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Intervals

Insert Interval

medium
DESCRIPTION (inspired by Leetcode.com)

Given a list of intervals intervals and an interval newInterval, write a function to insert newInterval into a list of existing, non-overlapping, and sorted intervals based on their starting points. The function should ensure that after the new interval is added, the list remains sorted without any overlapping intervals, merging them if needed.

Input:

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

Output:

[[1,5],[6,9]]

Explanation: The new interval [2,5] overlaps with [1,3], so they are merged into [1,5].

Explanation

We first want to create a new list merged to store the merged intervals we will return at the end.
This solution operates in 3 phases:
  1. Add all the intervals ending before newInterval starts to merged.
  2. Merge all overlapping intervals with newInterval and add that merged interval to merged.
  3. Add all the intervals starting after newInterval to merged.
Phase 1
In this phase, we add all the intervals that end before newInterval starts to merged. This involves iterating through the intervals list until the current interval no longer ends before newInterval starts (i.e. intervals[i][1] >= newInterval[0]).
Visualization
def insertIntervals(intervals, newInterval):
    merged = []
    i = 0
    n = len(intervals)
    while i < n and intervals[i][1] < newInterval[0]:
        merged.append(intervals[i])
        i += 1
    while i < n and intervals[i][0] <= newInterval[1]:
        newInterval[0] = min(intervals[i][0], newInterval[0])
        newInterval[1] = max(intervals[i][1], newInterval[1])
        i += 1
    merged.append(newInterval)
    for j in range(i, n):
        merged.append(intervals[j])
    return merged
][31][64][76][108][1511newInterval][85merged

initialize variables

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Phase 2
In this phase, we merge all the intervals that overlap with newInterval together into a single interval by updating newInterval to be the minimum start and maximum end of all the overlapping intervals. This involves iterating through the intervals list until the current interval starts after newInterval ends (i.e. intervals[i][0] > newInterval[1]). When that condition is met, we add newInterval to merged and move onto phase 3.
Visualization
def insertIntervals(intervals, newInterval):
    merged = []
    i = 0
    n = len(intervals)
    while i < n and intervals[i][1] < newInterval[0]:
        merged.append(intervals[i])
        i += 1
    while i < n and intervals[i][0] <= newInterval[1]:
        newInterval[0] = min(intervals[i][0], newInterval[0])
        newInterval[1] = max(intervals[i][1], newInterval[1])
        i += 1
    merged.append(newInterval)
    for j in range(i, n):
        merged.append(intervals[j])
    return merged
][31][64][76][108][1511newInterval][85merged][31

add intervals before newInterval

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Phase 3
Phase 3 involves adding all the intervals starting after newInterval to merged. This involves iterating through the intervals list until the end of the list, and adding each interval to merged.
After completing these 3 phases, we return merged as the final result.
Visualization
def insertIntervals(intervals, newInterval):
    merged = []
    i = 0
    n = len(intervals)
    while i < n and intervals[i][1] < newInterval[0]:
        merged.append(intervals[i])
        i += 1
    while i < n and intervals[i][0] <= newInterval[1]:
        newInterval[0] = min(intervals[i][0], newInterval[0])
        newInterval[1] = max(intervals[i][1], newInterval[1])
        i += 1
    merged.append(newInterval)
    for j in range(i, n):
        merged.append(intervals[j])
    return merged
][31][64][76][108][1511newInterval][104merged][31][104

j = 4

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Complexity Analysis

Time Complexity: O(n) where n is the number of intervals. We iterate through all intervals once to merge them.

Space Complexity: O(n) where n is the number of intervals. We need space for the merged output array.

Solution

|
list of intervals [start, end]
|
[start, end]
Try these examples:
Visualization
def insertIntervals(intervals, newInterval):
    merged = []
    i = 0
    n = len(intervals)
    while i < n and intervals[i][1] < newInterval[0]:
        merged.append(intervals[i])
        i += 1
    while i < n and intervals[i][0] <= newInterval[1]:
        newInterval[0] = min(intervals[i][0], newInterval[0])
        newInterval[1] = max(intervals[i][1], newInterval[1])
        i += 1
    merged.append(newInterval)
    for j in range(i, n):
        merged.append(intervals[j])
    return merged
][31][64][76][108][1511newInterval][85

insert interval

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