Merge Intervals

Write a C program to merge all overlapping intervals. Given an array of intervals where intervals[i] = [start_i, end_i], 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: 
    

      
Step 1: Sort the intervals based on the start time: [[1,3],[2,6],[8,10],[15,18]] (already sorted).
      
Step 2: Initialize the result with the first interval: [[1,3]].
      
Step 3: Consider interval [2,6]. Since 2 <= 3 (end of previous interval), these intervals overlap.
      
Step 4: Merge [1,3] and [2,6] into [1,6]. Result: [[1,6]].
      
Step 5: Consider interval [8,10]. Since 8 > 6 (end of previous merged interval), these intervals don't overlap.
      
Step 6: Add [8,10] to the result. Result: [[1,6],[8,10]].
      
Step 7: Consider interval [15,18]. Since 15 > 10 (end of previous interval), these intervals don't overlap.
      
Step 8: Add [15,18] to the result. Final result: [[1,6],[8,10],[15,18]].
    
  

Example 2

  
Input: intervals = [[1,4],[4,5]]
  
Output: [[1,5]]
  
Explanation: 
    

      
Step 1: Sort the intervals based on the start time: [[1,4],[4,5]] (already sorted).
      
Step 2: Initialize the result with the first interval: [[1,4]].
      
Step 3: Consider interval [4,5]. Since 4 <= 4 (end of previous interval), these intervals are touching and can be merged.
      
Step 4: Merge [1,4] and [4,5] into [1,5]. Final result: [[1,5]].
    
  

Constraints

  
1 <= intervals.length <= 10^4
  
intervals[i].length == 2
  
0 <= start_i <= end_i <= 10^4
  
Time Complexity: O(n log n), where n is the number of intervals (due to sorting)
  
Space Complexity: O(n), for storing the result