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Link for the Problem – Merge Intervals– LeetCode Problem
Merge Intervals– LeetCode Problem
Problem:
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] overlaps, 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 Solutions
Merge Intervals in C++:
class Solution { public: vector<vector<int>> merge(vector<vector<int>>& intervals) { vector<vector<int>> ans; sort(begin(intervals), end(intervals)); for (const auto& interval : intervals) if (ans.empty() || ans.back()[1] < interval[0]) ans.push_back(interval); else ans.back()[1] = max(ans.back()[1], interval[1]); return ans; } };
Merge Intervals in Java:
class Solution { public int[][] merge(int[][] intervals) { List<int[]> ans = new ArrayList<>(); Arrays.sort(intervals, (a, b) -> (a[0] - b[0])); for (int[] interval : intervals) if (ans.isEmpty() || ans.get(ans.size() - 1)[1] < interval[0]) ans.add(interval); else ans.get(ans.size() - 1)[1] = Math.max(ans.get(ans.size() - 1)[1], interval[1]); return ans.toArray(new int[ans.size()][]); } }
Merge Intervals in Python:
class Solution: def merge(self, intervals: List[List[int]]) -> List[List[int]]: ans = [] for interval in sorted(intervals): if not ans or ans[-1][1] < interval[0]: ans.append(interval) else: ans[-1][1] = max(ans[-1][1], interval[1]) return ans
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