I have an algorithm problem that I came across at work but have not been able to come up with a satisfactory solution for. I browsed this forum some and the closest I have come to the same problem is How to find a duplicate element in an array of shuffled consecutive integers?.
I have a list of N elements of integers which can contain the elements 1-M (M>N), further the list is unsorted. I want a function that will take this list as input and return a value in range 1-M not present in the list. The list contains no duplicates. I was hoping for an o(N) solution, with out using additional space
UPDATE: function cannot change the original list L
for instance N = 5 M = 10
List (L): 1, 2, 4, 8, 3
then f(L) = 5
To be honest i dont care if it returns an element other than 5, just so long as it meets the contraints above
The only solution I have come up with so far is using an additional array of M elements. Walking through the input list and setting the corresponding array elements to 1 if present in the list. Then iterating over this list again and returning the index of the first element with value 0. As you can see this uses additional o(M) space and has complexity 2*o(N).
Any help would we appreciated.
Thanks for the help everyone. The stack overflow community is definitely super helpful.
To give everyone a little more context of the problem I am trying to solve.
I have a set of M token that I give out to some clients (one per client). When a client is done with the token they get returned to my pile. As you can see the original order I give client a token is sorted.
so M = 3 Tokens
client1: 1 <2,3>
client2: 2 <3>
client1 return: 1 <1,3>
client 3: 3 <1>
Now the question is giving client4 token 1. I could at this stage give client 4 token 2 and sort the list. Not sure if that would help. In any case if I come up with a nice clean solution I will be sure to post it
Just realised I might have confused everyone. I do not have the list of free token with me when I am called. I could statically maintain such a list but I would rather not
You can have O(N) time and space. You can be sure there is an absent element within 1..N+1, so make an array of N+1 elements, and ignore numbers larger than N+1.
If M is large compared to N, say M>2N, generate a random number in 1..M and check if it is not on the list in O(N) time, O(1) space. The probability you will find a solution in a single pass is at least 1/2, and therefore (geometric distribution) the expected number of passes is constant, average complexity O(N).
If M is N+1 or N+2, use the approach described here.