# Find indices where runs of zeros and ones ends

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Awais Saeed on 3 Aug 2021
Edited: Stephen23 on 4 Aug 2021
Assuming that I have a vector
H = [0 0 0 0 1 1 1 1 0 1 0 1 0 0 0 0 0];
I want to know the indices where long runs of 1's and 0's ends. Let's say if a bit repeats 4 times or greater, then it is a run of length greater than 4. According to this, I want to know the indices where all runs ends. First run is of zeros which ends at index 4. Second run is of ones which ends at 8. Third run is of zeros again and it ends at 17.
So far, I am able to count the run length only as following
i = find(diff(H));
run_length = [i numel(H)] - [0 i];
run_length =
4 4 1 1 1 1 5
However, the needed answer is [4 8 17]. cumsum would not work because it will give cummulative sum of whole vector as
cumsum(run_length)
ans =
4 8 9 10 11 12 17

Jonas on 3 Aug 2021

i would use

```H = [0 0 0 0 1 1 1 1 0 1 0 1 0 0 0 0 0];
Hstr=erase(num2str(H),' ');
[~,onesEnds]=regexp(Hstr,{'0000+','1111+'});
sort(cell2mat(onesEnds))
```
Stephen23 on 4 Aug 2021
Simpler and more efficient:
H = [0,0,0,0,1,1,1,1,0,1,0,1,0,0,0,0,0];
S = sprintf('%u',H);
X = regexp(S,'(0{4,}|1{4,})','end') % only zeros and ones
X = 1×3
4 8 17
X = regexp(S,'(.)(??\$1{3,})','end') % any character
X = 1×3
4 8 17