Insert Interval

Insert Interval

Question:

You are given an array of non-overlapping intervals intervals where intervals[i] = [starti, endi] represent the start and the end of the ith interval and intervals is sorted in ascending order by starti. You are also given an interval newInterval = [start, end] that represents the start and end of another interval.

Insert newInterval into intervals such that intervals is still sorted in ascending order by starti and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary).

Return intervals after the insertion.

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with
[3,5],[6,7],[8,10].

How to Solve:

This problem can be solved by a greedy approach. Firstly, we add all the intervals whose start time is less than or equal to our newInterval's start time. Next, we push newInterval into our answer vector and merge it with the last interval in answer if needed. For the rest of the intervals, we do the same thing: we push it into our answer vector, and merge it with the most recently pushed interval if needed.

Optimization:

We can use binary search (upper_bound() in C++) to determine beforehand, the pivot at which the start times are no longer smaller or equal to newInterval's start time. However, empirical testing shows that a simple linear scan (worst case two-pass) is actually faster than binary searching for the pivot. Nevertheless, in insert.cpp you can find both solutions.

My C++ Solution:

// Linear scan (seemingly not optimized, but empirically faster)
class Solution {
 public:
  vector<vector<int>> insert(vector<vector<int>> &intervals, vector<int> &newInterval) {
    vector<vector<int>> ans;
    ans.reserve(intervals.size() + 1);
    for (auto &interval : intervals) {
      if (interval[0] <= newInterval[0]) {
        ans.push_back(interval);
      } else {
        break;
      }
    }
    if (ans.empty() || ans.back()[1] < newInterval[0]) {
      ans.push_back(newInterval);
    } else {
      ans.back()[1] = max(ans.back()[1], newInterval[1]);
    }
    for (auto &interval : intervals) {
      if (interval[0] <= newInterval[0]) {
        continue;
      }
      if (ans.back()[1] >= interval[0]) {
        ans.back()[1] = max(ans.back()[1], interval[1]);
      } else {
        ans.push_back(interval);
      }
    }
    return ans;
  }
};


// Binary Search (for some reason, it's actually slower)
class Solution {
 public:
  vector<vector<int>> insert(vector<vector<int>> &intervals, vector<int> &newInterval) {
    int n = intervals.size();
    vector<vector<int>> ans;
    ans.reserve(n + 1);
    auto comp = [](auto &v1, auto &v2) { return v1[0] < v2[0]; };
    int pivot = upper_bound(begin(intervals), end(intervals), newInterval, comp) - begin(intervals);
    for (int i = 0; i < pivot; ++i) {
      if (intervals[i][0] <= newInterval[0]) {
        ans.push_back(intervals[i]);
      }
    }
    if (ans.empty() || ans.back()[1] < newInterval[0]) {
      ans.push_back(newInterval);
    } else {
      ans.back()[1] = max(ans.back()[1], newInterval[1]);
    }
    for (int i = pivot; i < n; ++i) {
      if (ans.back()[1] >= intervals[i][0]) {
        ans.back()[1] = max(ans.back()[1], intervals[i][1]);
      } else {
        ans.push_back(intervals[i]);
      }
    }
    return ans;
  }
};