Remove Covered Intervals

Remove Covered Intervals

Given a list of intervals, remove all intervals that are covered by another interval in the list.

Interval [a,b) is covered by interval [c,d) if and only if c <= a and b <= d.

After doing so, return the number of remaining intervals.

Example 1:

Input: intervals = [[1,4],[3,6],[2,8]]
Output: 2
Explanation: Interval [3,6] is covered by [2,8], therefore it is removed.

Example 2:

Input: intervals = [[1,4],[2,3]]
Output: 1

Example 3:

Input: intervals = [[0,10],[5,12]]
Output: 2

Example 4:

Input: intervals = [[3,10],[4,10],[5,11]]
Output: 2

Example 5:

Input: intervals = [[1,2],[1,4],[3,4]]
Output: 1

Constraints:

• 1 <= intervals.length <= 1000

• intervals[i].length == 2

• 0 <= intervals[i][0] < intervals[i][1] <= 10^5

• All the intervals are unique.

Solutions

🧠 Cpp

class Solution
{
    enum {START, END};

public:

    //O(n log n) solution: sort + 1 pass
    int removeCoveredIntervals(vector<vector<int>>& intervals)
    {
        std::sort(begin(intervals), end(intervals),
                 [](auto a, auto b)
                 {
                   return a[END] == b[END] ? a[START] > b[START] : a[END] < b[END];  
                 }
                 );

        //copy data to list for O(1) erasure
        std::list<vector<int>> i_vals(begin(intervals), end(intervals));


        //ENDs is already sorted, so next element has end >= than prev
        //only thing is to check the start of the interval, if prev has START > next start
        //then it's covered
        for(auto iter = begin(i_vals); iter != prev(end(i_vals));)
        {
            if((*iter)[START] >= (*next(iter))[START])
            {
                iter = i_vals.erase(iter);
                if(iter != begin(i_vals))
                    advance(iter, -1);
            }
            else
                ++iter;
        }

        return i_vals.size();
    }
};