;
        int minPos = 0;

        for (int i = 0; i <= A.length - 2; i++) {
            float _min;
            if (i <= A.length - 3) {
                _min = Math.min((A[i] + A[i + 1] + A[i + 2]) / 3f, (A[i] + A[i + 1]) / 2f);

            } else {
                _min = Math.min(min, (A[i] + A[i + 1]) / 2f);
            }
            if (min > _min) {
                min = _min;
                minPos = i;
            }
        }
        return minPos;
    }
}

 

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