<p>You are given an array of <code>intervals</code>, where <code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code> and each <code>start<sub>i</sub></code> is <strong>unique</strong>.</p>

<p>The <strong>right interval</strong> for an interval <code>i</code> is an interval <code>j</code> such that <code>start<sub>j</sub> &gt;= end<sub>i</sub></code> and <code>start<sub>j</sub></code> is <strong>minimized</strong>. Note that <code>i</code> may equal <code>j</code>.</p>

<p>Return <em>an array of <strong>right interval</strong> indices for each interval <code>i</code></em>. If no <strong>right interval</strong> exists for interval <code>i</code>, then put <code>-1</code> at index <code>i</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> intervals = [[3,4],[2,3],[1,2]]
<strong>Output:</strong> [-1,0,1]
<strong>Explanation:</strong> There is no right interval for [3,4].
The right interval for [2,3] is [3,4] since start<sub>0</sub> = 3 is the smallest start that is &gt;= end<sub>1</sub> = 3.
The right interval for [1,2] is [2,3] since start<sub>1</sub> = 2 is the smallest start that is &gt;= end<sub>2</sub> = 2.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> intervals = [[1,4],[2,3],[3,4]]
<strong>Output:</strong> [-1,2,-1]
<strong>Explanation:</strong> There is no right interval for [1,4] and [3,4].
The right interval for [2,3] is [3,4] since start<sub>2</sub> = 3 is the smallest start that is &gt;= end<sub>1</sub> = 3.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= intervals.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>intervals[i].length == 2</code></li>
	<li><code>-10<sup>6</sup> &lt;= start<sub>i</sub> &lt;= end<sub>i</sub> &lt;= 10<sup>6</sup></code></li>
	<li>The start point of each interval is <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    int binary_search(vector<pair<pair<int, int>, int>>& sorted_intervals, int target) {
        int n = sorted_intervals.size();
        if(sorted_intervals[n-1].first.first < target)
            return -1;

        int start = 0, end = n - 1;
        int res = -1;
        while(start <= end) {
            int mid = start + (end - start) / 2;
            
            if(sorted_intervals[mid].first.first >= target) {
                res = mid;
                end = mid - 1;            // re-scan in the LEFT sub-section
            }
            else 
                start = mid + 1;             // scan in the RIGHT sub-section
        }

        return res == -1 ? -1 : sorted_intervals[res].second;
    }

    // using sorting + Binary search - T.C. O(n logn)        S.C. O(n)
    vector<int> findRightInterval(vector<vector<int>>& intervals) {
        int n = intervals.size();

        vector<pair<pair<int, int>, int>> sorted_intervals;
        for(int i = 0; i < n; i++) 
            sorted_intervals.push_back({{intervals[i][0], intervals[i][1]}, i});

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

        vector<int> res(n, -1);
        for(int i = 0; i < n; i++) {
            int curr_end = intervals[i][1];
            res[i] = binary_search(sorted_intervals, curr_end);
        }

        return res;
    }
};
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