23. Merge Intervals
mediumAsked at EtsyCollapse overlapping time ranges into non-overlapping intervals — the core algorithm Etsy uses to merge seller promotional sale windows before applying discounts.
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 the intervals in the input.
Constraints
1 <= intervals.length <= 10^4intervals[i].length == 20 <= start_i <= end_i <= 10^4
Examples
Example 1
intervals = [[1,3],[2,6],[8,10],[15,18]][[1,6],[8,10],[15,18]]Example 2
intervals = [[1,4],[4,5]][[1,5]]Approaches
1. Brute force (repeated passes)
Repeatedly scan the list, merging any pair of overlapping intervals, until no merges happen in a full pass. O(n^2) passes in the worst case.
- Time
- O(n^2)
- Space
- O(n)
function merge(intervals) {
let merged = [...intervals];
let changed = true;
while (changed) {
changed = false;
const next = [];
let used = new Array(merged.length).fill(false);
for (let i = 0; i < merged.length; i++) {
if (used[i]) continue;
let [s, e] = merged[i];
for (let j = i + 1; j < merged.length; j++) {
if (used[j]) continue;
if (merged[j][0] <= e) {
e = Math.max(e, merged[j][1]);
used[j] = true;
changed = true;
}
}
next.push([s, e]);
}
merged = next;
}
return merged;
}Tradeoff:
2. Sort then single scan
Sort intervals by start time. Walk through once: if the current interval overlaps the last merged interval (current.start <= last.end), extend last.end to the max of both ends. Otherwise push a new interval.
- 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];
const [start, end] = intervals[i];
if (start <= last[1]) {
last[1] = Math.max(last[1], end);
} else {
result.push([start, end]);
}
}
return result;
}Tradeoff:
Etsy-specific tips
Etsy's promotions engine merges overlapping sale windows so a listing isn't double-discounted. Walk the interviewer through why sort-then-scan reduces a 2D problem to 1D: after sorting by start, you only ever need to compare the current interval against the last merged one. Be ready to extend this to 'insert a new interval into an already-merged list' (LC 57).
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