Merging Ranges In C++

A simple algorithm would be:

• Sort the ranges by starting values
• Iterate over the ranges from beginning to end, and whenever you find a range that overlaps with the next one, merge them

Boost.Icl might be of use for you.

The library offers a few templates that you may use in your situation:

• interval_set — Implements a set as a set of intervals - merging adjoining intervals.
• separate_interval_set — Implements a set as a set of intervals - leaving adjoining intervals separate
• split_interval_set — implements a set as a set of intervals - on insertion overlapping intervals are split

There is an example for merging intervals with the library :

interval<Time>::type night_and_day(Time(monday,   20,00), Time(tuesday,  20,00));
interval<Time>::type day_and_night(Time(tuesday,   7,00), Time(wednesday, 7,00));
interval<Time>::type  next_morning(Time(wednesday, 7,00), Time(wednesday,10,00));
interval<Time>::type  next_evening(Time(wednesday,18,00), Time(wednesday,21,00));

// An interval set of type interval_set joins intervals that that overlap or touch each other.
interval_set<Time> joinedTimes;
joinedTimes.insert(night_and_day);
joinedTimes.insert(day_and_night); //overlapping in 'day' [07:00, 20.00)
joinedTimes.insert(next_morning);  //touching
joinedTimes.insert(next_evening);  //disjoint

cout << "Joined times  :" << joinedTimes << endl;

and the output of this algorithm:

Joined times  :[mon:20:00,wed:10:00)[wed:18:00,wed:21:00)

And here about complexity of their algorithms:

Time Complexity of Addition

What you need to do is:

1. Sort items lexicographically where range key is [r_start,r_end]

2. Iterate the sorted list and check if current item overlaps with next. If it does extend current item to be r[i].start,r[i+1].end, and goto next item. If it doesn't overlap add current to result list and move to next item.

Here is sample code:

    vector<pair<int, int> > ranges;
    vector<pair<int, int> > result;
    sort(ranges.begin(),ranges.end());
    vector<pair<int, int> >::iterator it = ranges.begin();
    pair<int,int> current = *(it)++;
    while (it != ranges.end()){
       if (current.second > it->first){ // you might want to change it to >=
           current.second = std::max(current.second, it->second); 
       } else {
           result.push_back(current);
           current = *(it);
       }
       it++;
    }
    result.push_back(current);

O(n*log(n)+2n):

• Make a mapping of r1_i -> r2_i,
• QuickSort upon the r1_i's,
• go through the list to select for each r1_i-value the largest r2_i-value,
• with that r2_i-value you can skip over all subsequent r1_i's that are smaller than r2_i