Codility---MinAvgTwoSlice

Task description

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:

class Solution { public int solution(int[] 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).

Elements of input arrays can be modified.

Solution

 

Programming language used: Java

Total time used: 8 minutes

Code: 04:06:50 UTC, java, final, score:  100

show code in pop-up

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// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int solution(int[] A) {
        // write your code in Java SE 8
        int minAvgIdx=0;
        double minAvg = (A[0]+A[1])/2;
        double currAvg;
        for(int i=0; i< A.length-2; i++) {
            //two number 
            currAvg = (double)(A[i] +A[i+1])/2;
            if(minAvg > currAvg) {
                minAvg = currAvg;
                minAvgIdx = i;
            }
            
            //three number 
            currAvg =(double) (A[i] +A[i+1] +A[i+2])/3;
            if(minAvg > currAvg) {
                minAvg = currAvg;
                minAvgIdx = i;
            }
        }
        currAvg =(double) (A[A.length-2] +A[A.length-1])/2;
        if(minAvg > currAvg) {
            minAvg = currAvg;
            minAvgIdx = A.length-2;
        }
        return  minAvgIdx;
    }
}


https://codility.com/demo/results/trainingP69B3H-F2P/