#### Approach #1: Opening and Closing Events [Accepted]

Intuition

We can think of the problem as drawing intervals on a number line. This gives us the idea of opening and closing events.

To illustrate this concept, say we have nums = [10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13], with no 9s and no 14s. We must have two sequences start at 10, two sequences start at 11, and 3 sequences end at 12.

In general, when considering a chain of consecutive integers x, we must have C = count[x+1] - count[x] sequences start at x+1 when C > 0, and -C sequences end at x if C < 0. Even if there are more endpoints on the intervals we draw, there must be at least this many endpoints.

With the above example, count[11] - count[10] = 2 and count[13] - count[12] = -3 show that two sequences start at 11, and three sequences end at 12.

Also, if for example we know some sequences must start at time 1 and 4 and some sequences end at 5 and 7, to maximize the smallest length sequence, we should pair the events together in the order they occur: ie., 1 with 5 and 4 with 7.

Algorithm

For each group of numbers, say the number is x and there are count of them. Furthermore, say prev, prev_count is the previous integer encountered and it's count.

Then, in general there are abs(count - prev_count) events that will happen: if count > prev_count then we will add this many t's to starts; and if count < prev_count then we will attempt to pair starts.popleft() with t-1.

More specifically, when we have finished a consecutive group, we will have prev_count endings; and when we are in a consecutive group, we may have count - prev_count starts or prev_count - count endings.

For example, when nums = [1,2,3,3,4,5], then the starts are at [1, 3] and the endings are at [3, 5]. As our algorithm progresses:

• When t = 1, count = 1: starts = [1]
• When t = 2, count = 1: starts = [1]
• When t = 3, count = 2: starts = [1, 3]
• When t = 4, count = 1: starts = [3], since prev_count - count = 1 we process one closing event, which is accepted as t-1 >= starts.popleft() + 2.
• When t = 5, count = 1: starts = [3]

And at the end, we process prev_count more closing events nums[-1].

Complexity Analysis

• Time Complexity: , where is the length of nums. We iterate over the array and every event is added or popped to starts at most once.

• Space Complexity: , the size of starts.

#### Approach #2: Greedy [Accepted]

Intuition

Call a chain a sequence of 3 or more consecutive numbers.

Considering numbers x from left to right, if x can be added to a current chain, it's at least as good to add x to that chain first, rather than to start a new chain.

Why? If we started with numbers x and greater from the beginning, the shorter chains starting from x could be concatenated with the chains ending before x, possibly helping us if there was a "chain" from x that was only length 1 or 2.

Algorithm

Say we have a count of each number, and let tails[x] be the number of chains ending right before x.

Now let's process each number. If there's a chain ending before x, then add it to that chain. Otherwise, if we can start a new chain, do so.

It's worth noting that our solution can be amended to take only additional space, since we could do our counts similar to Approach #1, and we only need to know the last 3 counts at a time.

Complexity Analysis

• Time Complexity: , where is the length of nums. We iterate over the array.

• Space Complexity: , the size of count and tails.

Analysis written by: @awice. Approach #2 inspired by @compton_scatter.