DSA: Interval Scheduling Problems
Solution to Last Week's Challenge
Last week's problem was: DSA: Tries Explained with Real Examples
The optimal solution uses dsa: tries explained with real examples pattern with careful state management.
Complexity Analysis:
- Time Complexity: O(n)
- Space Complexity: O(1)
Interval scheduling problems represent a fundamental class of greedy algorithms that every senior engineer must master. At their core, these problems involve managing a set of tasks, each defined by a start and end time, where the goal is to optimize a specific objective—such as maximizing the number of completed tasks or minimizing the resources required. In real-world systems, this translates directly to CPU task scheduling, cloud infrastructure orchestration, and even optimizing delivery windows in logistics software.
From an interview perspective, interval problems are high-signal questions. They test a candidate's ability to move beyond brute-force recursion and recognize when a local optimal choice (greedy) leads to a global optimal solution. Companies like Google and Amazon frequently use these patterns because they require a clean implementation of sorting logic and state tracking, which are essential skills for building scalable backend services.
Theory
The backbone of interval scheduling is the "Greedy Choice Property." For the classic problem of maximizing non-overlapping intervals, the optimal strategy is to always pick the interval that finishes earliest. This leaves as much room as possible for subsequent tasks.
The complexity of these problems is almost always dominated by the sorting step. While the iteration through the intervals is typically O(N), we must sort the input by either the start or end time to make greedy decisions, leading to a standard O(N log N) time complexity.
Pattern Visualization
To solve almost any interval-related problem, you can follow a standardized mental flowchart. The first question is always: "How do I order the data to make a definitive decision?"
Implementation
Let's look at the "Maximum Non-overlapping Intervals" problem. We want to find the maximum number of intervals we can schedule without any two overlapping.
Brute Force Approach (Recursive)
A brute force approach would involve exploring all subsets of intervals and checking for overlaps, resulting in O(2^N) complexity. This is rarely acceptable.
Optimized Greedy Approach By sorting by the end time, we ensure we always pick the task that "gets out of the way" the soonest.
def eraseOverlapIntervals(intervals):
if not intervals:
return 0
# Sort by end time: O(N log N)
intervals.sort(key=lambda x: x[1])
count = 0
# Track the end time of the last added non-overlapping interval
last_end = float('-inf')
for start, end in intervals:
if start >= last_end:
# No overlap, we can pick this interval
count += 1
last_end = end
else:
# Overlap occurs, skip this interval
pass
# To return the number of intervals to remove: len(intervals) - count
return count
# Time Complexity: O(N log N)
# Space Complexity: O(1) or O(N) depending on sorting implementationComplexity Analysis Table
| Problem Type | Sorting Strategy | Logic | Time Complexity | Space Complexity |
|---|---|---|---|---|
| Merge Intervals | Start Time | Merge if curr.start <= prev.end | O(N log N) | O(N) |
| Max Non-overlapping | End Time | Skip if curr.start < prev.end | O(N log N) | O(1) |
| Meeting Rooms II | Start Time | Use Min-Heap for end times | O(N log N) | O(N) |
| Insert Interval | Already Sorted | Binary search or linear scan | O(N) | O(N) |
Common Patterns
1. Merging Overlapping Intervals
Commonly used in calendar applications to find busy blocks.
def merge_intervals(intervals):
intervals.sort(key=lambda x: x[0])
merged = []
for interval in intervals:
if not merged or merged[-1][1] < interval[0]:
merged.append(interval)
else:
merged[-1][1] = max(merged[-1][1], interval[1])
return merged2. Meeting Rooms (Minimum Resources)
Using a heap to track the end times of rooms currently in use.
import heapq
def min_meeting_rooms(intervals):
if not intervals: return 0
intervals.sort(key=lambda x: x[0])
rooms = [] # Min-heap of end times
heapq.heappush(rooms, intervals[0][1])
for i in intervals[1:]:
if i[0] >= rooms[0]:
heapq.heappop(rooms)
heapq.heappush(rooms, i[1])
return len(rooms)Practice Problems
Interval problems range from straightforward sorting tasks to complex heap-based optimizations.
This Week's Interview Challenge
Problem: The Bandwidth Scheduler
You are building a monitoring tool for a network switch. You are given a list of transmissions, where each transmission is an interval [start, end] representing the time a data packet is occupying a specific channel. You are also given an integer k, representing the total number of channels available on the switch.
Determine the minimum number of additional channels required to ensure all transmissions can occur without delay. If the current k channels are sufficient, return 0.
Example 1:
- Input:
transmissions = [[0, 30], [5, 10], [15, 20]],k = 1 - Output:
1(You need 2 channels total, so 1 more than the 1 you have).
Example 2:
- Input:
transmissions = [[7, 10], [2, 4]],k = 3 - Output:
0(3 channels are more than enough).
Constraints:
1 <= transmissions.length <= 10^40 <= start < end <= 10^61 <= k <= 10^4
Hints:
- This is a variation of the "Meeting Rooms II" problem.
- Think about the maximum number of overlapping intervals at any single point in time.
- Consider using a Min-Heap or a Chronological Ordering (Sweep Line) approach.
- The result is
max_concurrent_transmissions - k, but ensure the result isn't negative.
Template:
def additionalChannelsNeeded(transmissions, k):
# Your logic here
pass
# Submit your solution and check back next week for the detailed answer!Conclusion
Interval scheduling is less about complex data structures and more about choosing the right sorting criteria. When faced with an interval problem:
- Sort immediately. Usually by start time for merging/resource counting, or end time for maximizing tasks.
- Visualize the timeline. Draw the intervals to see where the overlaps occur.
- Identify the constraint. Are you looking for the maximum number of items that don't touch, or the maximum number of items that do touch?
Mastering these three steps will turn these common interview hurdles into easy wins.