MinAvgTwoSlice

tab1

Task Score     :100
Correctness   :100
Performance : 100

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).

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].

Solution

public class MinAvgTwoSlice {

 public static void main(String[] args) {
  int[] A ={4,2,2,5,1,5,8};
  System.out.println(new MinAvgTwoSlice().solution(A));
 }
 public int solution(int[] A) {

  if(A.length == 0) {
   return 0;
  }
  if(A.length < 2) {
   return A[0];
  }
  
  int min = 0;
  float prevMin = Float.MAX_VALUE;
  for (int i = 0; i < A.length-1; i++) {
   float sum = Float.valueOf((A[i] + A[i+1]))/2; // Minimum will come only with 2 or 3 elements, average with more
                // Elements will be higher
   if(prevMin > sum) {
    prevMin = sum;
    min = i;
   }
   if((i < A.length-2) && Float.valueOf(A[i]+ A[i+1]+ A[i+2])/3 < prevMin) {
    prevMin = Float.valueOf((A[i]+ A[i+1]+ A[i+2]))/3;
    min = i;
   }
  }
  return min;
    }
}