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Let I have a 1D array A of real positive increasing integers of length N. I want to check if the elements in it have natural order (and not missing any natural number). let me explain it with exampls:

Example 1:

A = [ 2 5 7 8 9 14 16]

there are 3 ordered numbers from 7 to 9, [7 8 9]. I want to write a code which can show me these numbers.

Example 2:

B = [ 1 4 5 6 7 8 10 11 12 16 20 ]

in this array there are two ordered set of numbers, one [4 5 6 7 8] and other is [10 11 12]. I want my code to display both of them. hmm... I want to have power to decide the minimum length of ordered numbers.

Example 3 (final):

C = [ 1 2 3 4 5 11 14 15 16 17 18 22 25 26 30 31 32]

Let I say that minimum length to choose ordered number is 3 then, there are three such pairs in C, [1 2 3 4 5], [14 15 16 17 18] and [30 31 32].

Hope I have made my question clear.

Adam Danz
on 28 Feb 2019

Edited: Adam Danz
on 28 Feb 2019

In this solution, the vector C is differentiated and then we find the starting point of each consecutive segment of 1s using bwlabel() (image processing toolbox). The rest is pretty much straight forward. I'm not sure what you're going to do with these consecutive segments but I've stored them in a cell array 'consecs' and put them into a table 'T' to show you how you pull them out of the vector C and display their start/end indices. Set the variable 'm' to the minimum length of consecutive numbers accepted.

C = [ 1 2 3 4 5 11 14 15 16 17 18 22 25 26 30 31 32];

m = 3; % minimum length of consecutive numbers

% Identify the start/end index of all consecutive numbers

dC = bwlabel(diff(C)==1); % requires image processing toolbox

startIdx = arrayfun(@(x) find(dC == x,1), unique(dC));

endIdx = arrayfun(@(x,y)sum(dC == x)+y, unique(dC), startIdx);

% Get rid of start/stop indices of length '0' (they are meaningless)

startIdx(unique(dC)==0) = [];

endIdx(unique(dC)==0) = [];

% get rid of any that are below the minimum length

len = endIdx - startIdx +1;

startIdx(len < m) = [];

endIdx(len < m) = [];

% Pull out the consecutive numbers have have a length equal to or greater than m

consecs = arrayfun(@(x,y) C(x):C(y), startIdx, endIdx, 'UniformOutput', false)';

% Show results in a table

T = table(startIdx', endIdx', consecs, 'VariableNames', {'StartIndex', 'StopIndex', 'Consecs'})

And the result

T =

3×3 table

StartIndex StopIndex Consecs

__________ _________ ____________

1 5 [1×5 double]

7 11 [1×5 double]

15 17 [1×3 double]

>> T.Consecs{1}

ans =

1 2 3 4 5

>> T.Consecs{2}

ans =

14 15 16 17 18

>> T.Consecs{3}

ans =

30 31 32

Adam Danz
on 28 Feb 2019

Jos (10584)
on 28 Feb 2019

Edited: Jos (10584)
on 28 Feb 2019

I cannot resist to put the "simple" matlab version here. Note that this problem is closely related to Run-Length Encoding schemes.

% the data

C = [ 1 2 3 4 5 11 14 15 16 17 18 22 25 26 30 31 32]

MinLength = 3 ;

% the engine

dC = diff([false diff(C)==1 false]) % true for consecutive numbers in C ("natural order")

% now finding the sequences of trues in dC is easy

S1 = find(dC == 1) % start indices of these sequences

S2 = find(dC == -1) % end indices

r = find(S2-S1+1 >= MinLength) ; % select those sequences with minimum lengths

OUT = arrayfun(@(k) C(S1(k):S2(k)), r, 'un', 0) % get those sequences

Sven
on 28 Feb 2019

This example is even fast for large arrays and looks easier to understand. So depending on how large the array you are searching is this might be more appropriate.

function sets = getUniqueSetsOfOrderedNumbers(arr, minLength)

sets = {};

k = 0;

setLength = 1;

for i = 2:length(arr)

% Check the difference between neighbouring numbers

if arr(i) - arr(i-1) == 1

setLength = setLength + 1;

else % the natural number order broke

if setLength >= minLength % Yes? -> we found a long enough set

k = k + 1;

sets{k} = arr(i-setLength:i-1);

end

setLength = 1;

end

end

% Check again in case arr ended with ordered numbers

% In this case it didn't enter the else block to check num again

if setLength >= minLength

k = k + 1;

sets{k} = arr(i-setLength+1:i);

end

end

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