Task description

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).

Determine whether a given string of parentheses (multiple types) is properly nested

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 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].

```
function solution(A) {
let two = [];
let three = [];
let avg = Infinity;
let index = -1;
for(let i=1; i<A.length; i++) {
two[i] = (A[i] + A[i-1])/2;
if(two[i] < avg) {
avg = two[i];
index = i-1;
}
if(i>1) {
three[i] = (A[i] + A[i-1] + A[i-2])/3;
if(three[i] < avg) {
avg = three[i];
index = i-2;
}
}
}
return index;
}
```

Odd Occurrences In Array Find value odd number of elements.

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