DevExCode

56. Merge Intervals

ArraySorting

Explanation

To merge overlapping intervals, we first sort the intervals based on their start times. Then, we iterate through the sorted intervals and merge overlapping intervals by comparing the end time of the current interval with the start time of the next interval. If they overlap, we update the end time to be the maximum of the two end times. If they do not overlap, we add the current interval to the result and continue with the next interval.

Time Complexity: O(n log n) - Sorting the intervals Space Complexity: O(n) - Space required for the result

View Leetcode Problem
class Solution {
    public int[][] merge(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));
        List<int[]> merged = new ArrayList<>();
        for (int[] interval : intervals) {
            if (merged.isEmpty() || merged.get(merged.size() - 1)[1] < interval[0]) {
                merged.add(interval);
            } else {
                merged.get(merged.size() - 1)[1] = Math.max(merged.get(merged.size() - 1)[1], interval[1]);
            }
        }
        return merged.toArray(new int[merged.size()][]);
    }
}

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