Merge Overlapping Intervals in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort intervals by their start time.

DSA C

int compare(const void *a, const void *b) {
    int *interval1 = (int *)a;
    int *interval2 = (int *)b;
    return interval1[[1]] - interval2[0];
}

Drag options to blanks, or click blank then click option'

A1

B-1

C2

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using the end time (index 1) instead of start time.

Using an invalid index like 2 or -1.

We compare intervals by their start time, which is at index 0.

2fill in blank

medium

Complete the code to check if two intervals overlap.

DSA C

if (intervals[i][0] <= merged[mergedIndex][[1]]) {
    // Overlap detected
}

Drag options to blanks, or click blank then click option'

A2

B0

C1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing start with start instead of end.

Using wrong indices.

We check if the current interval's start is less or equal to the last merged interval's end (index 1).

3fill in blank

hard

Fix the error in updating the end time of the merged interval.

DSA C

merged[mergedIndex][1] = (intervals[i][1] > merged[mergedIndex][[1]]) ? intervals[i][1] : merged[mergedIndex][1];

Drag options to blanks, or click blank then click option'

A-1

B1

C2

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using start time index 0 instead of end time index 1.

Using invalid indices.

We compare the current interval's end (index 1) with the merged interval's end (index 1).

4fill in blank

hard

Fill both blanks to correctly copy the current interval to merged when no overlap.

DSA C

mergedIndex++;
merged[mergedIndex][[1]] = intervals[i][[2]];

Drag options to blanks, or click blank then click option'

A0

B1

Ci

DmergedIndex

Attempts:

3 left

💡 Hint

Common Mistakes

Using end time index 1 instead of start time 0.

Mixing up indices for source and destination.

We copy the start time (index 0) from intervals[i] to merged[mergedIndex].

5fill in blank

hard

Fill all three blanks to copy both start and end times when no overlap.

DSA C

merged[mergedIndex][[1]] = intervals[i][[2]];
merged[mergedIndex][[3]] = intervals[i][1];

Drag options to blanks, or click blank then click option'

A0

B1

CmergedIndex

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing start and end indices.

Using wrong indices for merged or intervals.

Copy the start time (index 0) from intervals[i][0] to merged[mergedIndex][0] for blanks 1 and 2, and end time (index 1) for blank 3.

Practice

1. What is the main goal of the Merge Overlapping Intervals problem?

easy

A. Find the maximum number of intervals without overlap

B. Sort intervals by their end times only

C. Combine all intervals that overlap into one continuous interval

D. Count how many intervals are in the list

Solution

1. Step 1: Understand the problem goal

   The problem asks to merge intervals that overlap, so they become one continuous interval without gaps.

2. Step 2: Identify the correct description

   Combine all intervals that overlap into one continuous interval correctly describes merging overlapping intervals into one combined interval.

3. Final Answer:

   Combine all intervals that overlap into one continuous interval -> Option C

4. Quick Check:

   Merge overlapping intervals = Combine overlaps [OK]

Hint: Think of joining overlapping time slots into one [OK]

Common Mistakes:

• Confusing merging with sorting only
• Counting intervals instead of merging
• Ignoring overlaps

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

easy

A. qsort(intervals, n, sizeof(Interval), compareStart);

B. qsort(intervals, n, sizeof(Interval), compareLength);

C. sort(intervals.begin(), intervals.end());

D. qsort(intervals, n, sizeof(Interval), compareEnd);

Solution

1. Step 1: Identify sorting criteria for merging

   Intervals must be sorted by their start time to merge correctly.

2. Step 2: Match sorting function

   qsort(intervals, n, sizeof(Interval), compareStart); uses compareStart which sorts by start time, correct for merging.

3. Final Answer:

   qsort(intervals, n, sizeof(Interval), compareStart); -> Option A

4. Quick Check:

   Sort by start time = qsort with compareStart [OK]

Hint: Sort intervals by start time before merging [OK]

Common Mistakes:

• Sorting by end time instead of start time
• Using C++ sort syntax in C code
• Sorting by interval length

3. What is the output of the merged intervals after running this code on input intervals: [[1,3], [2,6], [8,10], [15,18]]?

Intervals = [[1,3], [2,6], [8,10], [15,18]]
// After sorting by start time
// Merge overlapping intervals
// Print merged intervals

medium

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

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

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

D. [[1,6], [8,15], [18,18]]

Solution

1. Step 1: Sort intervals by start time

   Intervals are already sorted: [1,3], [2,6], [8,10], [15,18].

2. Step 2: Merge overlapping intervals

   [1,3] and [2,6] overlap, merge to [1,6]. Others do not overlap.

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 intervals if current start ≤ previous end [OK]

Common Mistakes:

• Not merging overlapping intervals
• Merging non-overlapping intervals
• Incorrectly merging intervals with gaps

4. Identify the bug in this C code snippet for merging intervals:

Interval merged[100];
int m = 0;
merged[0] = intervals[0];
for (int i = 1; i < n; i++) {
  if (intervals[i].start <= merged[m].end) {
    merged[m].end = intervals[i].end;
  } else {
    m++;
    merged[m] = intervals[i];
  }
}

medium

A. Loop starts from 0 instead of 1

B. It incorrectly updates the end of the previous interval without checking max end

C. Uses <= instead of < in the overlap condition

D. It compares start with start instead of end

Solution

1. Step 1: Analyze interval merging logic

   When intervals overlap, end should be max of both ends, not just assigned.

2. Step 2: Identify incorrect update

   Code sets previous end to current end without max check, causing wrong merges.

3. Final Answer:

   It incorrectly updates the end of the previous interval without checking max end -> Option B

4. Quick Check:

   Use max end when merging intervals [OK]

Hint: Always update end to max of overlapping intervals [OK]

Common Mistakes:

• Assigning end without max check
• Using strict < instead of <= for touching intervals
• Wrong loop start index

5. Given intervals [[1,4], [4,5], [5,6], [7,8]], what is the merged output if intervals that touch at boundaries are considered overlapping?

hard

A. [[1,8]]

B. [[1,4], [4,5], [5,6], [7,8]]

C. [[1,5], [5,6], [7,8]]

D. [[1,6], [7,8]]

Solution

1. Step 1: Understand touching intervals as overlapping

   Intervals that share boundary points merge: [1,4] and [4,5] merge to [1,5], then with [5,6] to [1,6].

2. Step 2: Merge intervals accordingly

   After merging first three intervals, [7,8] remains separate.

3. Final Answer:

   [[1,6], [7,8]] -> Option D

4. Quick Check:

   Touching intervals merge into one [OK]

Hint: Treat intervals with end == start as overlapping [OK]

Common Mistakes:

• Not merging touching intervals
• Merging all intervals incorrectly
• Ignoring boundary overlap condition