Merge Overlapping Intervals in DSA C - Build from Scratch

Merge Overlapping Intervals

📖 Scenario: You are working on a calendar app that manages meeting times. Sometimes meetings overlap, and you want to combine those overlapping meetings into one longer meeting to avoid confusion.

🎯 Goal: Build a program that merges overlapping intervals from a list of meeting times and prints the merged intervals.

📋 What You'll Learn

Create an array of intervals with exact values

Add a variable to store the number of intervals

Implement the logic to merge overlapping intervals

Print the merged intervals in the format: [start, end]

💡 Why This Matters

🌍 Real World

Merging overlapping intervals is useful in calendar apps, booking systems, and timeline visualizations to avoid conflicts and show clear schedules.

💼 Career

Understanding interval merging helps in software development roles involving scheduling, data processing, and algorithm design.

1

Create the initial intervals array

Create an array called intervals of type int[][2] with these exact intervals: {1, 3}, {2, 6}, {8, 10}, {15, 18}.

DSA C

int intervals[][2] = {
    // Your code here
};

Hint

Use curly braces to define each interval inside the array.

2

Add the number of intervals variable

Create an int variable called n and set it to 4, the number of intervals in intervals.

DSA C

int intervals[][2] = {{1, 3}, {2, 6}, {8, 10}, {15, 18}};
// Your code here

Hint

Count how many intervals are in the array and assign that number to n.

3

Implement merging logic

Write code to merge overlapping intervals from intervals using n. Store merged intervals in merged array and keep count in mergedCount. Use a simple sorting and merging approach.

DSA C

int intervals[][2] = {{1, 3}, {2, 6}, {8, 10}, {15, 18}};
int n = 4;
// Your code here

Hint

Sort intervals by start time, then merge by comparing current interval start with last merged interval end.

4

Print merged intervals

Print each merged interval from merged array using mergedCount. Format each interval as [start, end] on its own line.

DSA C

int intervals[][2] = {{1, 3}, {2, 6}, {8, 10}, {15, 18}};
int n = 4;

// Simple bubble sort by start time
for (int i = 0; i  intervals[j + 1][0]) {
            int temp0 = intervals[j][0];
            int temp1 = intervals[j][1];
            intervals[j][0] = intervals[j + 1][0];
            intervals[j][1] = intervals[j + 1][1];
            intervals[j + 1][0] = temp0;
            intervals[j + 1][1] = temp1;
        }
    }
}

int merged[10][2];
int mergedCount = 0;

merged[0][0] = intervals[0][0];
merged[0][1] = intervals[0][1];
mergedCount = 1;

for (int i = 1; i  merged[mergedCount - 1][1]) {
            merged[mergedCount - 1][1] = intervals[i][1];
        }
    } else {
        merged[mergedCount][0] = intervals[i][0];
        merged[mergedCount][1] = intervals[i][1];
        mergedCount++;
    }
}
// Your code here

Hint

Use a for loop to print each merged interval with printf.

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