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;
}

Merge Intervals LeetCode Solution Review:

In our experience, we suggest you solve this Merge Intervals LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the Merge Intervals LeetCode Solution

Find on LeetCode

Conclusion:

I hope this Merge Intervals LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

More Coding Solutions >>

LeetCode Solutions

Hacker Rank Solutions

CodeChef Solutions