How do you solve Interval List Intersections on LeetCode?

Interval List Intersections (LeetCode #986) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Both lists are sorted, allowing for efficient traversal with two pointers.

What is the time complexity of Interval List Intersections?

The optimal solution for Interval List Intersections runs in O(n) time and uses O(n) space. This complexity is linear because we traverse each list only once, making it efficient for larger inputs.

What is the brute force solution for Interval List Intersections?

The brute force approach for Interval List Intersections is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every interval in the first list against every interval in the second list to find intersections. This is straightforward but inefficient for larger lists. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Interval List Intersections?

Interval List Intersections uses the following concepts: Array, Two Pointers, Sweep Line. The recommended patterns to study are: Two Pointers, Sweep Line.

What are the interview tips for Interval List Intersections?

Always consider edge cases, such as empty lists or non-overlapping intervals. Explain your thought process clearly while coding; this helps interviewers understand your approach. Practice similar problems to get comfortable with interval manipulation.

What are common mistakes when solving Interval List Intersections?

Failing to handle cases where one interval completely overlaps another. Not correctly updating pointers based on which interval ends first.

#986

Interval List Intersections

Medium

ArrayTwo PointersSweep LineTwo PointersSweep Line

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses two pointers to traverse both lists simultaneously. This is efficient because both lists are sorted, allowing us to skip unnecessary comparisons.

⚙️

Algorithm

4 steps

1Step 1: Initialize two pointers, one for each list.
2Step 2: Compare the current intervals pointed to by both pointers.
3Step 3: If they overlap, calculate the intersection and add it to the result.
4Step 4: Move the pointer of the interval that ends first to the next interval.

solution.py13 lines

1# Full working Python code
2result = []
3i, j = 0, 0
4while i < len(firstList) and j < len(secondList):
5    a, b = firstList[i]
6    c, d = secondList[j]
7    if a <= d and c <= b:
8        result.append([max(a, c), min(b, d)])
9    if b < d:
10        i += 1
11    else:
12        j += 1
13return result

ℹ

Complexity note: This complexity is linear because we traverse each list only once, making it efficient for larger inputs.

1Both lists are sorted, allowing for efficient traversal with two pointers.
2The intersection of intervals can be calculated using max and min functions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.