Find Each Interval's Nearest Eligible Starting Interval

# Find Each Interval's Nearest Eligible Starting Interval You are given intervals `[start_i,end_i]` with unique starting points. For each interval `i`, find the original index `j` whose start is the smallest start satisfying `start_j >= end_i`. Return these indices in an array aligned with the original interval order. Use `-1` when no eligible start exists. An interval may select itself when its endpoints allow it. ### Function `vector<int> findRightInterval(vector<vector<int>>& intervals)` ### Example 1 ```text Input: intervals = [[4,4],[1,4],[6,8],[8,9]] Output: [0,0,3,-1] ``` The point interval at index `0` selects itself. It is also the earliest eligible start for index `1`. Index `2` selects the start at index `3`. ### Example 2 ```text Input: intervals = [[-4,-2],[-2,0],[0,0]] Output: [1,2,2] ``` ### Constraints - `1 <= intervals.length <= 20000`. - Each interval contains exactly two integers. - `-1000000 <= start_i <= end_i <= 1000000`. - All starting points are unique. ### Clarifications The comparison is against the current interval's end, and equality is allowed. A zero-length interval is valid and can be its own answer. Return original indices rather than sorted positions or endpoint values. Unique starts make the selected interval unique whenever one exists. ### What a Strong Answer Covers - The smallest eligible start and the inclusive endpoint comparison. - Original index preservation, negative endpoints, and no-answer cases. - Point intervals and permitted self-selection. - Correctness and resource bounds.