Merge Overlapping Intervals in DSA Javascript - Time & Space Complexity

Time Complexity: Merge Overlapping Intervals

O(n log n)

Understanding Time Complexity

We want to understand how the time needed to merge overlapping intervals changes as the number of intervals grows.

How does the code handle more intervals and how long does it take?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function mergeIntervals(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;
}
    

This code sorts intervals by start time, then merges any that overlap into one.

Identify Repeating Operations

• Primary operation: Sorting the intervals array.
• How many times: Sorting takes multiple comparisons, roughly proportional to n log n where n is the number of intervals.
• Secondary operation: One loop through all intervals to merge them, running n times.

How Execution Grows With Input

As the number of intervals grows, sorting takes more time, growing faster than just going through the list once.

Input Size (n)	Approx. Operations
10	About 10 * log(10) ≈ 33 operations for sorting + 10 for merging
100	About 100 * log(100) ≈ 664 operations for sorting + 100 for merging
1000	About 1000 * log(1000) ≈ 9966 operations for sorting + 1000 for merging

Pattern observation: Sorting dominates and grows faster than the simple loop, so total time grows roughly like n log n.

Final Time Complexity

Time Complexity: O(n log n)

This means the time to merge intervals grows a bit faster than just the number of intervals because of sorting, but still much better than checking every pair.

Common Mistake

[X] Wrong: "The merging loop is the slowest part, so the time is O(n)."

[OK] Correct: Sorting the intervals first takes more time than the loop, so the overall time depends mostly on sorting, not just the loop.

Interview Connect

Understanding how sorting affects the time helps you explain your solution clearly and shows you know what part of your code costs the most time.

Self-Check

"What if the intervals were already sorted? How would the time complexity change?"

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