codility - MinAvgTwoSlice ( javascript )

codility / traning center [MinAvgTwoSlice]

A non-empty zero-indexed 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:

function solution(A);

that, given a non-empty zero-indexed 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.

Assume that:

• 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].

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

이 문제는 N크기의 배열에서 두 개 이상의 요소로 이루어진 부분 배열의 최솟값을 찾아, 그 부분 배열의 첫번째 index를 구하는 문제이다.

시간복잡도를 확인해 보면 O(N)의 범위안으로 해결해야 하는데 for문을 한번 사용해서 최솟값을 찾아야 한다.

길이가 4인 배열을 생각해 볼 때, 최솟값은 언제나 두개 혹은 세개의 요소로 이루어진 부분 배열임을 알 수 있다.

for문을 통해 두 개, 혹은 세개의 부분배열의 최솟값을 계산해 주면 해결할 수 있다.

function solution(A) {
    var min = (A[0] + A[1]) / 2;
    var minIdx = 0;
    for ( var i = 1; i < A.length - 1 ; i ++ ) {
        var two = (A[i] + A[i + 1]) / 2;
        if (i > A.length - 2) {
            if ( two < min) {
                min = two;
                minIdx = i;
            }
        } else {
            var three = (A[i] + A [i + 1] + A[i + 2]) / 3;
            if (two < min || three < min) {
                min = two < three ? two : three;
                minIdx = i;
            }
        }
    }
    return minIdx;
}

Test result report