Day 37 — Merge Intervals

Day 37 — Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

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Day 37 — Merge Intervals 🏁


Given a collection of intervals, merge *all *overlapping intervals.


Input: [[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].


class Solution {
        public int[][] merge(int[][] intervals) {
             if (intervals.length == 0) {
                return new int[0][0];

            Arrays.sort(intervals, (i1, i2) -> i1[0] - i2[0]);
            List<int[]> list = new ArrayList<>();
            list.add(new int[] {intervals[0][0], intervals[0][1]});
            for (int i = 1; i < intervals.length; i++) {
                int[] lastItem = list.get(list.size() - 1);

                if (intervals[i][0] > lastItem[1]) {
                    list.add(new int[] {intervals[i][0], intervals[i][1]});
                } else if (intervals[i][1] > lastItem[1]) {
                    list.add(new int[] {lastItem[0], intervals[i][1]});

            int ans[][]= new int[list.size()][2];

            for(int i=0; i< list.size(); i++)

                int arr[] = list.get(i);
                ans[i][0] = arr[0];
                ans[i][1] = arr[1];

            // int ans[] = new in[2][2];
            return ans;

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