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56. Merge Intervals

Medium

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

Solution

function merge(intervals: number[][]): number[][] {
    let result = []
    intervals.sort((a, b) => a[0] - b[0])
    for (let i = 0; i < intervals.length; i++) { //NOSONAR
        let len = result.length
        if (len > 0 && intervals[i][0] <= result[len - 1][1]) {
            if (result[len - 1][0] >= intervals[i][0]) {
                result[len - 1][0] = intervals[i][0]
            }
            if (result[len - 1][1] <= intervals[i][1]) {
                result[len - 1][1] = intervals[i][1]
            }
        } else {
            result.push(intervals[i])
        }
    }
    return result
}

export { merge }

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