Question:
I need to solve an exercise with the following statement:
As input we have a vector of n elements, where each element has a value between 0 and k . In addition we have two integer values a and b . We must answer how many elements in the input vector have value in the interval [ a , b ]. Design an algorithm whose complexity is O(1) for this problem.
I can't imagine any algorithm with constant complexity to solve this. The fastest ones I know (linear sort) don't meet because they have O(n) complexity. Would it be possible to solve this problem with O(1) complexity? Or is there an error in the wording?
Answer:
I cannot give you a rigid mathematical proof, but I can safely say that what is asked is impossible. For any nontrivial case:

k < 0
(only admits empty vector with input, return is zero) 
a > k
(return is always zero) 
b < 0
(return is always zero) 
b < a
(return is always zero) 
a = 0 E b = k
(the return is alwaysn
– assuming the size of the vector can be determined in constant time)
It is possible to produce two distinct vectors:

V + [x]
, wherex ∈ [a,b]

V + [y]
, wherey ∉ [a,b]
which will produce different results if evaluated by this function.
Suppose there is an algorithm with O(1)
complexity. So, for vectors of size greater than or equal to n0
there is a constant K
such that the running time is less than or equal to K
. Let V
be the vector (one of the vectors?) whose running time is the longest (ie K
), and whose return value is R
. How would this algorithm evaluate V + [_]
?
 If it does not perform any additional operations to the ones it has already performed to evaluate
V
, then it cannot deterministically say whether the result isR
orR + 1
– since the additional element can be at leastx
ory
;  If it performs one or more additional operations, then its running time is greater than
K
(assuming an operation has a running time greater than zero). That is, the algorithm does not haveO(1)
complexity order, as a presupposition.
By the way, this also means that – contrary to the assumption – vector V
was not in fact the one with the "longest running time of all". In fact, such a vector does not exist – since the set of all possible vectors with elements from 0
to k
is infinite (although enumerable). Any vector you choose as "the largest" can – as demonstrated – give rise to an vector whose running time is at least one elementary operation greater than it.
The best algorithm that solves this problem will have a complexity of O(n)
. It is inevitable that all elements of the vector are visited at least once – either to compare, to include in a count, or to be the object of an arithmetic operation. Unless there's some "gotcha" I'm not seeing, this exercise has no solution…