Merge Overlapping Intervals in DSA Python - Execution Trace

Concept Flow - Merge Overlapping Intervals

Sort intervals by start

↓

Initialize merged list empty

↓

For each interval

↓

If merged empty or no overlap

↓

Add interval

↓

Continue loop

↓

Return merged

Sort intervals, then go through each. Add if no overlap, else merge with last. Return merged list.

Execution Sample

DSA Python

intervals = [[1,3],[2,6],[8,10],[15,18]]
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])
print(merged)

Merges overlapping intervals from a list and prints the merged intervals.

Execution Table

Step	Operation	Current Interval	Merged List Before	Action	Merged List After	Visual State
1	Sort intervals	[1,3],[2,6],[8,10],[15,18]	[]	Sort by start	[]	[1,3] -> [2,6] -> [8,10] -> [15,18] -> null
2	Check merged empty	[1,3]	[]	merged empty, append	[[1,3]]	[1,3] -> null
3	Check overlap	[2,6]	[[1,3]]	2 <= 3 overlap, merge	[[1,6]]	[1,6] -> null
4	Check overlap	[8,10]	[[1,6]]	8 > 6 no overlap, append	[[1,6],[8,10]]	[1,6] -> [8,10] -> null
5	Check overlap	[15,18]	[[1,6],[8,10]]	15 > 10 no overlap, append	[[1,6],[8,10],[15,18]]	[1,6] -> [8,10] -> [15,18] -> null
6	End loop	N/A	[[1,6],[8,10],[15,18]]	Return merged	[[1,6],[8,10],[15,18]]	[1,6] -> [8,10] -> [15,18] -> null

💡 All intervals processed, merged list returned

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	Final
intervals	[[1,3],[2,6],[8,10],[15,18]]	[[1,3],[2,6],[8,10],[15,18]]	[[1,3],[2,6],[8,10],[15,18]]	[[1,3],[2,6],[8,10],[15,18]]	[[1,3],[2,6],[8,10],[15,18]]	[[1,3],[2,6],[8,10],[15,18]]
merged	[]	[[1,3]]	[[1,6]]	[[1,6],[8,10]]	[[1,6],[8,10],[15,18]]	[[1,6],[8,10],[15,18]]
interval (current)	N/A	[1,3]	[2,6]	[8,10]	[15,18]	N/A

Key Moments - 3 Insights

Why do we merge intervals by updating the end of the last merged interval?

Because overlapping intervals share some range, we extend the last merged interval's end to cover the new interval's end if it's larger, as shown in step 3 where [1,3] and [2,6] merge to [1,6].

Why do we append the interval if merged is empty or no overlap?

If merged is empty, there's nothing to merge with, so we add the first interval (step 2). If the current interval starts after the last merged interval ends, they don't overlap, so we add it separately (steps 4 and 5).

Why must intervals be sorted before merging?

Sorting ensures intervals are in order by start time, so overlapping intervals come one after another. Without sorting, we might miss overlaps, as the algorithm relies on comparing current interval with last merged interval (step 1).

Visual Quiz - 3 Questions

Look at the execution table at step 3, what is the merged list after merging?

A [[1,6]]

B [[1,3]]

C [[2,6]]

D [[1,3],[2,6]]

💡 Hint

Check the 'Merged List After' column at step 3 in the execution table.

At which step does the condition 'no overlap' cause the interval to be appended to merged?

A Step 2

B Step 4

C Step 3

D Step 6

💡 Hint

Look for steps where the action says 'append' due to no overlap in the execution table.

If the intervals were not sorted, how would the merged list after step 2 change?

A It would be empty

B It would contain all intervals merged

C It would contain only the first interval unsorted

D It would cause an error

💡 Hint

Refer to the variable_tracker and concept_flow about sorting before merging.

Concept Snapshot

Merge Overlapping Intervals:
1. Sort intervals by start time.
2. Initialize empty merged list.
3. For each interval:
   - If merged empty or no overlap, append.
   - Else merge by updating last interval's end.
4. Return merged list with no overlaps.

Full Transcript

To merge overlapping intervals, first sort the intervals by their start times. Then create an empty list called merged. Go through each interval in order. If merged is empty or the current interval does not overlap with the last interval in merged, add it to merged. If it overlaps, update the end of the last interval in merged to cover the current interval's end if it is larger. Continue until all intervals are processed. The result is a list of intervals with no overlaps.

Practice

1. What is the main purpose of merging overlapping intervals in a list of intervals?

easy

A. To remove all intervals that have the same start and end points

B. To sort intervals by their start time only

C. To combine intervals that overlap into a single continuous interval

D. To split intervals into smaller parts

Solution

1. Step 1: Understand overlapping intervals

   Overlapping intervals share some common range, so merging combines them into one continuous interval.

2. Step 2: Purpose of merging

   Merging reduces multiple overlapping intervals into fewer intervals without overlaps.

3. Final Answer:

   To combine intervals that overlap into a single continuous interval -> Option C

4. Quick Check:

   Merging = Combining overlaps [OK]

Hint: Merging means joining overlapping intervals into one [OK]

Common Mistakes:

• Thinking merging means sorting only
• Confusing merging with removing duplicates
• Assuming intervals are split instead of combined

2. Which of the following Python code snippets correctly sorts a list of intervals by their start time before merging?

easy

A. intervals.sort(key=lambda x: x[0], reverse=True)

B. intervals.sort(key=lambda x: x[0])

C. sorted(intervals)

D. intervals.sort(key=lambda x: x[1])

Solution

1. Step 1: Identify sorting key for intervals

   We sort intervals by their start time, which is the first element in each interval.

2. Step 2: Check code correctness

   Using intervals.sort(key=lambda x: x[0]) sorts by start time ascending, which is correct.

3. Final Answer:

   intervals.sort(key=lambda x: x[0]) -> Option B

4. Quick Check:

   Sort by start time = key x[0] [OK]

Hint: Sort intervals by first element before merging [OK]

Common Mistakes:

• Sorting by end time instead of start time
• Using sorted(intervals) which returns new list without modifying original
• Sorting in reverse order before merging

3. What is the output of the following Python code?

intervals = [[1,3],[2,6],[8,10],[15,18]]
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])
print(merged)

medium

A. [[1, 6], [8, 10], [15, 18]]

B. [[1, 3], [2, 6], [8, 10], [15, 18]]

C. [[1, 6], [2, 6], [8, 10], [15, 18]]

D. [[1, 3], [8, 10], [15, 18]]

Solution

1. Step 1: Sort intervals by start time

   Intervals after sorting: [[1,3],[2,6],[8,10],[15,18]] (already sorted)

2. Step 2: Merge overlapping intervals

   First interval [1,3] added. Next [2,6] overlaps with [1,3], merged to [1,6]. Then [8,10] no overlap, added. Finally [15,18] no overlap, added.

3. Final Answer:

   [[1, 6], [8, 10], [15, 18]] -> Option A

4. Quick Check:

   Merged intervals = [[1,6],[8,10],[15,18]] [OK]

Hint: Merge if current start ≤ last merged end [OK]

Common Mistakes:

• Not updating the end of merged interval
• Appending intervals without merging
• Incorrectly sorting intervals

4. Find the error in the following code that attempts to merge overlapping intervals:

def merge_intervals(intervals):
    intervals.sort()
    merged = []
    for i in range(len(intervals)):
        if not merged or merged[-1][1] < intervals[i][0]:
            merged.append(intervals[i])
        else:
            merged[-1][1] = max(merged[-1][1], intervals[i][1])
    return merged

print(merge_intervals([(1,4),(2,3)]))

medium

A. Not checking if intervals list is empty before sorting

B. Sorting intervals without a key may cause incorrect order

C. Using range(len(intervals)) instead of iterating directly over intervals

D. Modifying merged[-1][1] directly causes an error because tuples are immutable

Solution

1. Step 1: Identify data type of intervals elements

   Intervals are list of tuples like [(1,4),(2,3)], tuples are immutable.

2. Step 2: Trace the merging process

   After sorting, append first tuple (1,4) to merged. Second overlaps, attempt merged[-1][1] = max(4,3) fails: 'tuple' object does not support item assignment.

3. Final Answer:

   Modifying merged[-1][1] directly causes an error because tuples are immutable -> Option D

4. Quick Check:

   Tuples immutable, can't modify [OK]

Hint: Check if intervals are tuples before modifying elements [OK]

Common Mistakes:

• Assuming all intervals are lists when they might be tuples
• Ignoring immutability errors
• Thinking sorting always needs a key

5. Given a list of intervals where some intervals are nested inside others, how would you modify the merge algorithm to handle intervals like [[1,10],[2,5],[6,9],[11,15]] to produce the correct merged output?

hard

A. Sort intervals by start time and merge by extending the end to max of overlapping intervals

B. Sort intervals by end time and merge by extending the start to min of overlapping intervals

C. Remove nested intervals before merging to avoid duplicates

D. Merge only intervals that have exactly the same start or end points

Solution

1. Step 1: Sort intervals by start time

   Sorting ensures intervals are processed in order, e.g. [[1,10],[2,5],[6,9],[11,15]].

2. Step 2: Merge by extending end to max of overlapping intervals

   When intervals overlap or are nested, update the end of merged interval to the maximum end value.

3. Step 3: Result

   Merged intervals become [[1,10],[11,15]] because [2,5] and [6,9] are inside [1,10].

4. Final Answer:

   Sort intervals by start time and merge by extending the end to max of overlapping intervals -> Option A

5. Quick Check:

   Sort + max end merge = correct nested handling [OK]

Hint: Always extend merged end to max overlapping end [OK]

Common Mistakes:

• Sorting by end time instead of start time
• Removing nested intervals instead of merging
• Merging only exact matches, missing nested cases