Removing repeating data points but keeping the first and last ones

Hello, say I have matrix with repeating parts in the frist and second row and all zeros in the third

A=1,2,3,4,5,6,7,8,9,1,1,1,1,1,1,1,1,1,3,4,5,6

9,9,9,9,9,9,9,9,9,2,3,4,5,6,7,8,9,8,2,2,2,2

0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0

I want to keep the columns where there is no repeating in the column before and after. Additonally, I want to keep the columns where the repeating starts and ends. The resulting matrix would look like this in the end

A=1,9,1,1,3,6

9,9,2,8,2,2

0,0,0,0,0,0

How would I get this

thank you

[Image Analyst]

What if your criteria would result in the first and second row having different numbers of elements?

[Adam Danz]

I wondered that, too. For example,

A=[  1,2,3,4,5,6,7,1,1,1,1,1,1,1,1,1,1,1,3,4,5,6;
     9,9,9,9,9,9,9,9,9,9,9,4,5,6,7,8,9,8,2,2,2,2;
     0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
 %                 ^ ^ ^ ^

Is that an example of what you were thinking, Image Analyst?

My answer removes columns in any of the rows have repeats (but keeps the first and last ones).

so,

Ap =      1     1     3     6
          9     8     2     2
          0     0     0     0

Answer 1

[Adam Danz]

A is your input matrix, Ap is the trimmed matrix.

isUnq = find(~any(diff(A([1,2],:),[],2)==0, 1))+1;
keepColIdx = unique([1,isUnq, isUnq-1, size(A,2)]);

Ap = A(:,keepColIdx);

Result

Ap =
     1     9     1     1     3     6
     9     9     2     8     2     2
     0     0     0     0     0     0