Why Insert Interval into Sorted List in DSA Go?

The Big Idea

What if you could add new time slots without worrying about overlaps or order?

The Scenario

Imagine you have a list of booked meeting times sorted by start time. Now, you want to add a new meeting without messing up the order or overlapping times.

The Problem

Manually checking where to insert the new meeting and adjusting overlapping times is slow and error-prone. You might miss overlaps or insert in the wrong place, causing confusion.

The Solution

Using the Insert Interval method, you can automatically find the right spot and merge overlapping intervals smoothly, keeping the list sorted and clean.

Before vs After

✗ Before

intervals := [][]int{{1,3},{6,9}}
newInterval := []int{2,5}
// Manually check and merge intervals one by one

✓ After

intervals := [][]int{{1,3},{6,9}}
newInterval := []int{2,5}
intervals = insertInterval(intervals, newInterval)

What It Enables

This lets you efficiently manage and update schedules or ranges without errors or extra effort.

Real Life Example

Adding a new appointment to a calendar app that automatically adjusts overlapping events and keeps the schedule tidy.

Key Takeaways

Manual insertion and merging of intervals is slow and error-prone.

Insert Interval automates finding the right place and merging overlaps.

It keeps interval lists sorted and clean for easy management.

Practice

1. What is the main goal when inserting a new interval into a sorted list of non-overlapping intervals?

easy

A. To merge overlapping intervals and keep the list sorted

B. To add the new interval at the end without merging

C. To remove all intervals that come before the new interval

D. To sort the list again after insertion

Solution

1. Step 1: Understand the problem context

   The list contains sorted, non-overlapping intervals. Inserting a new interval may cause overlaps.

2. Step 2: Determine the correct action

   We must merge any overlapping intervals with the new one to keep the list sorted and non-overlapping.

3. Final Answer:

   To merge overlapping intervals and keep the list sorted -> Option A

4. Quick Check:

   Insert interval = merge and keep sorted [OK]

Hint: Merge overlaps to keep intervals sorted and non-overlapping [OK]

Common Mistakes:

• Just appending without merging
• Removing intervals incorrectly
• Sorting again after insertion

2. Which Go syntax correctly declares a slice of intervals where each interval is a slice of two integers?

easy

A. intervals := [][2]int{{1, 3}, {6, 9}}

B. intervals := [][]int{1, 3, 6, 9}

C. intervals := []int{1, 3, 6, 9}

D. intervals := [][]int{{1, 3}, {6, 9}}

Solution

1. Step 1: Identify slice of slices syntax

   In Go, [][]int means a slice of slices of int, suitable for intervals like [start, end].

2. Step 2: Check each option

   intervals := [][]int{{1, 3}, {6, 9}} correctly uses [][]int with nested slices for intervals. intervals := [][2]int{{1, 3}, {6, 9}} uses array [2]int which is valid but less common. intervals := []int{1, 3, 6, 9} is a flat slice, not intervals. intervals := [][]int{1, 3, 6, 9} is invalid syntax.

3. Final Answer:

   intervals := [][]int{{1, 3}, {6, 9}} -> Option D

4. Quick Check:

   Slice of slices = [][]int{{start,end}} [OK]

Hint: Use [][]int for slice of intervals in Go [OK]

Common Mistakes:

• Using flat []int instead of [][]int
• Incorrect array vs slice syntax
• Missing curly braces for nested slices

3. What is the output of the following Go code snippet after inserting interval [2,5] into intervals [[1,2],[6,9]]?

intervals := [][]int{{1, 2}, {6, 9}}
newInterval := []int{2, 5}
// After insertion and merging
result := [][]int{{1, 5}, {6, 9}}
fmt.Println(result)

medium

A. [[1 2] [6 9] [2 5]]

B. [[1 2] [2 5] [6 9]]

C. [[1 5] [6 9]]

D. [[2 5] [6 9]]

Solution

1. Step 1: Insert and check overlap

   New interval [2,5] overlaps with [1,2] because 2 <= 2. Merge them to [1,5].

2. Step 2: Keep other intervals

   [6,9] does not overlap with [1,5], so it stays as is.

3. Final Answer:

   [[1 5] [6 9]] -> Option C

4. Quick Check:

   Merged intervals = [[1 5] [6 9]] [OK]

Hint: Merge overlapping intervals to combine ranges [OK]

Common Mistakes:

• Not merging overlapping intervals
• Appending new interval without sorting
• Incorrect interval order

4. Given this Go function snippet to insert an interval, what is the bug?

func insert(intervals [][]int, newInterval []int) [][]int {
    var result [][]int
    for _, interval := range intervals {
        if interval[1] < newInterval[0] {
            result = append(result, interval)
        } else if interval[0] > newInterval[1] {
            result = append(result, newInterval)
            newInterval = interval
        } else {
            newInterval[0] = min(newInterval[0], interval[0])
            newInterval[1] = max(newInterval[1], interval[1])
        }
    }
    result = append(result, newInterval)
    return result
}

medium

A. Appending newInterval multiple times causes duplicates

B. Not updating newInterval correctly during merge

C. Missing return statement

D. Incorrect comparison operators in if conditions

Solution

1. Step 1: Analyze appending logic

   When interval[0] > newInterval[1], newInterval is appended and then replaced by interval. This can cause multiple appends of newInterval.

2. Step 2: Identify duplicate insertion

   Because newInterval is appended inside the loop multiple times, duplicates appear in result.

3. Final Answer:

   Appending newInterval multiple times causes duplicates -> Option A

4. Quick Check:

   Multiple appends = duplicates [OK]

Hint: Append newInterval once after loop to avoid duplicates [OK]

Common Mistakes:

• Appending inside loop multiple times
• Not updating newInterval boundaries
• Forgetting to return result

5. You have intervals [[1,3],[6,9]] and want to insert [2,7]. What is the final list after insertion and merging?

hard

A. [[1,3],[2,7],[6,9]]

B. [[1,9]]

C. [[1,7],[6,9]]

D. [[1,9],[6,9]]

Solution

1. Step 1: Merge [2,7] with overlapping intervals

   [2,7] overlaps with [1,3] and [6,9]. Merge all three intervals into one covering from min start 1 to max end 9.

2. Step 2: Final merged interval

   The merged interval is [1,9], replacing the original two intervals.

3. Final Answer:

   [[1,9]] -> Option B

4. Quick Check:

   Merge all overlapping = [[1,9]] [OK]

Hint: Merge all overlapping intervals into one big interval [OK]

Common Mistakes:

• Partial merges leaving overlaps
• Appending without merging
• Incorrect interval boundaries