Merge Intervals - Array - Medium - LeetCode
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Merge Intervals - Array - Medium - LeetCode

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  • 1The problem involves merging overlapping intervals from a given collection of intervals.
  • 2The solution sorts the intervals and iteratively merges them based on their start and end points.
  • 3The time complexity of the merging algorithm is O(n log n) due to sorting, with a space complexity of O(n).

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"The problem involves merging overlapping intervals from a given collection of intervals."

Merge Intervals - Array - Medium - LeetCode

Given a collection of intervals, merge all overlapping intervals.

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. NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

Constraints:

intervals[i][0] <= intervals[i][1]

public class Solution {
 public int[][] Merge(int[][] intervals) {
 if(intervals.Length==0){
 return intervals;
 }
 Array.Sort(intervals, (a, b) => { return a[0] - b[0]; });
 var res = new List<int[]>();
 int start = intervals[0][0];
 int end = intervals[0][1];
 
 for(int i=1;i<intervals.Length;i++){ 
 if(end>=intervals[i][0] && end <= intervals[i][1]){
 end = intervals[i][1]; 
 }
 else if(end < intervals[i][0]){
 res.Add(new int[]{start,end});
 start = intervals[i][0];
 end = intervals[i][1];
 }
 }
 
 res.Add(new int[]{start, end});
 
 return res.ToArray();
 }
}

Time Complexity: O(nlogn)

Space Complexity: O(n)

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sumitc91

Published on 30 October 2020 · 1 min read · 143 words

Part of AskGif Blog · coding

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