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What is the time complexity of the optimal two pointers approach for finding the intersection of two interval lists, each of length m and n respectively?

medium🪤 Complexity Trap Q13 of Q15
Intervals - Interval List Intersections
What is the time complexity of the optimal two pointers approach for finding the intersection of two interval lists, each of length m and n respectively?
AO(max(m, n) * log(min(m, n))) due to sorting or binary search steps.
BO(m * n), since each interval in one list is compared to all intervals in the other list.
CO(m + n), because each pointer advances at most m or n times through the lists.
DO(m + n + k), where k is the number of intersections found.
Step-by-Step Solution

  1. Step 1: Identify pointer movements

    Each pointer i and j moves forward only, never backward, up to m and n times respectively.

  2. Step 2: Sum pointer advances

    Total steps ≤ m + n, so time complexity is O(m + n).

  3. Final Answer:

    Option C -> Option C

  4. Quick Check:

    Two pointers scan each list once without nested loops [OK]
Quick Trick: Two pointers each advance linearly, no nested iteration [OK]
Common Mistakes:
MISTAKES
  • Confusing with brute force O(m*n)
  • Assuming sorting is needed
  • Including output size k in complexity incorrectly
Trap Explanation:
PITFALL
  • O(m*n) looks plausible but ignores that pointers advance linearly; output size k is not dominant here.
Interviewer Note:
CONTEXT
  • Checks understanding of pointer-based linear scan complexity.
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