Insert new interval in the given intervals
Problem Statement - link #
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.
Examples #
Example 1:
Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]
Example 2:
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].
Example 3:
Input: intervals = [], newInterval = [5,7]
Output: [[5,7]]
Example 4:
Input: intervals = [[1,5]], newInterval = [2,3]
Output: [[1,5]]
Example 5:
Input: intervals = [[1,5]], newInterval = [2,7]
Output: [[1,7]]
Constraints #
0 <= intervals.length <= 10^4
intervals[i].length == 2
0 <= starti <= endi <= 10^5
intervals
is sorted bystarti
in ascending order.newInterval.length == 2
0 <= start <= end <= 10^5
Solution #
class Solution {
public:
vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
vector<vector<int>> mnk;
int i= newInterval[0], j= newInterval[1];
bool flag = false;
for(int k=0; k < intervals.size() ; k++){
if(intervals[k][1] < i){
mnk.push_back(intervals[k]);
}else if(intervals[k][0] > j){
if(!flag){
mnk.push_back({i,j});
flag = true;
}
mnk.push_back(intervals[k]);
}else{
i = min(i,intervals[k][0]);
j = max(j,intervals[k][1]);
}
}
if(!flag){
mnk.push_back({i,j});
}
return mnk;
}
};