Help with Cody's 'Problem 672. Longest run of consecutive numbers', I wrote 'if a(i)==a(i+1)' and it says error

Kurt on 30 Aug 2023

I found I didn't need the "vertcat" function.

Here is my code inside the function with Comments to explain the logic behind the "diff" & "find" functions.

% Convert the Vector or Matrix into a single column Vector to simplify indexing and Consecutive Run Counting

b = a(:);

% Compare the values of the next adjacent element to the current element, the original vector length is reduced by 1; Zeros "flag" where the same Consecutive Number starting location is, the next nonzero value where it stops

c = diff(b);

% Surround the Difference Calculation Vector with nonzero elements. The original Vector length is now increased by 1

d = [1;c;1];

% Locate the Indices of the nonzero elements using the "find" function. The first & last indices are the ones that were added to "wrap" the vector in nonzero elements.

f = find(d);

% Count the occurance of each vector element in order of it's appearance. Values > 1 indicate a consecutive "run" but not neccessarily the most. Summing all these element values will give you the original Vector length.

g = diff(f);

% Determine the most frequent reoccurance(s), "mode" function isn't enough, there could be multiple numbers reoccuring for the same amount of times.

h = max(g);

% Retrieve the starting indices for where the most element values occur in the original vector

k = f(g==h);

% Return the Element Values that have the Longest Run of Consecutive Numbers which matches the input style

val = a(k);

Agam Sharma on 11 Jun 2022

count=zeros(1,length(a))

for i=2:length(a)

if(a(i)==a(i-1))

count(i)=count(i-1)+1

end

[m e]=max(count)

a(e)

Hikmat Binyaminov on 27 Feb 2021

function val=longrun(a)
sz = size(a);
n = zeros(sz);
k = 1;
n(k)=1;
for i=2:length(a)
    if a(i)-a(i-1) == 0
       n(k)=n(k)+1;
    else
       k = k+1; 
       n(k)=1;
    end
end
n=n(n~=0);
I = find(n==max(n));
for j=1:length(I)
    val(j) = a(sum(n(1:I(j))));  
end
if sz(1)~=1;
  val=val';
end
end

Ciro Bermudez on 8 Dec 2020

In my opinion It's a great problem, this is the solution I came up with

function val=longrun(a)
count = 1; max_count = 1; val = a(1); idx = 1;
for i = 2:length(a) 
    if a(i) == a(i-1)
        count = count + 1;
    else 
        count = 1;
    end
    
    if max_count == count
%         if val(idx) ~=  a(i)
            idx = idx + 1;
            val(idx) = a(i);
%         end
    end
    
    if max_count < count 
        max_count = count;
        val = a(i);
        idx = 1;
    end
end
if size(a,2) < size(a,1)
    val = val.';
end
end

Something weird happens when you use the complete code, uncommenting the if statement, in my opinion gives a better solution, the down side is that it wouldn't pass Cody tests. Hopes it helps other people with ideas to solve this problem.

Peeyush Pankaj on 28 Oct 2020

While diff() approach gives you good scores, here's logic based solution for the ones who want to understand how it should be done

function val=longrun(a)
N=max(size(a,1),size(a,2));
count=1; max_count=1; reset=0; j=1;
for i=2:N
    if a(i) == a(i-1)
        count=count+1;
        % Enter this if loop for the first set of continuos numbers
        if reset==0
            max_count=count;
            val(reset+1)=a(i);
        % Elseif condition for next set of continuos numbers, but matching the max_count with previously detected     
        elseif reset>0 & count==max_count
            j=j+1;
            val(j)=a(i)
        elseif count>max_count
            max_count=count; j=1;
            val(j)=a(i);
        end
            
    elseif count>1  
        count=1; reset=reset+1
    else 
        count=1;
    end
end
% If there's no duplicate values in a row
if reset==0 & max_count==1
    val=a
% For outputing as column vector
elseif size(a,1)>size(a,2)
    val=val'
end
end

1 Reply

aziz muhammed akkaya on 15 Mar 2021

Thank you for the code, it really helped me understand the logic. However, there is a subtle gap in the code. In the condition count>max_count if val is not emptied, it repeats the max number. For example in a case like this one a = [1 1 1 2 2 2 2 3 3 3], the answer will be [2 2]. So, in that condition val should be emptied somewhere before it is added the new value.

Again thank you for the code, it helped a lot.

David Hill on 23 Mar 2020

function ans = longrun(a)
    find(vertcat(1,diff(a(:)),1));
    a(ans(diff(ans)==max(diff(ans))));
end

Anand H Lokapur on 23 Mar 2020

could you please provide solution or the snippet of the main part, I have tried to figure this out but can't and now I really want to know how to .

N/A on 13 Jan 2020

Thank you so much, David! That function came really useful.

David Hill on 11 Jan 2020

By cheating in my opinion. The real best solution is about 32. You should look at the function diff(). The use of diff() provides the best solution.

N/A on 11 Jan 2020

Thanks a lot, guys! John, I'm sorry about the other post. I thought I couldn't add more info and new questions on here.

Could you guys help me with this problem? Now that I've figured out the errors, it just doesn't behave as expected. I know there is stuff that I haven't addressed still (like the inclusion of 0s and negative numbers), but I don't know how to go on.

Here's my try:

[x,y] = size(a);
counter = zeros(1,10);
  if x == 1
      for i=1:1:y-1
          if a(i) == a(i+1)
              counter(a(i)) = counter(a(i))+1
          end
      end
  else
            for i=1:1:x-1
          if a(i) == a(i+1)
              counter(a(i)) = counter(a(i))+1
          end
      end
  end
  
  if counter ~= zeros(1,10) %if there is no repeated number anywhere, like in [1:5]
    res = find(counter == max(counter))
  else
      res = []; %the output is empty
  end
  
  if x == 1
      val = res;
  elseif y == 1
      val = res'; %if the input is a column, the output should also be a column
  end

Also, I saw that the leading solution size is like 9. How is that even possible?

Aquatris on 10 Jan 2020

One thing is you are trying to access a(12) where a has only 11 variables. This can be solved by changing "for i = 1:1:y" to "for i = 1:1:(y-1)". Because you dont want to compare a(y) with a(y+1) since a(y+1) does not exist,

If you make that change, it seems like it works.

John D'Errico on 10 Jan 2020

For a vector of length 10, what happens when i is 10? I.e., what will happen here:

if a(i) == a(i+1)

what would you expect matlab to do when it tries to access a(11)? Remember, it only has 10 elements.

N/A on 10 Jan 2020

Oh, I didn't know that, thank you! But still, why isn't it working?

Aquatris on 10 Jan 2020

What "for i = 1:1:11" does is not to assign the vector to i but individual elements in the vector to the i in each iteration. First iteration, i will be 1, second iteration i will be 2 and so on.

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Help with Cody's 'Problem 672. Longest run of consecutive numbers', I wrote 'if a(i)==a(i+1)' and it says error

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