LC 57 - 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.

Note that you don’t need to modify intervals in-place. You can make a new array and return it.

Example 1:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]

Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]

Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].

Constraints:

• 0 <= intervals.length <= 104
• intervals[i].length == 2
• 0 <= starti <= endi <= 105
• intervals is sorted by starti in ascending order.
• newInterval.length == 2
• 0 <= start <= end <= 105

Links

Question here and solution here

Solution

concept

The greedy solution requires us to iterate through intervals and check the interval for 3 conditions:

1. if the newInterval is non-overlapping and before the current interval, then we just append the rest of the intervals and return. check condition newInterval[1] < intervals[i][0]
2. if the newInterval is non-overlapping and after the current interval, then we append the current interval into ans first, but not returning, since newInterval could still overlap with intervals afterwards. check condition newInterval[0] > intervals[i][1]
3. if either the condition above is not true, then there is an overlap between newInterval and the current interval, we need to merge and modify newInterval = [min(newInterval[0], intervals[i][0]), max(newInterval[1], intervals[i][1])]. Take note at the end we need to add ans.append(newInterval) since if the modified newInterval is the last interval, it wont be added. so we need to manually append it.

code

class Solution:
	"""
	greedy
	"""
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        ans = []

        for i in range(len(intervals)):
            # if non-overlapping
            # the rest of the intervals wont be overlapping with newInterval, just add the rest and return
            if newInterval[1] < intervals[i][0]:
                ans.append(newInterval)
                return ans + intervals[i:]
            # current interval is not overlapping with newInterval so it is safe to add to ans, but not return yet
            elif newInterval[0] > intervals[i][1]:
                ans.append(intervals[i])
            # handle merging
            else:
                newInterval = [min(newInterval[0], intervals[i][0]), max(newInterval[1], intervals[i][1])]

        # if the abv return nv executre, means newInterval is the last one
        ans.append(newInterval)

        return ans
        
class Solution:
	"""
	linear search
	"""
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        n = len(intervals)
        i = 0
        res = []
		
		# locate the start of the merge
        while i < n and intervals[i][1] < newInterval[0]:
            res.append(intervals[i])
            i += 1
		# merge
        while i < n and newInterval[1] >= intervals[i][0]:
            newInterval[0] = min(newInterval[0], intervals[i][0])
            newInterval[1] = max(newInterval[1], intervals[i][1])
            i += 1
        res.append(newInterval)
		# add the rest
        while i < n:
            res.append(intervals[i])
            i += 1

        return res
        
class Solution:
	"""
	binary search to locate the start of the merging
	search on the start index
	"""
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        if not intervals:
            return [newInterval]

        n = len(intervals)
        target = newInterval[0]
        left, right = 0, n - 1

        while left <= right:
            mid = (left + right) // 2
            if intervals[mid][0] < target:
                left = mid + 1
            else:
                right = mid - 1

        intervals.insert(left, newInterval)

        res = []
        for interval in intervals:
            if not res or res[-1][1] < interval[0]:
                res.append(interval)
            else:
                res[-1][1] = max(res[-1][1], interval[1])
        return res

Complexity

time: $O(n)$
space: $O(n)$