## Insert Interval

06/27/2016 Array Sort

## Question

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:

Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:

Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

## Solution

Result: Accepted Time: 16 ms

Here should be some explanations.

class Solution {
public:
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
auto it = lower_bound(intervals.begin(), intervals.end(), newInterval,
[](const Interval& a, const Interval &b ){return a.start < b.start;});
vector<Interval> ret( intervals.begin(),it);
if(ret.size()&&ret.back().end >= newInterval.start)
ret.back().end = max(newInterval.end,ret.back().end);
else
ret.push_back(newInterval);
while(it!=intervals.end())
{
auto & tmp = ret.back();
if(tmp.end >= it->start)
tmp.end = max(tmp.end,it->end);
else
ret.push_back(*it);
++it;
}
return std::move(ret);
}
};


Complexity Analytics

• Time Complexity: $O(n)$
• Space Complexity: $O(1)$