day-149.cpp

/*

  Find Right Interval



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.
NOTE: input types have been changed on April 15, 2019. Please reset to default
code definition to get new method signature.


*/

// Using binary search to solve this. O(NlogN) time complexity

class Solution {
   public:
    int binarySearch(vector<pair<int, int>>& intervalDS, int target) {
        int low = 0;
        int high = intervalDS.size() - 1;

        if (intervalDS[high].first < target) return -1;

        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (intervalDS[mid].first >= target) {
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }

        return intervalDS[low].second;
    }

    vector<int> findRightInterval(vector<vector<int>>& intervals) {
        int N = intervals.size();
        if (N <= 1) return {-1};

        vector<pair<int, int>> intervalDS;

        int idx = 0;
        for (int idx = 0; idx < N; idx++) {
            intervalDS.push_back({intervals[idx][0], idx});
        }

        sort(intervalDS.begin(), intervalDS.end());

        vector<int> answers(N, -1);
        for (int idx = 0; idx < N; idx++) {
            answers[idx] = binarySearch(intervalDS, intervals[idx][1]);
        }

        return answers;
    }
};