20. Merge Intervals
mediumAsked at KlarnaMerge all overlapping intervals into a minimal disjoint set.
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 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]]Example 2
intervals = [[1,4],[4,5]][[1,5]]Approaches
1. Pairwise merge until stable
Loop merging any overlapping pair found, until no more merges are possible.
- Time
- O(n^2)
- Space
- O(n)
function merge(intervals) {
let list = intervals.map(x => x.slice());
let changed = true;
while (changed) {
changed = false;
outer: for (let i = 0; i < list.length; i++)
for (let j = i+1; j < list.length; j++)
if (Math.max(list[i][0], list[j][0]) <= Math.min(list[i][1], list[j][1])) {
list[i] = [Math.min(list[i][0], list[j][0]), Math.max(list[i][1], list[j][1])];
list.splice(j, 1);
changed = true;
break outer;
}
}
return list;
}Tradeoff:
2. Sort by start, sweep
Sort intervals by start time, then sweep and extend the most recent interval whenever the next one overlaps; otherwise push a new interval.
- Time
- O(n log n)
- Space
- O(n)
function merge(intervals) {
if (!intervals.length) return [];
intervals.sort((a, b) => a[0] - b[0]);
const out = [intervals[0].slice()];
for (let i = 1; i < intervals.length; i++) {
const last = out[out.length - 1];
if (intervals[i][0] <= last[1]) last[1] = Math.max(last[1], intervals[i][1]);
else out.push(intervals[i].slice());
}
return out;
}Tradeoff:
Klarna-specific tips
Klarna's merchant-settlement engine literally runs this sweep to collapse payout windows; they will grade you on whether you handle touching-but-not-overlapping intervals consistently with their billing semantics.
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