What if you could instantly see your true busy times without confusion or mistakes?
Why Merge Overlapping Intervals in DSA C#?
Imagine you have a list of meeting times written on sticky notes scattered on your desk. Some meetings overlap, but you want to find the total time you are busy without double counting overlapping meetings.
Manually checking each meeting and combining overlapping times is slow and confusing. You might miss overlaps or count some times twice, leading to mistakes and wasted effort.
Merge Overlapping Intervals groups all overlapping meetings into one combined time slot. This way, you get a clear, simple list of busy times without overlaps, making your schedule easy to understand.
List<(int start, int end)> meetings = new List<(int, int)>{(1,3), (2,6), (8,10)};
// Manually check and merge overlaps with many if-else checksList<(int start, int end)> merged = MergeIntervals(meetings);
// Use a function that sorts and merges intervals automaticallyThis lets you quickly find combined busy times, freeing you from juggling overlapping schedules and helping you plan better.
When booking rooms for events, merging overlapping bookings helps avoid double bookings and shows exactly when rooms are occupied.
Manual merging is error-prone and slow.
Merging intervals simplifies overlapping times into clear blocks.
This helps manage schedules and resources efficiently.
Practice
Merge Overlapping Intervals algorithm?Solution
Step 1: Understand the problem goal
The algorithm aims to merge intervals that share common parts so they become one continuous interval.Step 2: Identify the correct purpose
Among the options, only combining overlapping intervals matches the goal.Final Answer:
To combine all intervals that overlap into a single continuous interval -> Option CQuick Check:
Merge overlapping intervals = combine overlaps [OK]
- Confusing merging with sorting
- Thinking it splits intervals instead of combining
- Assuming it counts intervals
Solution
Step 1: Identify sorting by start time
We need to sort intervals by their start property in ascending order.Step 2: Check each option
intervals.Sort((a, b) => a.Start.CompareTo(b.Start)); uses Sort with CompareTo on Start ascending, which is correct. Others sort by end descending, start descending, or length, which are incorrect for merging.Final Answer:
intervals.Sort((a, b) => a.Start.CompareTo(b.Start)); -> Option DQuick Check:
Sort by start ascending = intervals.Sort((a, b) => a.Start.CompareTo(b.Start)); [OK]
- Sorting by end time instead of start
- Sorting descending instead of ascending
- Sorting by interval length
var intervals = new List<(int Start, int End)>{ (1,3), (2,6), (8,10), (15,18) };
intervals.Sort((a,b) => a.Start.CompareTo(b.Start));
var merged = new List<(int, int)>();
foreach(var interval in intervals) {
if(merged.Count == 0 || merged[^1].End < interval.Start) {
merged.Add(interval);
} else {
var last = merged[^1];
merged[^1] = (last.Start, Math.Max(last.End, interval.End));
}
}
foreach(var i in merged) Console.Write($"({i.Start},{i.End}) ");Solution
Step 1: Sort intervals by start
Sorted intervals: (1,3), (2,6), (8,10), (15,18)Step 2: Merge overlapping intervals
(1,3) and (2,6) overlap, merged to (1,6). Others do not overlap, so final merged list is (1,6), (8,10), (15,18).Final Answer:
(1,6) (8,10) (15,18) -> Option AQuick Check:
Merged intervals = (1,6) (8,10) (15,18) [OK]
- Not merging overlapping intervals
- Merging intervals incorrectly by extending start
- Printing original intervals without merging
var intervals = new List<(int Start, int End)>{ (1,4), (4,5) };
intervals.Sort((a,b) => a.Start.CompareTo(b.Start));
var merged = new List<(int, int)>();
foreach(var interval in intervals) {
if(merged.Count == 0 || merged[^1].End <= interval.Start) {
merged.Add(interval);
} else {
var last = merged[^1];
merged[^1] = (last.Start, Math.Max(last.End, interval.End));
}
}
foreach(var i in merged) Console.Write($"({i.Start},{i.End}) ");Solution
Step 1: Analyze the merge condition
The condition uses merged[^1].End <= interval.Start, which treats touching intervals as non-overlapping.Step 2: Identify correct condition for merging
To merge intervals that touch (like (1,4) and (4,5)), use < instead of <= so they merge into (1,5).Final Answer:
The condition should use < instead of <= to merge touching intervals -> Option AQuick Check:
Use < to merge touching intervals [OK]
- Using <= causes touching intervals to not merge
- Forgetting to sort intervals
- Ignoring empty input cases
[(1,5), (2,3), (4,8), (10,12), (9,10)], what is the merged list after applying the merge overlapping intervals algorithm?Solution
Step 1: Sort intervals by start
Sorted intervals: (1,5), (2,3), (4,8), (9,10), (10,12)Step 2: Merge overlapping intervals step-by-step
Merge (1,5) and (2,3) -> (1,5); merge (1,5) and (4,8) -> (1,8); merge (9,10) and (10,12) -> (9,12). Final merged intervals: (1,8), (9,12).Final Answer:
[(1,8), (9,12)] -> Option BQuick Check:
Merged intervals = [(1,8), (9,12)] [OK]
- Not merging intervals that overlap inside others
- Missing to merge touching intervals
- Returning original intervals without merging
Ready to prove it?
Pass the test and this pattern moves unknown
