Problem
Given an array of intervals where intervals[i] = [start, end], merge all overlapping intervals.
Example
Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]Solution
Sort by start. Walk through; if current overlaps with last in result, extend the end. Otherwise add new interval.
def merge(intervals):
intervals.sort(key=lambda x: x[0])
result = [intervals[0]]
for start, end in intervals[1:]:
if start <= result[-1][1]:
result[-1][1] = max(result[-1][1], end)
else:
result.append([start, end])
return resultfunction merge(intervals) {
intervals.sort((a, b) => a[0] - b[0]);
const result = [intervals[0]];
for (let i = 1; i < intervals.length; i++) {
const [start, end] = intervals[i];
if (start <= result[result.length-1][1]) {
result[result.length-1][1] = Math.max(result[result.length-1][1], end);
} else {
result.push([start, end]);
}
}
return result;
}vector<vector<int>> merge(vector<vector<int>>& intervals) {
sort(intervals.begin(), intervals.end());
vector<vector<int>> result = {intervals[0]};
for (int i = 1; i < intervals.size(); i++) {
if (intervals[i][0] <= result.back()[1]) {
result.back()[1] = max(result.back()[1], intervals[i][1]);
} else {
result.push_back(intervals[i]);
}
}
return result;
}public int[][] merge(int[][] intervals) {
Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
List<int[]> result = new ArrayList<>();
result.add(intervals[0]);
for (int i = 1; i < intervals.length; i++) {
int[] last = result.get(result.size() - 1);
if (intervals[i][0] <= last[1]) {
last[1] = Math.max(last[1], intervals[i][1]);
} else {
result.add(intervals[i]);
}
}
return result.toArray(new int[0][]);
}Complexity
- Time: O(n log n)
- Space: O(n)
Explanation
After sorting, only adjacent intervals can overlap. Compare each new interval’s start with the previous interval’s end.
Comments
Join the discussion. Got a question, found an issue, or want to share your experience?