non-overlapping-intervals.js

/**
 * Non-overlapping Intervals
 *
 * Given a collection of intervals, find the minimum number of intervals you need to remove to make
 * the rest of the intervals non-overlapping.
 *
 * Actually this problem is the same as "Given a collection of intervals, find the maximum number of
 * intervals that are non-overlapping." see:
 * https://en.wikipedia.org/wiki/Interval_scheduling#Interval_Scheduling_Maximization
 *
 * Note:
 * You may assume the interval's end point is always bigger than its start point.
 * Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.
 * Example 1:
 * Input: [ [1,2], [2,3], [3,4], [1,3] ]
 *
 * Output: 1
 *
 * Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.
 *
 * Example 2:
 *
 * Input: [ [1,2], [1,2], [1,2] ]
 * Output: 2
 *
 * Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
 *
 * Example 3:
 *
 * Input: [ [1,2], [2,3] ]
 * Output: 0
 *
 * Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
 */

/**
 * Definition for an interval.
 * function Interval(start, end) {
 *     this.start = start;
 *     this.end = end;
 * }
 */

/**
 * @param {Interval[]} intervals
 * @return {number}
 */
const eraseOverlapIntervals = intervals => {
  if (intervals.length === 0) {
    return 0;
  }

  intervals.sort((a, b) => a.end - b.end);

  let end = intervals[0].end;
  let count = 1;

  for (let i = 1; i < intervals.length; i++) {
    if (intervals[i].start >= end) {
      end = intervals[i].end;
      count++;
    }
  }

  return intervals.length - count;
};

export { eraseOverlapIntervals };