Index the first maximum value in a sequence of consecutive numbers

I have a column vector with the following sequences:

x = [0,0,1,1,1,2,0,1,2,2,3,3,3,3,3,0,0,0]';

I would like a logical vector a, equalizing 1 if it is equal to the first maximum value of the particular sequence of consecutive numbers (the sequences are separated by zeros), ie:

x = [0,0,1,1,1,2,0,1,2,2,3,3,3,3,3,0,0,0]'; 
a = [0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0]';

The code should be as efficient as possible (No loops).

[Jos (10584)]

Homework? What have you tried so far?

Why are loops excluded? Often loops are most efficient ;)

Answer 1

[Birdman]

Without for loop:

a=zeros(1,numel(x));
idx=find([0 diff(x)~=0]);
idxx=find(x(idx)==0)-1;
a(idx(idxx))=1

[Jos (10584)]

this will critically fail for inputs like

x = [0 1 2 1 2 0]

and even worse, for:

x = [0 99 1 2 0]

[Birdman]

@Jos: it says consecutive.

[Jos (10584)]

Consecutive is not the same as monotonically increasing.

It's up to the OP if this solves his problem. Apparently so, because your nice(!) answer was accepted, but I just wanted to point out some issues.

[Jan]

@Birdman: For large x, the code find([0 diff(x)~=0]) can be slightly improved: If you concatenate two vectors, the resulting type is determined by the "larger" (concerning the class) one. diff(x)~=0 is a logical vector, but the initial 0 is a double. Then [0, diff(x)~=0] is a double vector, while [false, diff(x)~=0] would be a logical vector, which needs an 8th of the memory only.

The code fails for the input [3, 3, 0] due to a negative subscript. For [1, 2, 3] the output is [0,0,0], but this may be wanted.

[Birdman]

Thanks Jan for your attention.