23. Merge Intervals
mediumAsked at NotionCollapse overlapping time ranges into non-overlapping spans — the algorithm Notion's calendar view runs every render to merge back-to-back event blocks into single visual bars without double-counting covered time.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
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 input intervals.
Constraints
1 <= intervals.length <= 10^40 <= start_i <= end_i <= 10^4
Examples
Example 1
intervals = [[1,3],[2,6],[8,10],[15,18]][[1,6],[8,10],[15,18]]Explanation: Intervals [1,3] and [2,6] overlap; merged to [1,6].
Example 2
intervals = [[1,4],[4,5]][[1,5]]Explanation: Intervals [1,4] and [4,5] are considered overlapping.
Approaches
1. Sort then sweep
Sort by start time; iterate and either extend the last merged interval's end or push a new interval — one pass after sorting.
- Time
- O(n log n)
- Space
- O(n)
function merge(intervals) {
intervals.sort((a, b) => a[0] - b[0]);
const result = [intervals[0]];
for (let i = 1; i < intervals.length; i++) {
const last = result[result.length - 1];
if (intervals[i][0] <= last[1]) {
last[1] = Math.max(last[1], intervals[i][1]);
} else {
result.push([...intervals[i]]);
}
}
return result;
}Tradeoff:
2. Sort then sweep (immutable output)
Same sweep but builds fresh arrays rather than mutating result entries — safer in functional pipelines.
- Time
- O(n log n)
- Space
- O(n)
function merge(intervals) {
const sorted = [...intervals].sort((a, b) => a[0] - b[0]);
const result = [];
let [curStart, curEnd] = sorted[0];
for (let i = 1; i < sorted.length; i++) {
const [start, end] = sorted[i];
if (start <= curEnd) {
curEnd = Math.max(curEnd, end);
} else {
result.push([curStart, curEnd]);
[curStart, curEnd] = [start, end];
}
}
result.push([curStart, curEnd]);
return result;
}Tradeoff:
Notion-specific tips
Notion calendar and timeline views must merge overlapping time blocks for clean rendering. Interviewers watch whether you sort first without prompting — if you don't, they'll ask why adjacent-only comparisons are insufficient. Also handle the edge case where intervals[i][1] < last[1] (one interval fully contained in another).
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