56. Merge Intervals

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] overlap, 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:

class Solution {
    public int[][] merge(int[][] intervals) {
        // base case
        if (intervals.length <= 1){
            return intervals;
        }

        List<int[]> result = new ArrayList<>();

        Arrays.sort(intervals, (a,b) -> Integer.compare(a[0], b[0]));

        int[] newInterval = intervals[0];
        for (int i = 1; i < intervals.length; i++){
            int[] interval = intervals[i];
            if (newInterval[1] < interval[0]){
                result.add(newInterval);
                newInterval = interval;
            }else{
                // overlapping
                newInterval[0] = Math.min(newInterval[0], interval[0]);
                newInterval[1] = Math.max(newInterval[1], interval[1]);
            }
        }

        result.add(newInterval);



        return result.toArray(new int[result.size()][]);
    }
}
// TC: O(nlogn)
// SC: O(n) /O(1)

class Solution {
    public int[][] merge(int[][] intervals) {
        if (intervals.length <= 1) {
            return intervals;
        }

        // Sort intervals by the start time
        Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));

        List<int[]> result = new ArrayList<>();
        int[] newInterval = intervals[0];
        result.add(newInterval); // store address

        for (int[] interval : intervals) {
            if (interval[0] <= newInterval[1]) {  // Check if there is an overlap
                newInterval[1] = Math.max(newInterval[1], interval[1]); // Merge the interval
            } else { // No overlap
                newInterval = interval; // Move to the next interval
                result.add(newInterval);
            }
        }

        return result.toArray(new int[result.size()][]);
    }
}

// TC: O(nlogn)
// SC: O(n) / O(1)