Extract certain data from array where repeats are present (but constant values)

Hello, I have some data where I my relevant column (Y position) is as this yellow graph

I want to pull out just the data that is between the green arrows - I do know the actual starting and finishing Y positions (101 and 121 respectively) , but there are other values at these values - but they are constant

My data is actually in an array, and the Yposition is column 2, but I delete the 1st column, so the column becomes 1

this was my attempt to remove all rows of my array outside of these green arrows.

            % Get user input to the real starting and finishing Y values
            prompt = {'Col:','start:','finish:'};
            dlgtitle = 'Keep Rows Between';
            dims = [1 35];
            definput = {'1','101','121'};     % definput = {'Region:',num2str(X),num2str(Y)};
            answer = inputdlg(prompt,dlgtitle,dims,definput);       
            col=str2num(answer{1}); low=str2num(answer{2}); high=str2num(answer{3});

            % Get Current table data
            t=app.UITable;
            d=t.Data;     

            d=rmmissing(d);   % One wat Remove Nans
           
            Y=table2array(d);
            Y=Y(:,col);

            low
            head(Y)
            class(Y)

            % Get index of 
            idx1=find(Y<=low);
            idx2=find(Y>=high);
            size(idx1)
            size(idx2)
            id=[idx1;idx2];

            d(id,:)=[];
            t.Data=d;

However, this keeps all values not just within the greena rrows.

I have included my data

Answer 1

[Matt J]

Why not as follows,

low= find(Y==min(Y),1,'last');
high=find(Y==max(Y),1,'first');

d=d(low:high,:);

[Walter Roberson]

It is not immediately obvious to me that the data will definitely end up higher than than the lower bump. Perhaps that will always be the case, but I don't know that for certain.

[Jason]

Yes thats my concern. what is true, is for the values I want (in between the green lines, the value is always higher than the previous one

[Matt J]

Then, perhaps,

dY=gradient(Y);
low= find(Y==min(Y),1,'last');
high=find(dY==max(dY),1,'last');

d=d(low:high,:);

[Jason]

Thanks Matt