arrays - Find shortest subarray containing all elements

Question

Suppose you have an array of numbers, and another set of numbers. You have to find the shortest subarray containing all numbers with minimal complexity.

The array can have duplicates, and let's assume the set of numbers does not. It's not ordered - the subarray may contain the set of number in any order.

For example:

Array: 1 2 5 8 7 6 2 6 5 3 8 5
Numbers: 5 7

Then the shortest subarray is obviously Array[2:5] (python notation).

Also, what would you do if you want to avoid sorting the array for some reason (a la online algorithms)?

Answer 1

Proof of a linear-time solution

I will write right-extension to mean increasing the right endpoint of a range by 1, and left-contraction to mean increasing the left endpoint of a range by 1. This answer is a slight variation of Aasmund Eldhuset's answer. The difference here is that once we find the smallest j such that [0, j] contains all interesting numbers, we thereafter consider only ranges that contain all interesting numbers. (It's possible to interpret Aasmund's answer this way, but it's also possible to interpret it as allowing a single interesting number to be lost due to a left-contraction -- an algorithm whose correctness has yet to be established.)

The basic idea is that for each position j, we will find the shortest satisfying range ending at position j, given that we know the shortest satisfying range ending at position j-1.

EDIT: Fixed a glitch in the base case.

Base case: Find the smallest j' such that [0, j'] contains all interesting numbers. By construction, there can be no ranges [0, k < j'] that contain all interesting numbers so we don't need to worry about them further. Now find the smallestlargest i such that [i, j'] contains all interesting numbers (i.e. hold j' fixed). This is the smallest satisfying range ending at position j'.

To find the smallest satisfying range ending at any arbitrary position j, we can right-extend the smallest satisfying range ending at position j-1 by 1 position. This range will necessarily also contain all interesting numbers, though it may not be minimal-length. The fact that we already know this is a satisfying range means that we don't have to worry about extending the range "backwards" to the left, since that can only increase the range over its minimal length (i.e. make the solution worse). The only operations we need to consider are left-contractions that preserve the property of containing all interesting numbers. So the left endpoint of the range should be advanced as far as possible while this property holds. When no more left-contractions can be performed, we have the minimal-length satisfying range ending at j (since further left-contractions clearly cannot make the range satisfying again) and we are done.

Since we perform this for each rightmost position j, we can take the minimum-length range over all rightmost positions to find the overall minimum. This can be done using a nested loop in which j advances on each outer loop cycle. Clearly j advances by 1 n times. Since at any point in time we only ever need the leftmost position of the best range for the previous value of j, we can store this in i and just update it as we go. i starts at 0, is at all times <= j <= n, and only ever advances upwards by 1, meaning it can advance at most n times. Both i and j advance at most n times, meaning that the algorithm is linear-time.

In the following pseudo-code, I've combined both phases into a single loop. We only try to contract the left side if we have reached the stage of having all interesting numbers:

# x[0..m-1] is the array of interesting numbers.
# Load them into a hash/dictionary:
For i from 0 to m-1:
    isInteresting[x[i]] = 1

i = 0
nDistinctInteresting = 0
minRange = infinity
For j from 0 to n-1:
    If count[a[j]] == 0 and isInteresting[a[j]]:
        nDistinctInteresting++
    count[a[j]]++

    If nDistinctInteresting == m:
        # We are in phase 2: contract the left side as far as possible
        While count[a[i]] > 1 or not isInteresting[a[i]]:
            count[a[i]]--
            i++

        If j - i < minRange:
            (minI, minJ) = (i, j)

count[] and isInteresting[] are hashes/dictionaries (or plain arrays if the numbers involved are small).