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Given $n$ segments on a line, each described by a pair of coordinates $(a_{i1}, a_{i2})$. We have to find the length of their union.

The following algorithm was proposed by Klee in 1977. It works in $O(n\log n)$ and has been proven to be the asymptotically optimal.

## 1. Solution

We store in an array $x$ the endpoints of all the segments sorted by their values. And additionally we store whether it is a left end or a right end of a segment. Now we iterate over the array, keeping a counter $c$ of currently opened segments. Whenever the current element is a left end, we increase this counter, and otherwise we decrease it. To compute the answer, we take the length between the last to $x$ values $x_i – x_{i-1}$, whenever we come to a new coordinate, and there is currently at least one segment is open.

## 2. Implementation

int length_union(const vector<pair<int, int>> &a) { int n = a.size(); vector<pair<int, bool>> x(n*2); for (int i = 0; i < n; i++) { x[i*2] = {a[i].first, false}; x[i*2+1] = {a[i].second, true}; } sort(x.begin(), x.end()); int result = 0; int c = 0; for (int i = 0; i < n * 2; i++) { if (i > 0 && x[i].first > x[i-1].first && c > 0) result += x[i].first - x[i-1].first; if (x[i].second) c--; else c++; } return result; }