Merge Overlapping Intervals in DSA Javascript - Build from Scratch

Merge Overlapping Intervals

📖 Scenario: You work in a calendar app team. Users add meeting times as intervals. Sometimes meetings overlap. You want to merge overlapping meetings to show a clean schedule.

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

📋 What You'll Learn

Create an array of intervals where each interval is an array of two numbers [start, end].

Create a variable to hold the merged intervals.

Write logic to merge overlapping intervals.

Print the merged intervals as an array.

💡 Why This Matters

🌍 Real World

Merging overlapping intervals is useful in calendar apps, booking systems, and resource scheduling to avoid conflicts.

💼 Career

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

1

Create the intervals array

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

DSA Javascript

// Your code here

Hint

Use const intervals = [[1, 3], [2, 6], [8, 10], [15, 18]]; to create the array.

2

Create the merged array

Create an empty array called merged to store merged intervals.

DSA Javascript

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

Hint

Use const merged = []; to create an empty array.

3

Merge overlapping intervals

Sort intervals by start time using intervals.sort((a, b) => a[0] - b[0]). Then use a for loop with variable interval to iterate over intervals. Inside the loop, if merged is empty or merged[merged.length - 1][1] < interval[0], push interval to merged. Otherwise, update the end of the last interval in merged to the maximum of merged[merged.length - 1][1] and interval[1].

DSA Javascript

const intervals = [[1, 3], [2, 6], [8, 10], [15, 18]];
const merged = [];
// Your code here

Hint

Sort intervals first. Then loop through each interval. If no overlap, add to merged. If overlap, update the end time.

4

Print the merged intervals

Print the merged array using console.log(merged).

DSA Javascript

const intervals = [[1, 3], [2, 6], [8, 10], [15, 18]];
const merged = [];
intervals.sort((a, b) => a[0] - b[0]);
for (const interval of intervals) {
    if (merged.length === 0 || merged[merged.length - 1][1] < interval[0]) {
        merged.push(interval);
    } else {
        merged[merged.length - 1][1] = Math.max(merged[merged.length - 1][1], interval[1]);
    }
}
// Your code here

Hint

Use console.log(merged) to print the merged intervals.

Practice

1. What is the main purpose of the Merge Overlapping Intervals algorithm?

easy

A. To sort intervals by their length

B. To combine all intervals that overlap into one continuous interval

C. To find the maximum number of intervals

D. To split intervals into smaller parts

Solution

1. Step 1: Understand the problem goal

   The algorithm aims to merge intervals that share any overlapping range into a single interval.

2. Step 2: Identify the correct purpose

   Among the options, only combining overlapping intervals matches the goal.

3. Final Answer:

   To combine all intervals that overlap into one continuous interval -> Option B

4. Quick Check:

   Merge overlapping intervals = combine overlaps [OK]

Hint: Think about joining intervals that touch or overlap [OK]

Common Mistakes:

• Confusing merging with sorting intervals
• Thinking it splits intervals instead of merging
• Assuming it counts intervals instead of merging

2. Which of the following is the correct way to sort intervals by their start time in JavaScript before merging?

easy

A. intervals.sort((a, b) => a[0] - b[0])

B. intervals.sort((a, b) => b[1] - a[1])

C. intervals.sort((a, b) => b[0] - a[0])

D. intervals.sort((a, b) => a[1] - b[1])

Solution

1. Step 1: Identify sorting key for merging

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

2. Step 2: Choose correct sort function

   The correct syntax is intervals.sort((a, b) => a[0] - b[0]) to sort ascending by start time.

3. Final Answer:

   intervals.sort((a, b) => a[0] - b[0]) -> Option A

4. Quick Check:

   Sort by start time = a[0] - b[0] [OK]

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

Common Mistakes:

• Sorting by end time instead of start time
• Sorting in descending order
• Using wrong array indices for sorting

3. What is the output of the following code?

const intervals = [[1,3],[2,6],[8,10],[15,18]];
function merge(intervals) {
  intervals.sort((a,b) => a[0] - b[0]);
  const merged = [];
  for (const interval of intervals) {
    if (!merged.length || merged[merged.length-1][1] < interval[0]) {
      merged.push(interval);
    } else {
      merged[merged.length-1][1] = Math.max(merged[merged.length-1][1], interval[1]);
    }
  }
  return merged;
}
console.log(merge(intervals));

medium

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

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

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

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

Solution

1. Step 1: Sort intervals by start time

   Sorted intervals: [[1,3],[2,6],[8,10],[15,18]] (already sorted)

2. Step 2: Merge overlapping intervals

   Merge [1,3] and [2,6] into [1,6]. Others remain separate.

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 by comparing end of last merged with start of current [OK]

Common Mistakes:

• Not merging overlapping intervals
• Merging intervals incorrectly by extending wrong boundaries
• Returning original intervals without merging

4. The following code is intended to merge overlapping intervals but has a bug. What is the bug?

function merge(intervals) {
  intervals.sort((a,b) => a[0] - b[0]);
  const merged = [];
  for (const interval of intervals) {
    if (merged.length === 0 || merged[merged.length-1][1] <= interval[0]) {
      merged.push(interval);
    } else {
      merged[merged.length-1][1] = Math.max(merged[merged.length-1][1], interval[1]);
    }
  }
  return merged;
}

medium

A. The merged array is not initialized correctly

B. The intervals are not sorted before merging

C. The condition should use < instead of <= to detect overlap

D. The max function is used incorrectly

Solution

1. Step 1: Analyze the overlap condition

   The condition uses <= which treats touching intervals as non-overlapping.

2. Step 2: Correct the condition for merging

   Using < instead of <= allows intervals that touch at the boundary to merge correctly.

3. Final Answer:

   The condition should use < instead of <= to detect overlap -> Option C

4. Quick Check:

   Overlap check needs <, not <= [OK]

Hint: Use <, not <=, to merge touching intervals [OK]

Common Mistakes:

• Using <= causes missing merges for touching intervals
• Not sorting intervals before merging
• Incorrectly updating merged intervals

5. Given intervals [[1,4],[4,5],[7,9],[8,10]], what is the merged output if intervals that touch at boundaries are considered overlapping?

hard

A. [[1,4],[5,7],[8,10]]

B. [[1,4],[4,5],[7,9],[8,10]]

C. [[1,10]]

D. [[1,5],[7,10]]

Solution

1. Step 1: Sort intervals by start time

   Sorted intervals: [[1,4],[4,5],[7,9],[8,10]] (already sorted)

2. Step 2: Merge touching and overlapping intervals

   [1,4] and [4,5] touch, so merge to [1,5]. [7,9] and [8,10] overlap, merge to [7,10].

3. Final Answer:

   [[1,5],[7,10]] -> Option D

4. Quick Check:

   Touching intervals merge = [[1,5],[7,10]] [OK]

Hint: Treat touching intervals as overlapping to merge boundaries [OK]

Common Mistakes:

• Not merging intervals that only touch at boundaries
• Merging all intervals incorrectly into one
• Leaving intervals separate despite overlap