huffman code impelementation manual not use the built in function

hi guys any one can help me in huffman code (data compression) to be more efficiency

this code is already working but i need it to be general and more efficiency i already use the built in code to insure

clc;
clear all;
close all;
l=6;
o=5;
matrix=[0.3 0.3 0.3 0.43 0.57 1;
0.25 0.25 0.27 0.3 0.43 0;
0.15 0.18 0.25 0.27 0 0
0.12 0.15 0.18 0 0 0;
0.1 0.12 0 0 0 0;
0.08 0 0 0 0 0;];
code=cell(6,6);
index=[3,2,1,1];

code{1,6}=1;              %%%
code{1,5}=0;              %%%
code{2,5}=1;              %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
number1=2;
number2=3;


num1=2;
number=5;  % number of colum 


 
prob=[0.3 0.25  0.15 0.12 0.1 0.08 ];
[dict,avglen] = huffmandict(1:6,prob);




  rr=[]; 
  %%%% clear the zero in matrix %%%
 for i=5:-1:2
a=intersect(matrix(:,i),matrix(:,i-1));%%  intersect the the element 
code{num1,number-1}=[0,code{index(i-1),number}];
code{num1+1,number-1}=[1,code{index(i-1),number}];
number=number-1;
num1=num1+1;

[ll,~]=size(a);                       %% size of  output matrix 
for ii=1:ll
 a(a==0)=[];                          %% clear zero in array 
end

[l_new,~]=size(a);                    %% the new size of matrix after  the clear operation 

 for iu=1:l_new                        
 y=find(a(iu)==matrix(:,i));          %% find the element 
 rr=[rr y]; 
 rr=sort(rr,'descend');
 rr=flip(rr);
 end
 b=cell(1,l_new);                      %% create new cell have the index of element 
 
  %%%%%this is cmulative function according the index value%%%%
 for iy=1:l_new            
   b{1,iy}=code{rr(iy),i}; 
    for iv=1:l_new  
     code{iv,i-1}=b{1,iv}; 
    end
 end 
 rr=[];
 b={};
 end
 
 %%%%%%%%%%%%%%%%%%%final answer to calculate the average length%%%%%%%%%%%
 code=code(:,1);
 length=[];
 matrix=[];
 for i=1:l
  code{i}=flip(code{i});
 
  [~,c]=size(code{i});
 
 length=[length c];
  s=code{i};
  matrix=[matrix s];
 end 
 avl=0;
 for i=1:l
      average_length=prob(i)*length(i);
      avl=avl+average_length;
     
  end 

[Walter Roberson]

 for ii=1:ll
 a(a==0)=[];                          %%clear zero in array 
end

Why is that a loop? After the first execution there would be no more 0 left.

[Shehab Tarek]

to clear the zeros which are generate from intersect the colum 5 with colum 4 in matrix variable

LL not 11

[Walter Roberson]

a=[7 0 3 1 0 9]
a(a==0) = []

How many 0 are left in a?

a(a==0)=[]

how many more 0 did executing the statement again delete?

[Shehab Tarek]

yes delete any zero number in array

final result

a=[7 3 1 9];

[Walter Roberson]

so how many loop iteration?

[Shehab Tarek]

According to the size of array

[Walter Roberson]

no, after one iteration all of the remaining zeros are gone and you can stop.