Min Average Two Slice Codility

Question

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.

---

This is my best solution, but obviously not optimal in terms of time complexity.
Any ideas?

public int solution(int[] A) {     int result = 0;     int N = A.length;     int [] prefix = new int [N+1];     for (int i = 1; i < prefix.length; i++) {         prefix[i] = prefix[i-1] + A[i-1];     }     double avg = Double.MAX_VALUE;     for (int i = 1; i < N; i++) {         for (int j = i+1; j <=N; j++) {             double temp = (double)(prefix[j]-prefix[i-1]) /(double)(j-i+1);             if (temp < avg) {                 avg = temp;                 result = i-1;             }         }     }     return result; } 

https://codility.com/demo/results/demo65RNV5-T36/

Answer 1

100% score with C++ based on Kadane's algorithm

int solution(vector<int> &A) {      // Find prefix sum.     int N = A.size();     vector<int> ps(N + 1, 0);          for (int i = 1; i <= N; i++) {         ps[i] = A[i - 1] + ps[i - 1];     }      int lft_idx, min_lft_idx;     double avg_here, min_avg, avg_of_two, avg_with_prev;          // Initialize variables at the first possible slice (A[0:1]).     lft_idx = min_lft_idx = 0;     avg_here = min_avg = (A[0] + A[1]) / 2.0;          // Find min average of every slice that ends at ith element,     // starting at i = 2.     for (int i = 2; i < N; i ++) {          // average of A[lft_idx : i]         avg_with_prev = ((double) ps[i + 1] - ps[lft_idx]) /                          (i - lft_idx + 1);          // average of A[i - 1 : i]         avg_of_two = (A[i - 1] + A[i]) / 2.0;                  // Find minimum and update lft_idx of slice         // (previous lft_idx or i - 1).         if (avg_of_two < avg_with_prev) {             avg_here = avg_of_two;             lft_idx = i - 1;         }         else             avg_here = avg_with_prev;                  // Keep track of minimum so far and its left index.         if (avg_here < min_avg) {             min_avg = avg_here;             min_lft_idx = lft_idx;         }     }              return min_lft_idx; } 

I wasn't satisfied with the 2/3-element subslices solution (come on! who would think of that in the duration of an interview?), nor with the explanations (or lack of them), so I went on searching for other approaches. Then I found about Kadane's algorithm for the maximum subarray problem (MSP) which led me to a different solution.

The basic question is (similar to that of the MSP): What is the minimal average of a slice that includes the i-th element?

In order to answer it, we'll look for slices that end in the i-th element, only updating their left index. That is, we have to check for the slices A[lft_idx : i].

Assuming we know the lft_idx of the slice A[lft_idx : i - 1] with a minimal average, then we have two possibilities:

1. The average of A[lft_idx : i] is minimal.
2. The average of A[i - 1 : i] is minimal (the shortest possible slice has 2 elements).

What happens in case 1 is that we continue growing the slice that starts at lft_idx.

In case 2, however, we find that growing the previous slice actually increases the average. So we start over again and "reset" the start of the slice (lft_idx) to the previous element (i - 1). Now we have a new best slice of size 2 to grow from this point.

Finally, we want the global minimal average, so we need to keep track of the minimum so far and where it started (the question only asks for this, but we could as well save the right index).

Note: I'm using prefix sums here to calculate slice averages because that's where the question appears in Codility, but it could easily be replaced by a variable with the size of the previous slice and another multiplication.

Answer 2

I had posted this some days ago:

Check this out:

http://codesays.com/2014/solution-to-min-avg-two-slice-by-codility/

In there, they explain with great detail why their solution works. I haven't implemented it myself yet, but I will definitely try it.

Hope it helps!

but I just saw it was deleted by a moderator. They say the link is dead, but I just tried it and it works fine. I'm posting it once again, hoping it can be validated that the link is good.

And now I can also provide my implementation, based on the codesays link that I provided before: https://codility.com/demo/results/demoERJ4NR-ETT/

class Solution {     public int solution(int[] A) {         int minAvgIdx=0;         double minAvgVal=(A[0]+A[1])/2; //At least two elements in A.         double currAvg;         for(int i=0; i<A.length-2; i++){             /**              * We check first the two-element slice              */             currAvg = ((double)(A[i] + A[i+1]))/2;             if(currAvg < minAvgVal){                 minAvgVal = currAvg;                 minAvgIdx = i;             }             /**              * We check the three-element slice              */             currAvg = ((double)(A[i] + A[i+1] + A[i+2]))/3;             if(currAvg < minAvgVal){                 minAvgVal = currAvg;                 minAvgIdx = i;             }         }         /**          * Now we have to check the remaining two elements of the array          * Inside the for we checked ALL the three-element slices (the last one          * began at A.length-3) and all but one two-element slice (the missing          * one begins at A.length-2).          */         currAvg = ((double)(A[A.length-2] + A[A.length-1]))/2;         if(currAvg < minAvgVal){             minAvgVal = currAvg;             minAvgIdx = A.length-2;         }         return minAvgIdx;     } }