Merge Intervals LeetCode Solution

Problem – 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

Merge Intervals LeetCode Solution in Java

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

		// Sort by ascending starting point
		Arrays.sort(intervals, (i1, i2) -> Integer.compare(i1[0], i2[0]));

		List<int[]> result = new ArrayList<>();
		int[] newInterval = intervals[0];
		result.add(newInterval);
		for (int[] interval : intervals) {
			if (interval[0] <= newInterval[1]) // Overlapping intervals, move the end if needed
				newInterval[1] = Math.max(newInterval[1], interval[1]);
			else {                             // Disjoint intervals, add the new interval to the list
				newInterval = interval;
				result.add(newInterval);
			}
		}

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

Merge Intervals LeetCode Solution in Python

def merge(self, intervals):
    out = []
    for i in sorted(intervals, key=lambda i: i.start):
        if out and i.start <= out[-1].end:
            out[-1].end = max(out[-1].end, i.end)
        else:
            out += i,
    return out

Merge Intervals LeetCode Solution in C++

vector<Interval> merge(vector<Interval>& ins) {
    if (ins.empty()) return vector<Interval>{};
    vector<Interval> res;
    sort(ins.begin(), ins.end(), [](Interval a, Interval b){return a.start < b.start;});
    res.push_back(ins[0]);
    for (int i = 1; i < ins.size(); i++) {
        if (res.back().end < ins[i].start) res.push_back(ins[i]);
        else
            res.back().end = max(res.back().end, ins[i].end);
    }
    return res;
}
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