436.hpp

#ifndef LEETCODE_436_HPP
#define LEETCODE_436_HPP

#include <iostream>
#include <queue>
#include <algorithm>
#include <vector>
#include <unordered_map>
#include <unordered_set>
#include <set>
#include <numeric>
#include <stack>
#include <string>
#include <map>
#include "../common/leetcode.hpp"

using namespace std;

/*
Given a set of intervals, for each of the interval i,
 check if there exists an interval j whose start point is bigger than
 or equal to the end point of the interval i,
 which can be called that j is on the "right" of i.

For any interval i, you need to store the minimum interval j's index,
 which means that the interval j has the minimum start point to build
 the "right" relationship for interval i.
 If the interval j doesn't exist, store -1 for the interval i.
 Finally, you need output the stored value of each interval as an array.

Note:
You may assume the interval's end point is always bigger than its start point.
You may assume none of these intervals have the same start point.
Example 1:
Input: [ [1,2] ]

Output: [-1]

Explanation: There is only one interval in the collection, so it outputs -1.
Example 2:
Input: [ [3,4], [2,3], [1,2] ]

Output: [-1, 0, 1]

Explanation: There is no satisfied "right" interval for [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point;
For [1,2], the interval [2,3] has minimum-"right" start point.
Example 3:
Input: [ [1,4], [2,3], [3,4] ]

Output: [-1, 2, -1]

Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point.
 */


class Solution {
public:
    vector<int> findRightInterval(vector<Interval> &intervals) {
        map<int, int> s2idx;
        for (int i = 0; i < intervals.size(); ++i) {
            if (s2idx.count(intervals[i].start) == 0) {
                s2idx[intervals[i].start] = i;
            }
        }
        vector<int> res(intervals.size());
        for (int j = 0; j < res.size(); ++j) {
            auto cur = s2idx.lower_bound(intervals[j].end);
            res[j] = cur == s2idx.end() ? -1 : cur->second;
        }
        return res;
    }
};


#endif //LEETCODE_436_HPP