CODILITY - LESSON 3 - MinAvgTwoSlice

PROBLEM:

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.

MY SOLUTION (100%, time complexity O(N))
class Solution { public int solution(int[] A) { int minI=0; double minValue = 100001.0; for (int i=0; i<A.Length-1; i++) { if (((A[i]+A[i+1])/2.0) < minValue) { minI=i; minValue=(A[i]+A[i+1])/2.0; } if (i < A.Length-2) { if (((A[i] +A[i+1]+A[i+2])/3.0)< minValue) { minI=i; minValue= (A[i] +A[i+1]+A[i+2]) / 3.0; } } } return minI; } }

It checks the average of all subarrays formed with 2 elements and 3 elements. Any group of more elements will have an average >= to one of these groups, so the smaller group would be optimal.

PR

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