Find Right Interval
<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>
Array; Binary Search; Sorting
class Solution {
    public int[] findRightInterval(int[][] intervals) {
        int n = intervals.length;
        int[][] arr = new int[n][0];
        for (int i = 0; i < n; ++i) {
            arr[i] = new int[] {intervals[i][0], i};
        }
        Arrays.sort(arr, (a, b) -> a[0] - b[0]);
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            int j = search(arr, intervals[i][1]);
            ans[i] = j < n ? arr[j][1] : -1;
        }
        return ans;
    }

    private int search(int[][] arr, int x) {
        int l = 0, r = arr.length;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (arr[mid][0] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}