Merge Intervals
Given an array of intervals where `intervals[i] = [start, end]`, merge all overlapping intervals and return the result.
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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.
Input
An array of intervals `intervals` where each interval is `[start, end]`.
Output
An array of non-overlapping intervals that cover all the intervals in the input.
Examples
Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
[1,3] and [2,6] overlap (2 ≤ 3), so they merge into [1,6]. The other two don't overlap with anything.
Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
[1,4] and [4,5] share endpoint 4 — they are considered overlapping and merge into [1,5].
The brute-force approach
For each interval, check every other interval to see if they overlap. If yes, merge them and restart. Repeat until no more merges are possible.
changed = True
while changed:
changed = False
for i in range(len(intervals)):
for j in range(i+1, len(intervals)):
if overlaps(intervals[i], intervals[j]):
intervals[i] = merge(intervals[i], intervals[j])
intervals.pop(j)
changed = True
breakO(n²) per pass and potentially many passes. On inputs where all intervals merge into one, this is O(n³) or worse.
Spotting the pattern
This is a Intervals problem. The key question to ask yourself:
After sorting by start time, when does the current interval overlap the last merged interval?
Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.
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