README.md

436. Find Right Interval

You are given an array of intervals, where intervals[i] = [starti, endi] and each starti is unique.

The right interval for an interval i is an interval j such that startj >= endi and startj is minimized.

Return an array of right interval indices for each interval i. If no right interval exists for interval i, then put -1 at index i.

Example 1:

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

Example 2:

Input: intervals = [[3,4],[2,3],[1,2]]
Output: [-1,0,1]
Explanation: There is no right interval for [3,4].
The right interval for [2,3] is [3,4] since start0 = 3 is the smallest start that is >= end1 = 3.
The right interval for [1,2] is [2,3] since start1 = 2 is the smallest start that is >= end2 = 2.

Example 3:

Input: intervals = [[1,4],[2,3],[3,4]]
Output: [-1,2,-1]
Explanation: There is no right interval for [1,4] and [3,4].
The right interval for [2,3] is [3,4] since start2 = 3 is the smallest start that is >= end1 = 3.

Constraints:

• 1 <= intervals.length <= 2 * 104
• intervals[i].length == 2
• -106 <= starti <= endi <= 106
• The start point of each interval is unique.

Solutions (Ruby)

1. Sort

# @param {Integer[][]} intervals
# @return {Integer[]}
def find_right_interval(intervals)
  starts = []
  ends = []
  i = 0
  ret = [-1] * intervals.size

  (0...intervals.size).each do |j|
    starts.push([intervals[j][0], j])
    ends.push([intervals[j][1], j])
  end
  starts.sort!
  ends.sort!

  (0...ends.size).each do |j|
    i += 1 while i < ends.size && ends[j][0] > starts[i][0]
    break if i >= ends.size

    ret[ends[j][1]] = starts[i][1]
  end

  ret
end

Solutions (Rust)

1. Sort

impl Solution {
    pub fn find_right_interval(intervals: Vec<Vec<i32>>) -> Vec<i32> {
        let mut starts = vec![];
        let mut ends = vec![];
        let mut i = 0;
        let mut ret = vec![-1; intervals.len()];

        for j in 0..intervals.len() {
            starts.push((intervals[j][0], j));
            ends.push((intervals[j][1], j));
        }
        starts.sort_unstable();
        ends.sort_unstable();

        for j in 0..ends.len() {
            while i < ends.len() && ends[j].0 > starts[i].0 {
                i += 1;
            }
            if i >= ends.len() {
                break;
            }
            ret[ends[j].1] = starts[i].1 as i32;
        }

        ret
    }
}