What is the time complexity of the bottom-up DP solution with binary search for the minimum cost tickets problem, given n travel days?

medium🪤 Complexity Trap Q13 of Q15

Dynamic Programming: Knapsack - Minimum Cost for Tickets

AO(n^2) because for each day we scan all future days

BO(n log n) because for each day we perform 3 binary searches over the days array

CO(n) because we only iterate once over the days array

DO(n * 3 * n) due to nested loops over days and durations

Step-by-Step Solution

Step 1: Identify loops and operations
Outer loop runs n times (for each day). Inner loop runs 3 times (for each ticket duration). Each inner iteration does a binary search (log n) to find next uncovered day.
Step 2: Calculate total complexity
Total time = n * 3 * log n = O(n log n). Linear scan or nested loops over days would be O(n^2), but binary search reduces it.
Final Answer:
Option B -> Option B
Quick Check:
Binary search reduces inner loop from O(n) to O(log n) [OK]

Quick Trick: Binary search reduces complexity from quadratic to n log n [OK]

Common Mistakes:

MISTAKES

Trap Explanation:

PITFALL

Option A looks plausible because naive approach is quadratic, but binary search optimizes it.

Interviewer Note:

CONTEXT

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