September LeetCoding Challenge, Day 13: Insert Interval

I got mildly bored of writing these blog posts for the September LeetCoding Challenge, hence this huge gap in days between the last post and this one. I continued solving the problems, and LeetCode continued to put up challenges for the following months. Let’s see if I can at least complete the series of posts for September.

The problem for September 13 is Insert Interval. We are given a set of non-overlapping intervals, represented by their start and end points, and we are asked to return a new set of non-overlapping intervals that results from merging a new given interval to the existing set. We are also told that the original set of non-overlapping intervals is sorted according to their start point.

This problem can be solved in \(O(n)\) by iterating through the list of intervals and keeping track of an interval to merge (which originally is the new interval). Whenever we see a new interval, we decide whether we want to include it as is in the final set or merge with the new interval (if it overlaps). The following is an implementation of that idea:

class Solution {
public:
  vector<vector<int>>
  insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
    vector<vector<int>> result;
    for (vector<int> interval : intervals) {
      if (interval[1] < newInterval[0]) {
        result.push_back(interval);
      } else if (interval[0] > newInterval[1]) {
        result.push_back(newInterval);
        newInterval = interval;
      } else if (interval[1] >= newInterval[0] ||
                 interval[0] <= newInterval[1]) {
        newInterval = {min(interval[0], newInterval[0]),
                       max(interval[1], newInterval[1])};
      }
    }
    result.push_back(newInterval);
    return result;
  }
};

In order to simplify the logic of handling the fact that we’ve gone past the new interval to insert, the previous solution keeps replacing the interval to merge with the current interval once the starting point of the intervals exceeds the ending point of the interval to merge. This allows us to always push the interval to merge at the end, without having special considerations on whether or not it should be included. An alternative approach, but probably more error-prone to implement, would be to keep track if the merged interval had already been inserted or not.