Merge Overlapping Intervals in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Merge Overlapping Intervals algorithm?

To combine all intervals that overlap into a single continuous interval, so that the final list has no overlapping intervals.

beginner

Why do we sort intervals before merging them?

Sorting intervals by their start time helps us easily find overlapping intervals by comparing each interval with the previous one in order.

intermediate

In Go, how do you sort a slice of intervals by their start time?

Use sort.Slice with a comparison function: sort.Slice(intervals, func(i, j int) bool { return intervals[i][0] < intervals[j][0] })

beginner

What condition indicates two intervals overlap?

Two intervals [a, b] and [c, d] overlap if c <= b, meaning the start of the second interval is less than or equal to the end of the first.

beginner

How do you merge two overlapping intervals [a, b] and [c, d]?

The merged interval is [min(a, c), max(b, d)], combining the earliest start and latest end.

What is the first step in merging overlapping intervals?

A Sort intervals by start time

B Reverse the intervals

C Remove all intervals

D Merge without sorting

Sorting intervals by start time helps to easily identify overlaps in order.

If intervals [1, 5] and [4, 8] overlap, what is the merged interval?

A [1, 4]

B [4, 5]

C [5, 8]

D [1, 8]

Merged interval covers from earliest start 1 to latest end 8.

Which Go package is used to sort slices?

A math

B sort

C fmt

D strings

The sort package provides functions to sort slices and arrays.

When do two intervals NOT overlap?

A When they share any points

B When the second interval starts before the first

C When the start of the second interval is greater than the end of the first

D When they have the same start

If the second interval starts after the first ends, they do not overlap.

What data structure is commonly used to store intervals in Go for this problem?

A Slice of slices ([][]int)

B Map

C Linked list

D Channel

Intervals are stored as slices of two integers inside a slice.

Explain step-by-step how to merge overlapping intervals given a list of intervals.

Think about ordering and comparing intervals one by one.

Describe how sorting helps in merging overlapping intervals and what happens if intervals are not sorted.

Consider how order affects overlap detection.

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