Merge Overlapping Intervals [Medium]
- Method 1
- Method 2

Merge Overlapping Intervals [Medium]

Write a function that takes in a non-empty array of arbitrary intervals, merges any overlapping intervals, and returns the new intervals in no particular order.
Each interval interval is an array of two integers,with interval[0]as the start of the interval and interval[1] as the end of the interval.
Note that back-to-back intervals aren’t considered to be overlapping. For example, [1,5] and [6,7]aren’t overlapping; however, [1,6]and[6,7]are indeed overlapping.
Also note that the start of any particular interval will always be less than or equal to the end of that interval.

Sample Input #1

intervals=[[1,2],[3,5],[4,7],[6,8],[9,10]]

Sample Output #1

[[1,2],[3,8],[9,10]]

// Merge the intervals [3,5],[4,7],and [6,8].

// The intervals could be ordered differently.

Hints

Hint 1

The problem asks you to merge overlapping intervals.How can you determine if two intervals are overlapping?

Hint 2

Sort the intervals with respect to their starting values.This will allow you to merge all overlapping intervals in a single traversal through the sorted intervals.

Hint 3

After sorting the intervals with respect to their starting values,traverse them,and at each iteration,compare the start of the next interval to the end of the current interval to look for an overlap.If you find an overlap,mutate the current interval so as to merge the next interval into it.

Method 1

【 O(nlog(n)) time | O(n) space 】

import java.util.*;

class Program {

  public int[][] mergeOverlappingIntervals(int[][] intervals) {
     if (intervals.length <= 1) {
            return intervals;
        }

        //Array sorting
        Arrays.sort(intervals, (a, b) -> (a[0] - b[0]));

        List<int[]> result = new ArrayList<>();
        for (int[] interval : intervals) {
            // if list is empty or does not overlap with the previous, just append
            if (result.isEmpty() || result.get(result.size() - 1)[1] < interval[0]) {
                result.add(interval);
            } else {
                // if overlap, merge the current and previous intervals
                result.get(result.size() - 1)[1] = Math.max(result.get(result.size() - 1)[1], interval[1]);
            }
        }

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

Method 2

【 O(nlog(n)) time | O(n) space 】

package Array;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/**
 * @author zhengstars
 * @date 2023/02/05
 */
public class MergeOverlappingIntervals {

    public int[][] mergeOverlappingIntervals(int[][] intervals) {
        if(intervals.length <= 1){
            return intervals;
        }

        Arrays.sort(intervals,(a,b) -> a[0] - b[0]);


        List<int[]> result = new ArrayList<>();
        int[] currentInterval = intervals[0];
        result.add(currentInterval);

        for (int[] interval: intervals) {
            int currentEnd = currentInterval[1];
            int nextBegin = interval[0];
            int nextEnd = interval[1];

            if(nextBegin <= currentEnd){
                currentInterval[1] = Math.max(currentEnd, nextEnd);
            }else{
                currentInterval = interval;
                result.add(currentInterval);
            }
        }


        return result.toArray(new int[result.size()][2]);
    }
}

Merge Overlapping Intervals [Medium]

Method 1

Method 2

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