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# [其他] Regular expressions to solve programming interview riddles

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Acknowledgements

• Live : The code is executed in your browser
• Interactive : You can modify the code and it is evaluated as you type
In his repo, Mark shows how to use regular expressions and automatas to solve programming riddles. In this article, we will focus on regular expressions.

Prelude

For the purposes of this code, it is useful to replace Clojure’s max / max-key with versions that return nil when passed no inputs to maximize. Also, we are going to use clojure.combinatorics by Mark Engelberg ) for generating cartesian products of sequences:
[code](ns my.combinatorics
(:require [clojure.math.combinatorics :refer [cartesian-product]]))

(defn max ([] nil) ([& s] (apply clojure.core/max s)))
(defn max-key ([k] nil) ([k & s] (apply clojure.core/max-key k s)))[/code]  The classic interview problem - maximum segment sum

A popular problem is to find an O(n) algorithm for computing the maximum sum achievable by adding up some contiguous subsequence (aka segment) of a sequence of numbers (typical input is a mix of positive and negative integers).
For example, (maximum-segment-sum [-1 2 3 -4 5 -8 4]) should return 6 because 2+3+-4+5 is 6 .
If you’ve never seen this problem before, I encourage you to go try to solve it right now. It’s a fun problem.
The trick is to keep a running sum as you traverse the sequence, never letting the running sum dip below 0. This has the effect of ignoring negative numbers that make things “too negative”. Return the highest sum you saw.
This strategy can be implemented concisely in Clojure:
[code](defn maximum-segment-sum
(apply max (reductions (comp #(max 0 %) +) 0 s)))[/code]   Let’s the results of the reductions with [-1 2 3 -4 5 -8 4] :
[code](reductions (comp #(max 0 %) +) 0 [-1 2 3 -4 5 -8 4])[/code]   The max is 6 .
A harder problem - maximum non-segment sum

But we’re going to do something harder, we’re looking for the maximum sum among subsequences that are not a continguous segment.
For example, (maximum-non-segment-sum [-1 4 5 -3 -4]) should be 5 because 4+5+-4 = 5 , and those three numbers don’t form a segment.
We can’t choose just 4 , or just 5 , or 4+5 , because singletons and adjacent pairs are considered a segment. We can’t even choose the “empty” subsequence with a value of 0 , because that is also considered a segment. We could have chosen things like -1+5 or 5+-4 or 4+-3 , but they happen to be not as good.
Unfortunately, there’s no clever trick for solving this problem. We’re going to have to look for a more principled approach.
(If you don’t believe me, go spend a while trying to solve it, just so you can appreciate how hard this problem really is.)
Brute force with Regular expressions

Our strategy is going to be brute force:
[code](defn maximum-non-segment-sum
(->> (all-non-segment-subsequences s)
(map (partial apply +))
(apply max)))[/code]   But how to write all-non-segment-subsequences ?
First key insight is that you can represent a subsequence by applying a bitmask of 0 s and 1 s to the sequence.
"Takes a sequence and a bitmask, and returns the correpsonding subsequence"
(for [[item bit] (map vector s bitmask) :when (= bit 1)] item))[/code]  Let’s see how it works:
[code](apply-bitmask [1 2 3 4 5] [0 1 1 0 1])[/code]  We can describe the satisfactory bitmasks with a regex
[code](def non-segment-regex #"0*1+0+1(0|1)*")[/code]  What this regex says is that a non segment bitmask is a sequence of:

• 0 or 0 s
• 1 or more 1 s
• 1 or more 0 s
• a single 1
• 0 s or 1 s freely
And indeed, this regex recognizes whether a bitmask represents a non-segment
[code](re-matches non-segment-regex "011010" )[/code]  or not
[code](re-matches non-segment-regex "011110")[/code]  Now, let’s make a function that receives a identifies non-segment bitmasks:
"Takes a sequence of 0's and 1's and determines whether this represents a subsequence that is not a contiguous segment"

(not (nil? (re-matches non-segment-regex (clojure.string/join s)))))[/code]  It works as expected:
[code](defn maximum-segment-sum
(apply max (reductions (comp #(max 0 %) +) 0 s)))0[/code]   Now, we are ready to write our all-non-segment-subsequences : we will generate all the 0 s and 1 s sequences of the desired length and filter with non-segment-bitmask? .
We will use the cartesian-product from clojure.combinatorics :
[code](defn maximum-segment-sum
(apply max (reductions (comp #(max 0 %) +) 0 s)))1[/code]   Let’s take a look at all the non-segment subsequences of [1 2 -3 4 -5] :
[code](defn maximum-segment-sum
(apply max (reductions (comp #(max 0 %) +) 0 s)))2[/code]   And now, all the pieces of the puzzle are in place in order to run the maximum-non-segment-sum that we wrote above:
[code](defn maximum-segment-sum
(apply max (reductions (comp #(max 0 %) +) 0 s)))3[/code]  [code](defn maximum-segment-sum
(apply max (reductions (comp #(max 0 %) +) 0 s)))4[/code]  Please don’t try to run it with too long sequences!
(On my browser it starts to take too much time with 15 elements).
In our next article, we will show how to make our alorithm much more efficient using automatas.

dds709635 发表于 2016-10-1 03:44:12
 你觉得该怎么做呢？

 医生叫我进行光合作用，不要熬夜了。

lkbhjh..; 发表于 2016-11-3 20:01:40
 向楼主学习

 我抢，我抢，抢沙发，不加倍，好了，沙发是我的了！

 想污染一个地方有两种方法：垃圾,或是钞票.

nhix6919 发表于 2016-11-6 12:44:45
 帮顶个帖，攒人品，说不定我就会升职加薪、当上总经理、出任CEO、迎娶白富美、走上人生巅峰，嘿嘿，想想还有点小激动。

 专业抢沙发的！哈哈

 经典，收藏了！

pdgz5965 发表于 2016-11-15 15:08:33
 请你吃包辣条

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