## 757. Set Intersection Size At Least Two

An integer interval `[a, b]` (for integers `a < b`) is a set of all consecutive integers from `a` to `b`, including `a` and `b`.

Find the minimum size of a set S such that for every integer interval A in `intervals`, the intersection of S with A has size at least 2.

Example 1:

```Input: intervals = [[1, 3], [1, 4], [2, 5], [3, 5]]
Output: 3
Explanation:
Consider the set S = {2, 3, 4}.  For each interval, there are at least 2 elements from S in the interval.
Also, there isn't a smaller size set that fulfills the above condition.
Thus, we output the size of this set, which is 3.
```

Example 2:

```Input: intervals = [[1, 2], [2, 3], [2, 4], [4, 5]]
Output: 5
Explanation:
An example of a minimum sized set is {1, 2, 3, 4, 5}.
```

Note:

1. `intervals` will have length in range `[1, 3000]`.
2. `intervals[i]` will have length `2`, representing some integer interval.
3. `intervals[i][j]` will be an integer in `[0, 10^8]`.

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#### Approach #1: Greedy [Accepted]

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Intuition

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Let\'s try to solve a simpler problem: what is the answer when the set intersection size is at least one?

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Sort the points. Take the last interval `[s, e]`, which point on this interval will be in `S`? Since every other interval has start point `<= s`, it is strictly better to choose `s` as the start. So we can repeatedly take `s` in our set `S` and remove all intervals containing `s`.

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We will try to extend this solution to the case when we want an intersection of size two.

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Algorithm

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For each interval, we will perform the algorithm described above, storing a `todo` multiplicity which starts at `2`. As we identify points in `S`, we will subtract from these multiplicities as appropriate.

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One case that is important to handle is the following:\n`[[1, 2], [2, 3], [2, 4], [4, 5]]`. If we put `4, 5` in `S`, then we put `2` in `S`, when handling `[2, 3]` we need to put `3` in `S`, not `2` which was already put.

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We can handle this case succinctly by sorting intervals `[s, e]` by `s` ascending, then `e` descending. This makes it so that any interval encountered with the same `s` has the lowest possible `e`, and so it has the highest multiplicity. When at interval `[s, e]` and choosing points to be included into `S`, it will always be the case that the start of the interval (either `s` or `s, s+1`) will be unused.

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Complexity Analysis

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Time Complexity: , where is the length of `intervals`.

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Space Complexity: .

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Analysis written by: @awice.

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