0057 Insert Interval
Solved at: 2023-01-28 Insert Interval
Question
You are given an array of non-overlapping intervals intervals where intervals[i] = [starti, endi] represent the start and the end of the ith interval and intervals is sorted in ascending order by starti. You are also given an interval newInterval = [start, end] that represents the start and end of another interval.
Insert newInterval into intervals such that intervals is still sorted in ascending order by starti and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary).
Return intervals after the insertion.
Solution
class Solution {
func overlaps(_ interval1: [Int], _ interval2: [Int]) -> Bool {
if interval2[0] > interval1[0] {
return interval2[0] <= interval1[1]
}
else {
return interval1[0] <= interval2[1]
}
}
func merge(_ interval1: [Int], _ interval2: [Int]) -> [Int] {
if !overlaps(interval1, interval2) { fatalError("does not overlap") }
var low = min(interval1[0], interval2[0])
var high = max(interval1[1], interval2[1])
print("merging", interval1, interval2, [low, high])
return [low, high]
}
func insert(_ intervals: [Int](/r/int), _ newInterval: [Int]) -> [Int](/r/int) {
var intervalsCopy = intervals
var overlapped = false;
for (idx, interval) in intervalsCopy.enumerated() {
if overlaps(interval, newInterval) {
intervalsCopy.remove(at: idx)
intervalsCopy.insert(merge(interval, newInterval), at: idx)
overlapped = true
break;
}
}
if !overlapped {
intervalsCopy.append(newInterval)
}
intervalsCopy = intervalsCopy.sorted {$0[0] < $1[0]}
var idx = 0
while idx < intervalsCopy.count - 1 {
if overlaps(intervalsCopy[idx], intervalsCopy[idx+1]) {
var intervals1 = intervalsCopy.remove(at: idx)
var intervals2 = intervalsCopy.remove(at: idx)
intervalsCopy.insert(merge(intervals1, intervals2), at: idx)
idx -= 1
idx = max(0, idx)
} else {
idx += 1
}
}
return intervalsCopy
}
}
Results
- Runtime 159 ms, Beats 8.5%
- Memory 15 MB, Beats 28.83%
Complexity
Time:
Space: but in this case the problem gave the original array immutable, so