Posts published in “Interval Merge”

Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

Example 1:

Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

Constraints:

• 1 <= intervals.length <= 104
• intervals[i].length == 2
• 0 <= starti <= endi <= 104

Idea:

• First step (Important): sort the intervals by the first number
• Use result array to store the first interval
• Compare the last interval of the result array vs. new coming interval
  • new interval > last interval of the result array : store the new interval into result array
  • new interval < last interval of the result array : store the interval that has longer range

Solution:

/**
 * @param {number[][]} intervals
 * @return {number[][]}
 */
var merge = function(intervals) {
    intervals.sort(([a, b], [c, d]) => a - c);
    let res = [];
    for (let interval of intervals) {
        if (res.length === 0 || interval[0] > res[res.length - 1][1]) {
            res.push(interval.concat());
        } else {
            res[res.length - 1][1] = Math.max(interval[1], res[res.length - 1][1]);
        }
    }
    return res;
};