HackerRank: Bear and Steady Gene algorithm (IV)

March 5, 2016


Problem statement:

https://www.hackerrank.com/contests/hourrank-6/challenges/bear-and-steady-gene  

A gene is represented as a string of length n (where n is divisible by 4), composed of the letters A, C, T, and G. It is considered to be steady if each of the four letters occurs exactly n4 times. For example, GACT and AAGTGCCTare both steady genes.

Julia did 3 practice code using C#, but she likes one of solutions using Java, and then, she spent 5 minutes to convert it to C#.

study code:
Java code implementation to study:
https://gist.github.com/jianminchen/d52ced38de0bffa2d9e8

Julia's fourth practice:
use two pointers, sliding window, linear time solution

https://gist.github.com/jianminchen/0892107fc0f6b6bec7a0

Julia's comment:
Julia learned a few things through the code:
1.  Declare cnt array using size of 256, ascii code is 256. So, ACGT shoud be ok.
2.  Declare a string "ACGT", and then the code uses 'A', 'C', 'G', 'T' only once.
3. Only two loops, first loop, variable l - left pointer from 0 to n-1.
    second while loop, move r pointer if cnt arrray is bigger than mx - minimum
  while (cnt['A'] > mx || cnt['C'] > mx || cnt['T'] > mx || cnt['G'] > mx) {

How to paraphrase this conditional checking?
Let us go through the test case "GAAATAAA", 
cnt['A'] = 6, cnt['C'] = 0, cnt['G']= 1, cnt['T'] = 1
mx = 2, 
starting from G, and then, while loop <- get in, 
G-> GA->...->GAAATA, cnt['A'] = 2, cnt['G'] = 0,

now, it is time to exit while loop. <- making sense. Because cnt array contains the count 
for rest of substring "GAAATA", requirement is "none of 'ACGT" is more than 2".



April 14, 2016

I put the for loop into a standalone function, and then, add some comment for the function, explain the design. Basically, it is using sliding window. 

Here is the link of code, which passes HackerRank as well. 
https://gist.github.com/jianminchen/9b02beab326b2bfcd4b524f219d2946f

statistics:
This post is one of most viewed so far, over 130 views, from March 5 to April 13, 2016.

HackerRank: Bear and Steady Gene algorithm (III)

March 5, 2016

Problem statement: 

Bear and steady gene

  

A gene is represented as a string of length n (where n is divisible by 4), composed of the letters A, C, T, and G. It is considered to be steady if each of the four letters occurs exactly n4 times. For example, GACT and AAGTGCCT are both steady genes.

Practice 

Julia's 1st practice:

Code is here. 

Score 15 out of 50, there are 2 run time error, failed a few of test cases.

The algorithm ends up in time complexity O(n²), close to brute force solution.

Code study 

C# code implementation to study:

Study code is here. 

Work on two pointers, sliding window, so time complexity is O(n).

Let us go over two pointers algorithm here using example:

Test case string "GAAATAAA", gene string "ACTG". 
Go over C# program a few lines of statements here:

Line 40: 
var a = ReadToken().Select(ch=>"ACTG".IndexOf(ch)).ToArray();

so the array a = [3, 0, 0, 0, 2, 0, 0, 0] is representing the sample string "GAAATAAA", since 'G' is the index of 3 in the string "ACTG", 'A' is 0, 'C' is 1, 'T' is 2. 

Line 44 - 45:
for(int i = 0; i < 4; i++)
   target[i] = Math.Max(0, a.Count(aa => aa == i) - n / 4); 

So the Target[] = [4, 0, 0, 0]. In other words, 4 of A is needed, but C, T, G do not matter.

Line 47 - 51:
if (target.All(t => t == 0))
{
      Write(0);
      return; 
}

If all element in Target is 0, then return 0. Learn to use Array.All method. 

Left, right two pointer (l, r), starting from 0

Declare a sum array

Start a loop, loop on r

   

Add a[r]’s char into sum array,

r ++, r moves to next one

Inside the first loop, another loop for r, continue to move forward

if sums do not reach target 4 of them >= target

 

The test case "GAAATAAA", find substring: "GAAATA". 

Inside the first loop, another loop for left pointer – l

condition: l  <  r

If sums array is bigger than target array, then remove left pointer char count,

then move left pointer to next one, then the substring is reaching the target,

"GAAATA" => "AAATA"

and compare to existing smallest length

"AAATA" is the smallest one.

This algorithm is much easy to follow than the one in the blog - bear and steady gene.   

 

Reference: 

C# code implementation to study, code is here. 

HackerRank: Bear Steady Gene (II)

March 4, 2016

  Problem statement:

https://www.hackerrank.com/contests/hourrank-6/challenges/bear-and-steady-gene  

A gene is represented as a string of length n (where n is divisible by 4), composed of the letters A, C, T, and G. It is considered to be steady if each of the four letters occurs exactly n4 times. For example, GACT and AAGTGCCTare both steady genes.


Julia's 1st practice:

https://gist.github.com/jianminchen/80723bae951328a690bb

score 15 out of 50, there are 2 run time error, failed a few of test cases.

The algorithm ends up in time complexity O(n^2), close to brute force solution - O(n^2)

C# code implementation to study:

Readable code, with some analysis.

https://gist.github.com/jianminchen/fae9142eff9a6f9643fc

Work on two pointers, sliding window, so time complexity is O(2n) = O(n).

Let us go over two pointers algorithm here using example:
GAAATAAA,
A - count of A, denoted as cA = 6, n/4 = 2, so we have to change 4 of A to other thing.
A - 6 -2  = 4 , 4 of change
C - 0 - 2 = -2
T - 1 -2  = -1
G - 1-2  = -1
So, at least minimum is 4, but a substring containing 4 of A, shortest one is AAAA. But "AAAA" is not a substring of "GAAATAAA"

G  A A A T A A A
0
start
end
Let us find the substring starting from 0, but will include all 4 of A, and then, rest of string will not include any char of "ACGT" more than n/4.
GAAATA, string length is 6.
start - 0, index of 0
end - A, index of 5

continue to move start to next one, 1, then, G is moved out from substring, adjust count of chars.
AAATA can be the substring, so the length is min (6, 5) = 5. Do not need to move end pointer.

next step, start = 2, missing one of A, and then, end has to move to next one until the string fits into requirement.

Every thing should be covered in 20 minutes, problem reading, the design of algorithm, the coding.

So, Julia had second practice, and the code scores 50 out of 50 this time.

https://gist.github.com/jianminchen/61dfe437f82edb9793fc

Conclusion:
After coding, Julia understands the algorithm much better.

1. count of array for substring, cB=[4, -2, -1, -1], so in other words, substring has to contain at least 4 'A', C, G, T does not matter.

So, using CB array, sliding windows of substring GAAATA, each time, end index is moving forward, tracking the substring's char by taking off from CB; in other words, GAAATA, CB will be
[0, -2, -2, -2].

Now, it is easy to understand that GAAATA will be added to the solution set, the length is 6.
line 51, if (cBAllLessThan1(cB)) 
<-add substring to solution set, compare length<- easy to understand now.

2. the code in https://gist.github.com/jianminchen/fae9142eff9a6f9643fc has discussion:

switch(line[0]) {
	case 'A': A--; break;
	case 'C': C--; break;
	case 'G': G--; break;
	case 'T': T--; break;
	}

Julia removed those detail discussion, but her code takes more time. Because she created a bug in her writing. Debugging takes time.

3. Julia's practice in 2015, sliding windows, two pointers:
http://juliachencoding.blogspot.ca/search/label/slide%20window

April 27, 2016
change C# program to make variables more meaningful, matching the design:
code -> GENES <- global string array
function name matches design of two pointers moving.

https://gist.github.com/jianminchen/b8263048c297473319c23836e9468c14

searchStrArray -> searchStrNumber, more meaningful.
https://gist.github.com/jianminchen/b23c4f606a101b9aeec71eff3268db32