Merge Overlapping Intervals in DSA Go - 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 program's work increase when we add more intervals?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func mergeIntervals(intervals [][]int) [][]int {
    sort.Slice(intervals, func(i, j int) bool {
        return intervals[i][0] < intervals[j][0]
    })
    merged := [][]int{}
    for _, interval := range intervals {
        if len(merged) == 0 || merged[len(merged)-1][1] < interval[0] {
            merged = append(merged, interval)
        } else {
            merged[len(merged)-1][1] = max(merged[len(merged)-1][1], interval[1])
        }
    }
    return merged
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}
    

This code sorts intervals by start time, then merges overlapping ones by comparing each with the last merged interval.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting the intervals and then looping through them once.
• How many times: Sorting takes multiple comparisons depending on number of intervals (n), then the loop runs once for each interval (n times).

How Execution Grows With Input

As the number of intervals grows, sorting takes more time, and then merging checks each interval once.

Input Size (n)	Approx. Operations
10	About 10 log 10 + 10 = 40 operations
100	About 100 log 100 + 100 = 700 operations
1000	About 1000 log 1000 + 1000 = 10,000 operations

Pattern observation: The work grows a bit faster than the number of intervals because sorting is more expensive than just looping.

Final Time Complexity

Time Complexity: O(n log n)

This means the time needed grows a little faster than the number of intervals because sorting takes extra steps, but merging after sorting is just a simple pass.

Common Mistake

[X] Wrong: "Merging intervals is just O(n) because we only loop once."

[OK] Correct: Sorting the intervals first takes more time, which dominates the total work, so the overall time is not just linear.

Interview Connect

Understanding this time complexity helps you explain how sorting affects your solution and shows you can analyze combined steps clearly.

Self-Check

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

Practice

1. What is the main goal when merging overlapping intervals in a list like [[1,3],[2,6],[8,10]]?

easy

A. Remove all intervals that overlap

B. Sort intervals by their end values only

C. Combine intervals that overlap into one continuous interval

D. Split intervals into smaller parts

Solution

1. Step 1: Understand overlapping intervals

   Intervals overlap if one starts before the other ends. For example, [1,3] and [2,6] overlap because 2 is between 1 and 3.

2. Step 2: Merge overlapping intervals

   We combine overlapping intervals into one that covers the full range, like merging [1,3] and [2,6] into [1,6].

3. Final Answer:

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

4. Quick Check:

   Merging = Combine overlaps [OK]

Hint: Merge intervals by combining overlapping ranges [OK]

Common Mistakes:

• Only sorting by end values
• Removing instead of merging overlaps
• Splitting intervals unnecessarily

2. Which Go code snippet correctly sorts intervals by their start value before merging?

easy

A. sort.Slice(intervals, func(i, j int) bool { return intervals[i][0] < intervals[j][0] })

B. sort.Slice(intervals, func(i, j int) bool { return intervals[i][0] > intervals[j][0] })

C. sort.Slice(intervals, func(i, j int) bool { return intervals[i][1] < intervals[j][1] })

D. sort.Slice(intervals, func(i, j int) bool { return intervals[i][1] > intervals[j][1] })

Solution

1. Step 1: Identify sorting by start value

   We want to sort intervals by their first element (start), so the comparison should be intervals[i][0] < intervals[j][0].

2. Step 2: Match with Go sort.Slice syntax

   sort.Slice(intervals, func(i, j int) bool { return intervals[i][0] < intervals[j][0] }) uses sort.Slice with the correct comparison function for start values.

3. Final Answer:

   sort.Slice(intervals, func(i, j int) bool { return intervals[i][0] < intervals[j][0] }) -> Option A

4. Quick Check:

   Sort by start = sort.Slice(intervals, func(i, j int) bool { return intervals[i][0] < intervals[j][0] }) [OK]

Hint: Sort intervals by start using intervals[i][0] < intervals[j][0] [OK]

Common Mistakes:

• Sorting by end value instead of start
• Using greater than instead of less than
• Mixing start and end indices

3. What is the output of the following Go code that merges intervals?

package main
import (
  "fmt"
  "sort"
)
func merge(intervals [][]int) [][]int {
  if len(intervals) == 0 {
    return intervals
  }
  sort.Slice(intervals, func(i, j int) bool { return intervals[i][0] < intervals[j][0] })
  merged := [][]int{intervals[0]}
  for i := 1; i < len(intervals); i++ {
    last := merged[len(merged)-1]
    if intervals[i][0] <= last[1] {
      if intervals[i][1] > last[1] {
        last[1] = intervals[i][1]
      }
    } else {
      merged = append(merged, intervals[i])
    }
  }
  return merged
}
func main() {
  intervals := [][]int{{1,3},{2,4},{5,7},{6,8}}
  fmt.Println(merge(intervals))
}

medium

A. [[1 3] [2 4] [5 7] [6 8]]

B. [[1 3] [5 8]]

C. [[1 4] [5 7]]

D. [[1 4] [5 8]]

Solution

1. Step 1: Sort intervals by start

   Intervals sorted: [[1,3],[2,4],[5,7],[6,8]] (already sorted).

2. Step 2: Merge overlapping intervals

   [1,3] and [2,4] overlap, merged to [1,4]. [5,7] and [6,8] overlap, merged to [5,8]. Final merged list: [[1,4],[5,8]].

3. Final Answer:

   [[1 4] [5 8]] -> Option D

4. Quick Check:

   Merged intervals = [[1 4] [5 8]] [OK]

Hint: Sort then merge by comparing current start with last merged end [OK]

Common Mistakes:

• Not updating the end value when merging
• Appending intervals without checking overlap
• Incorrect sorting order

4. Find the error in this Go code snippet for merging intervals:

func merge(intervals [][]int) [][]int {
  sort.Slice(intervals, func(i, j int) bool { return intervals[i][0] < intervals[j][0] })
  merged := [][]int{}
  for _, interval := range intervals {
    if len(merged) == 0 || merged[len(merged)-1][1] < interval[0] {
      merged = append(merged, interval)
    } else {
      merged[len(merged)-1][1] = interval[1]
    }
  }
  return merged
}

medium

A. Does not sort intervals before merging

B. Does not check if new interval end is greater before updating

C. Appends intervals without checking overlap

D. Returns original intervals instead of merged

Solution

1. Step 1: Review merging logic

   The code updates the end of last merged interval directly with interval[1] without checking if it's greater.

2. Step 2: Identify missing condition

   It should update only if interval[1] > merged[len(merged)-1][1] to avoid shrinking the interval.

3. Final Answer:

   Does not check if new interval end is greater before updating -> Option B

4. Quick Check:

   Update end only if greater [OK]

Hint: Update merged end only if new interval end is larger [OK]

Common Mistakes:

• Overwriting end without comparison
• Skipping sorting step
• Appending without overlap check

5. Given intervals [[1,4],[4,5],[7,9],[8,10],[12,15]], what is the merged output if intervals that touch at endpoints are considered overlapping?

hard

A. [[1 5] [7 10] [12 15]]

B. [[1 4] [4 5] [7 10] [12 15]]

C. [[1 5] [7 9] [8 10] [12 15]]

D. [[1 4] [7 10] [12 15]]

Solution

1. Step 1: Understand touching intervals as overlapping

   Intervals like [1,4] and [4,5] touch at 4, so they merge into [1,5]. Similarly, [7,9] and [8,10] overlap and merge into [7,10].

2. Step 2: Merge all overlapping/touching intervals

   After merging, intervals become [[1,5],[7,10],[12,15]].

3. Final Answer:

   [[1 5] [7 10] [12 15]] -> Option A

4. Quick Check:

   Touching intervals merge = [[1 5] [7 10] [12 15]] [OK]

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

Common Mistakes:

• Not merging intervals that touch at endpoints
• Leaving touching intervals separate
• Incorrectly merging non-overlapping intervals