Insert Interval into Sorted List in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert the new interval into the list.

DSA Javascript

function insertInterval(intervals, newInterval) {
  let result = [];
  let i = 0;
  while (i < intervals.length && intervals[i][1] < newInterval[[1]]) {
    result.push(intervals[i]);
    i++;
  }
  return result;
}

Drag options to blanks, or click blank then click option'

A0

B1

Clength

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using newInterval[1] instead of newInterval[0] causes wrong comparisons.

Comparing intervals[i][0] instead of intervals[i][1].

We compare the end of intervals[i] with the start of newInterval, which is at index 0.

2fill in blank

medium

Complete the code to merge overlapping intervals with the new interval.

DSA Javascript

while (i < intervals.length && intervals[i][0] <= newInterval[[1]]) {
  newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
  newInterval[1] = Math.max(newInterval[1], intervals[i][[2]]);
  i++;
}

Drag options to blanks, or click blank then click option'

A0

B1

Clength

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up indices for start and end of intervals.

Using newInterval[0] instead of newInterval[1] in the while condition.

We merge intervals where intervals[i][0] <= newInterval[1]. We update newInterval[1] with the max end, which is intervals[i][1].

3fill in blank

hard

Fix the error in adding the merged newInterval to the result list.

DSA Javascript

result.[1](newInterval);

Drag options to blanks, or click blank then click option'

Aappend

Badd

Cpush

Dinsert

Attempts:

3 left

💡 Hint

Common Mistakes

Using append() which is not a JavaScript array method.

Using add() or insert() which do not exist on arrays.

In JavaScript, arrays use push() to add elements at the end.

4fill in blank

hard

Fill both blanks to add the remaining intervals after merging.

DSA Javascript

while (i [1] intervals.length) {
  result.[2](intervals[i]);
  i++;
}

Drag options to blanks, or click blank then click option'

A<

B<=

Cpush

Dappend

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= causes index out of range error.

Using append() which is not a JavaScript array method.

We loop while i is less than intervals.length and use push() to add elements to the result.

5fill in blank

hard

Fill all three blanks to complete the insertInterval function.

DSA Javascript

function insertInterval(intervals, newInterval) {
  let result = [];
  let i = 0;
  while (i < intervals.length && intervals[i][1] < newInterval[[1]]) {
    result.push(intervals[i]);
    i++;
  }
  while (i < intervals.length && intervals[i][0] <= newInterval[[2]]) {
    newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
    newInterval[1] = Math.max(newInterval[1], intervals[i][[3]]);
    i++;
  }
  result.push(newInterval);
  while (i < intervals.length) {
    result.push(intervals[i]);
    i++;
  }
  return result;
}

Drag options to blanks, or click blank then click option'

A0

B1

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping start and end indices.

Using wrong indices in comparisons causing incorrect merges.

We compare intervals[i][1] with newInterval[0] in the first loop, then intervals[i][0] with newInterval[1] in the second. We update newInterval[1] with intervals[i][1].

Practice

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

easy

A. To add the new interval at the end without checking overlaps

B. To merge overlapping intervals and keep the list sorted

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

D. To sort the intervals again after insertion without merging

Solution

1. Step 1: Understand the problem context

   We have a sorted list of intervals with no overlaps, and we want to insert a new interval.

2. Step 2: Identify the correct approach

   We must merge any intervals that overlap with the new one and keep the list sorted.

3. Final Answer:

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

4. Quick Check:

   Insert interval = merge and sort [OK]

Hint: Merge overlapping intervals to keep list sorted after insertion [OK]

Common Mistakes:

• Inserting without merging overlaps
• Appending without sorting
• Removing unrelated intervals

2. Which of the following is the correct syntax to insert an interval [3,5] into a JavaScript array named intervals?

easy

A. intervals.push([3,5])

B. intervals.insert([3,5])

C. intervals.add([3,5])

D. intervals.append([3,5])

Solution

1. Step 1: Recall JavaScript array methods

   JavaScript arrays use push() to add elements at the end.

2. Step 2: Match method to syntax

   push() is correct; insert(), add(), and append() are not valid JavaScript array methods.

3. Final Answer:

   intervals.push([3,5]) -> Option A

4. Quick Check:

   Use push() to add elements in JS arrays [OK]

Hint: Use push() to add elements to JavaScript arrays [OK]

Common Mistakes:

• Using insert() which is not a JS array method
• Using add() or append() from other languages
• Forgetting to add as an array element

3. What is the output of the following code?

const intervals = [[1,2],[5,6]];
const newInterval = [3,5];
function insertInterval(intervals, newInterval) {
  const result = [];
  let i = 0;
  while (i < intervals.length && intervals[i][1] < newInterval[0]) {
    result.push(intervals[i]);
    i++;
  }
  while (i < intervals.length && intervals[i][0] <= newInterval[1]) {
    newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
    newInterval[1] = Math.max(newInterval[1], intervals[i][1]);
    i++;
  }
  result.push(newInterval);
  while (i < intervals.length) {
    result.push(intervals[i]);
    i++;
  }
  return result;
}
console.log(insertInterval(intervals, newInterval));

medium

A. [[1,2],[5,6],[3,5]]

B. [[1,2],[3,5],[5,6]]

C. [[1,6]]

D. [[1,2],[3,6]]

Solution

1. Step 1: Process intervals before newInterval

   Intervals before newInterval are [[1,2]] because 2 < 3, so push [1,2].

2. Step 2: Merge overlapping intervals

   Intervals overlapping with newInterval are [5,6] because 5 <= 5. Merge [3,5] and [5,6] to get [3,6].

3. Step 3: Add merged interval and remaining intervals

   Push merged [3,6]. No intervals left after merging.

4. Final Answer:

   [[1,2],[3,6]] -> Option D

5. Quick Check:

   Merged intervals = [[1,2],[3,6]] [OK]

Hint: Merge intervals overlapping newInterval, keep others as is [OK]

Common Mistakes:

• Not merging overlapping intervals
• Adding newInterval multiple times
• Incorrectly ordering intervals after merge

4. Identify the bug in this code snippet that inserts a new interval into a sorted list:

function insert(intervals, newInterval) {
  let result = [];
  let i = 0;
  while (i < intervals.length && intervals[i][1] < newInterval[0]) {
    result.push(intervals[i]);
    i++;
  }
  while (i < intervals.length && intervals[i][0] < newInterval[1]) {
    newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
    newInterval[1] = Math.max(newInterval[1], intervals[i][1]);
    i++;
  }
  result.push(newInterval);
  while (i < intervals.length) {
    result.push(intervals[i]);
    i++;
  }
  return result;
}

medium

A. The newInterval is pushed before merging

B. The first while condition should use < instead of <=

C. The second while condition should use <= instead of <

D. The result array is not initialized

Solution

1. Step 1: Analyze the second while loop condition

   The second loop merges overlapping intervals. It uses intervals[i][0] < newInterval[1], which misses intervals starting exactly at newInterval[1].

2. Step 2: Correct the condition

   Change < to <= to include intervals starting at newInterval[1] for proper merging.

3. Final Answer:

   The second while condition should use <= instead of < -> Option C

4. Quick Check:

   Use <= to merge intervals starting at newInterval[1] [OK]

Hint: Use <= to include intervals starting exactly at newInterval end [OK]

Common Mistakes:

• Using < instead of <= in merge condition
• Pushing newInterval too early
• Not initializing result array

5. Given intervals = [[1,3],[6,9]] and newInterval = [2,5], what is the correct output after inserting and merging intervals?

hard

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

B. [[1,3],[2,5],[6,9]]

C. [[2,5],[6,9]]

D. [[1,9]]

Solution

1. Step 1: Identify intervals before newInterval

   Interval [1,3] overlaps with newInterval [2,5] because 3 >= 2, so no intervals come before without overlap.

2. Step 2: Merge overlapping intervals

   Merge [1,3] and [2,5] to get [1,5]. Interval [6,9] does not overlap and stays as is.

3. Final Answer:

   [[1,5],[6,9]] -> Option A

4. Quick Check:

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

Hint: Merge all overlapping intervals including newInterval [OK]

Common Mistakes:

• Not merging overlapping intervals
• Adding newInterval without merging
• Merging unrelated intervals