436. Find Right Interval.cpp

You are given an array of intervals, where 
intervals[i] = [starti, endi] and each starti is unique.

The right interval for an interval i is an interval j 
such that startj >= endi and startj is minimized. 
Note that i may equal j.

Return an array of right interval indices for each 
interval i. If no right interval exists for 
interval i, then put -1 at index i.

 

Example 1:

Input: intervals = [[1,2]]
Output: [-1]
Explanation: There is only one interval in the collection, so it outputs -1.
Example 2:

Input: intervals = [[3,4],[2,3],[1,2]]
Output: [-1,0,1]
Explanation: There is no right interval for [3,4].
The right interval for [2,3] is [3,4] since start0 = 3 is the smallest start that is >= end1 = 3.
The right interval for [1,2] is [2,3] since start1 = 2 is the smallest start that is >= end2 = 2.
Example 3:

Input: intervals = [[1,4],[2,3],[3,4]]
Output: [-1,2,-1]
Explanation: There is no right interval for [1,4] and [3,4].
The right interval for [2,3] is [3,4] since start2 = 3 is the smallest start that is >= end1 = 3.
 

Constraints:

1 <= intervals.length <= 2 * 104
intervals[i].length == 2
-106 <= starti <= endi <= 106
The start point of each interval is unique.











// TC -> O(nlogn)
// SC -> O(n)

class Solution {
public:
    vector<int> findRightInterval(vector<vector<int>>& intervals) {
        int n = intervals.size();
        vector <int> res(n);
        map <int, int> mp;
        for (int i = 0; i < n; i++)
            mp[intervals[i][0]] = i;

        for (int i = 0; i < n; i++) {
            int e = intervals[i][1];
            auto index = mp.lower_bound(e);
            if (index != mp.end())
                res[i] = (*index).second; 
            else
                res[i] = -1;
        }
        return res;
    }
};