Problem 55. Counting Sequence

Created by Cody Team

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3513 Solutions
1346 Solvers

Last Solution submitted on May 24, 2022

Last 200 Solutions

Problem Comments

12 Comments

12 Comments

Show 9 older comments

Yuri K on 16 Feb 2012

The problem does not define correct solution if same number is not in consecutive order, like x = [5 5 2 1 1 5 3]; Should it be three 5's, or two 5's then one 5? If it's an encoding problem the correct solution should be the last, right?

Richard Zapor on 19 May 2012

Typical DPCM encoding removes the DC and thus starts at 0, not 1. An 11% enhancement in encoding.

Jean-Marie Sainthillier on 22 Jul 2012

Good problem.

Jean-Marie Sainthillier on 13 Mar 2013

Most of the solutions with UNIQUE don't work when same number is not in consecutive order.

Alfonso Nieto-Castanon on 13 Mar 2013

in the new test case 5, with x=[1 2 2 1], the test suite answer is [2 1 2 2] (two 1's, two 2's). I would expect the correct answer to be [1 1 2 2 1 1] (one 1, two 2's, one 1). I guess I am misinterpreting the question?

Jean-Marie Sainthillier on 14 Mar 2013

No. you are right. The test case 5 is wrong.

Ned Gulley on 14 Mar 2013

My mistake guys. Sorry about that. By the way, these comments are very helpful for adding (and fixing) bad test cases. Thanks for taking the time to comment.

Santosh Kumar Choudhary on 28 Sep 2017

Test case 5 answer is wrong. the correct answer is [2 1 2 2]. Change the test case 5 answer

Deepthi Palagati on 19 Sep 2018

test case 5 answer is wrong

Hans Kramer on 23 Oct 2018

Just to clarify: test case 5 isnt wrong.

Sanzhar Askaruly on 7 Jul 2019

wow, last test case is so fun!

Bryan Lambo on 6 Jan 2020

This only took me a few minutes until I saw Case 5 haha! Fun and mind boggling

Solution Comments

Solution 6616294

Solution 2487947

1 Comment

1 Comment

hang on 7 Jun 2020

yes!

Solution 2302540

1 Comment

1 Comment

Haoyu Dai on 20 May 2020

function y = CountSeq(x)
ele=[];
cnt=[];
ele=[ele x(1)];
cnt=ones(1,20);
L=length(x);
k=1;
for i=2:L
if x(i)==x(i-1)
cnt(k)=cnt(k)+1;
else
ele=[ele x(i)];
k=k+1;
end
end
L1=length(ele);
y=zeros(1,2*L1);
for i=1:L1
y(2*i-1)=cnt(i);
y(2*i)=ele(i);
end
end

Solution 1639939

2 Comments

2 Comments

hang on 7 Jun 2020

can't understandable

andrea on 22 Jun 2021

I can not understand this solution, can someone explain it to me ?

Solution 98447

4 Comments

4 Comments

Show 1 older comment

Jean-Marie Sainthillier on 13 Mar 2013

Smart !
Doesn't work with [1 2 2 1] for example.

Ned Gulley on 13 Mar 2013

Thanks for the additional test question!

Santosh Kumar Choudhary on 28 Sep 2017

Test case 5 answer is wrong. the correct answer is [2 1 2 2]. Change the test case 5 answer

VISHWATEJ N on 2 Aug 2019

i need the code part of 3rd test case ..
can anyone help

Solution 17837

2 Comments

2 Comments

Aurelien Queffurust on 21 Feb 2012

does not work if numbers have 2 digits , for example : if x = [1 5 8 9 15 14 8 15]

Peter Wittenberg on 29 Mar 2012

Yes, it only works for 1 to 9, but that is what the problem specified. It is a good solution.

Solution 10498

1 Comment

1 Comment

Andrew Newell on 13 Jun 2013

Solutions like this look much more elegant to me than the regexp expressions; and although using 'ans' would shorten it slightly, who would do that in serious code?

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Problem 55. Counting Sequence

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