Merge Overlapping Intervals in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of merging intervals with overlapping pairs

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

DSA Go

package main
import (
	"fmt"
	"sort"
)
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 {
			if merged[len(merged)-1][1] < interval[1] {
				merged[len(merged)-1][1] = interval[1]
			}
		}
	}
	return merged
}
func main() {
	intervals := [][]int{{1,3},{2,6},{8,10},{15,18}}
	result := merge(intervals)
	fmt.Println(result)
}

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

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

C [[1,3],[8,10],[15,18]]

D [[1,6],[2,6],[8,10],[15,18]]

💡 Hint

Look at how intervals are sorted and merged when overlapping.

The code sorts intervals by start time, then merges overlapping intervals by extending the end time. Intervals [1,3] and [2,6] overlap and merge into [1,6]. Others remain separate.

🧠 Conceptual

intermediate

Number of intervals after merging

Given intervals [[1,4],[4,5],[7,9],[8,10]], how many intervals remain after merging overlapping ones?

A 4

B 3

C 2

D 1

💡 Hint

Check which intervals overlap or touch at edges.

Intervals [1,4] and [4,5] merge into [1,5]. Intervals [7,9] and [8,10] merge into [7,10]. So total 2 merged intervals.

🔧 Debug

advanced

Identify the error in merging intervals code

What error does this Go code produce when merging intervals? func merge(intervals [][]int) [][]int { merged := [][]int{} for _, interval := range intervals { if len(merged) == 0 || merged[len(merged)-1][1] < interval[0] { merged = append(merged, interval) } else { if merged[len(merged)-1][1] < interval[1] { merged[len(merged)-1][1] = interval[1] } } } return merged } Note: The input intervals are not sorted.

A Syntax error: missing import statement

B No error, code runs correctly

C Runtime panic: index out of range

D Output is incorrect because intervals are not sorted first

💡 Hint

Think about the importance of sorting intervals before merging.

Without sorting, intervals may not be in order, so merging logic fails to combine all overlapping intervals correctly, producing wrong output.

❓ Predict Output

advanced

Output of merging intervals with nested overlaps

What is the output of this Go code merging intervals with nested overlaps?

DSA Go

package main
import (
	"fmt"
	"sort"
)
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 {
			if merged[len(merged)-1][1] < interval[1] {
				merged[len(merged)-1][1] = interval[1]
			}
		}
	}
	return merged
}
func main() {
	intervals := [][]int{{1,10},{2,6},{8,10},{15,18}}
	result := merge(intervals)
	fmt.Println(result)
}

A [[1,10],[15,18]]

B [[1,6],[8,10],[15,18]]

C [[1,18]]

D [[2,6],[8,10],[15,18]]

💡 Hint

Nested intervals merge into the largest containing interval.

Intervals [1,10], [2,6], and [8,10] overlap and merge into [1,10]. Interval [15,18] remains separate.

🚀 Application

expert

Minimum number of meeting rooms needed

Given intervals representing meeting times [[0,30],[5,10],[15,20]], what is the minimum number of meeting rooms required to hold all meetings without overlap?

A 3

B 2

C 1

D 4

💡 Hint

Count how many meetings overlap at the same time.

Meetings [0,30] and [5,10] overlap, so at least 2 rooms are needed. Meeting [15,20] overlaps with [0,30] but not with [5,10] after it ends.

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