Codility Lesson3 1. 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, was called a slice of array A (notice that the slice con Tains at least-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 is, 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 within the range [−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. classSolution { Public intSolution (int[] A) {//Write your code in Java SE 8 intleft = 0;//Use of pointers left and right to traverse the array intright = 1; intLen =a.length; intsum = a[0] + a[1]; intSumleft = 0; DoubleAVG = 0; DoubleMin = 10001; intres = 0; while(Left < len-1 && Right <Len) {avg= (sum-sumleft)/(Double) (Right-left + 1);//Int/int can not get double if donot convert if(Avg <min) {min=avg; Res=Left ; } if(A[left] > Avg && left < right-1) {//When move left pointer can decrease the AVG then move it.left++; Sumleft+ = A[left-1]; Continue //must have this, or may appear the right grow to Len in next step and left have no chance to update. } if(Right < len-1 && A[right+1] < avg)//When move right pointer can decrease the AVG then move it.|| left = = right-1) {//Make sure threr are not "piece/0"right++; if(Right <len) sum+=A[right]; } Else{ Left++;//if not move leftSumleft + = A[left-1]; } } returnRes; }}

Codility Lesson3 1. Minavgtwoslice