Solution to Min-Avg-Two-Slice

Solution to Min-Avg-Two-Slice by codility - 

It IS a mathematical problem

6 FEB

February 6, 2014

Question: https://codility.com/demo/take-sample-test/min_avg_two_slice

Question Name: MinAvgTwoSlice

1 MinAvgSlice其子片段的Avg都是相等的，不然就可以只保留Avg较小的子片段

2  最小的不可分割的子片段必然是2Slice或者3Slice(4Slice可以分割为两个2Slice)

观察到这两点之后，只需要构建Prefix Sums数组，再遍历一遍寻找最小的2Slice/3Slice即可，有个家伙对这两点给出了证明

实现上只需注意在除2除3的时候变为除2.0和除3.0

The key to solve this task is these two patterns:  
(1) There must be some slices, with length of two or three, having the minimal average value among all the slices. 
(2) And all the longer slices with minimal average are built up with these 2-element and/or 3-element small slices.

Firstly we will prove the statement (1). In all the following discussion, we assume the slices have two or more element. In every array, there must be at lease one slice, to say S, having the Minimal Average (MA). And PLEASE PAY ATTENTION, the following proof is done with the S, which HAS the global minimal average.

1. If the length of S is two or three, it follows our conclusion.
2. If the length of S is odd, we could divide it into a 3-element sub-slice and some 2-element sub-slice. For example, for the slice [1, 2, 3, 4, 5], we could get a 3-element sub-slice [1, 2, 3] and a 2-element sub-slice [4, 5]. Or differently we could get [1, 2] and [3, 4, 5]. But the split does not matter in the following prove.
   
   • If the sub-slices have different averages, some of them will have smaller average than MA. But it conflicts with the condition that, the MA is known as the global minimal average for all slices.  By this contradictory, it’s proved that all the sub-slices MUST have the same average.
   • If all the sub-slices have the same average, the average of them must be MA. So all these sub-slices have overall minimal average and length of two or three.
3. If the length of S is even, we could divide it into some 2-element sub-slice. For the slice [1, 2, 3, 4], the only possible split is [1, 2] and [3, 4]. Then, similar with the previous case, all these 2-element sub-slices have the global minimal average.

And in the construction in step b, we have already proven the statement (2).

UPDATE 03-13-2014: We are NOT proving that, 4-or-more-element slices cannot have the global minimal average. We just proved that, there MUST be some 2-element and/or 3-element slices, having the global minimal average. In other words, we are NOT SURE whether there are some 4-or-more-element slices holding global minimal average. But we are ONE HUNDRED PERCENT SURE about the 2-element and/or 3-element slices.

UPDATE 03-02-2015: There is another excellent discussion in the comments by Kim. Thanks!

UPDATE 03-14-2016: If you had any question about the statement “If the sub-slices have different averages, some of them will have smaller average than MA”, please read this comment and this comment. Thanks!

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def solution ( A ) :      min_avg_value = ( A [ 0 ] + A [ 1 ] ) / 2.0    # The mininal average      min_avg_pos = 0      # The begin position of the first                          # slice with mininal average        for index in xrange ( 0 , len ( A ) - 2 ) :          # Try the next 2-element slice          if ( A [ index ] + A [ index + 1 ] ) / 2.0 < min_avg_value :              min_avg_value = ( A [ index ] + A [ index + 1 ] ) / 2.0              min_avg_pos = index          # Try the next 3-element slice          if ( A [ index ] + A [ index + 1 ] + A [ index + 2 ] ) / 3.0 < min_avg_value :              min_avg_value = ( A [ index ] + A [ index + 1 ] + A [ index + 2 ] ) / 3.0              min_avg_pos = index        # Try the last 2-element slice      if ( A [ - 1 ] + A [ - 2 ] ) / 2.0 < min_avg_value :          min_avg_value = ( A [ - 1 ] + A [ - 2 ] ) / 2.0          min_avg_pos = len ( A ) - 2        return min_avg_pos