[Codility] Lesson-05.3: MinAvgTwoSlice
This post handles extracting the slice, subset that shows a minimum average, from the numeric list.
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:
def solution(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].
Copyright 2009–2021 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
Key Point
- In the Codility lesson, minimizing computational complexity is important.
- As much as possible, avoiding the loop command is helpful.
- Also, mathematical knowledge helps to make solutions more powerful.
Solution (using Python)
def min_check(avg_min, avg_tmp, idx_min, idx_tmp):
if(avg_tmp < avg_min): return idx_tmp, avg_tmp
else: return idx_min, avg_mindef solution(A):
idx_max = len(A)+1
idx_min, avg_min = 0, sum(A[0:2])/2
if(idx_max == 2): return idx_minfor idx in range(3, idx_max):
idx2, idx3 = idx-2, idx-3
avg_tmp2 = sum(A[idx2:idx])/2
avg_tmp3 = sum(A[idx3:idx])/3
idx_min, avg_min = min_check(avg_min, avg_tmp2, idx_min, idx2)
idx_min, avg_min = min_check(avg_min, avg_tmp3, idx_min, idx3)
return idx_min
Please use the above solution for reference.
I recommend you to write your own source code.
