Find the Smallest Integer Not in a List

Since the OP has now specified that the original list is held in RAM and that the computer has only, say, 1GB of memory, I'm going to go out on a limb and predict that the answer is zero.

1GB of RAM means the list can have at most 134,217,728 numbers in it. But there are 264 = 18,446,744,073,709,551,616 possible numbers. So the probability that zero is in the list is 1 in 137,438,953,472.

In contrast, my odds of being struck by lightning this year are 1 in 700,000. And my odds of getting hit by a meteorite are about 1 in 10 trillion. So I'm about ten times more likely to be written up in a scientific journal due to my untimely death by a celestial object than the answer not being zero.


As pointed out in other answers you can do a sort, and then simply scan up until you find a gap.

You can improve the algorithmic complexity to O(N) and keep O(N) space by using a modified QuickSort where you eliminate partitions which are not potential candidates for containing the gap.

  • On the first partition phase, remove duplicates.
  • Once the partitioning is complete look at the number of items in the lower partition
  • Is this value equal to the value used for creating the partition?
    • If so then it implies that the gap is in the higher partition.
      • Continue with the quicksort, ignoring the lower partition
    • Otherwise the gap is in the lower partition
      • Continue with the quicksort, ignoring the higher partition

This saves a large number of computations.


Here's a simple O(N) solution that uses O(N) space. I'm assuming that we are restricting the input list to non-negative numbers and that we want to find the first non-negative number that is not in the list.

  1. Find the length of the list; lets say it is N.
  2. Allocate an array of N booleans, initialized to all false.
  3. For each number X in the list, if X is less than N, set the X'th element of the array to true.
  4. Scan the array starting from index 0, looking for the first element that is false. If you find the first false at index I, then I is the answer. Otherwise (i.e. when all elements are true) the answer is N.

In practice, the "array of N booleans" would probably be encoded as a "bitmap" or "bitset" represented as a byte or int array. This typically uses less space (depending on the programming language) and allows the scan for the first false to be done more quickly.

This is how / why the algorithm works.

Suppose that the N numbers in the list are not distinct, or that one or more of them is greater than N. This means that there must be at least one number in the range 0 .. N - 1 that is not in the list. So the problem of find the smallest missing number must therefore reduce to the problem of finding the smallest missing number less than N. This means that we don't need to keep track of numbers that are greater or equal to N ... because they won't be the answer.

The alternative to the previous paragraph is that the list is a permutation of the numbers from 0 .. N - 1. In this case, step 3 sets all elements of the array to true, and step 4 tells us that the first "missing" number is N.

The computational complexity of the algorithm is O(N) with a relatively small constant of proportionality. It makes two linear passes through the list, or just one pass if the list length is known to start with. There is no need to represent the hold the entire list in memory, so the algorithm's asymptotic memory usage is just what is needed to represent the array of booleans; i.e. O(N) bits.

(By contrast, algorithms that rely on in-memory sorting or partitioning assume that you can represent the entire list in memory. In the form the question was asked, this would require O(N) 64-bit words.)

@Jorn comments that steps 1 through 3 are a variation on counting sort. In a sense he is right, but the differences are significant:

  • A counting sort requires an array of (at least) Xmax - Xmin counters where Xmax is the largest number in the list and Xmin is the smallest number in the list. Each counter has to be able to represent N states; i.e. assuming a binary representation it has to have an integer type (at least) ceiling(log2(N)) bits.
  • To determine the array size, a counting sort needs to make an initial pass through the list to determine Xmax and Xmin.
  • The minimum worst-case space requirement is therefore ceiling(log2(N)) * (Xmax - Xmin) bits.

By contrast, the algorithm presented above simply requires N bits in the worst and best cases.

However, this analysis leads to the intuition that if the algorithm made an initial pass through the list looking for a zero (and counting the list elements if required), it would give a quicker answer using no space at all if it found the zero. It is definitely worth doing this if there is a high probability of finding at least one zero in the list. And this extra pass doesn't change the overall complexity.

EDIT: I've changed the description of the algorithm to use "array of booleans" since people apparently found my original description using bits and bitmaps to be confusing.


If the datastructure can be mutated in place and supports random access then you can do it in O(N) time and O(1) additional space. Just go through the array sequentially and for every index write the value at the index to the index specified by value, recursively placing any value at that location to its place and throwing away values > N. Then go again through the array looking for the spot where value doesn't match the index - that's the smallest value not in the array. This results in at most 3N comparisons and only uses a few values worth of temporary space.

# Pass 1, move every value to the position of its value
for cursor in range(N):
    target = array[cursor]
    while target < N and target != array[target]:
        new_target = array[target]
        array[target] = target
        target = new_target

# Pass 2, find first location where the index doesn't match the value
for cursor in range(N):
    if array[cursor] != cursor:
        return cursor
return N

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