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What is the overall time complexity of a bottom-up dynamic programming solution that uses binary search to find the next valid travel day when computing the minimum cost tickets for n travel days?

medium📊 Complexity Q5 of Q15

Dynamic Programming: Knapsack - Minimum Cost for Tickets

What is the overall time complexity of a bottom-up dynamic programming solution that uses binary search to find the next valid travel day when computing the minimum cost tickets for n travel days?

AO(n log n)

BO(n^2)

CO(n)

DO(log n)

Step-by-Step Solution

Solution:

Step 1: Understand DP iteration over n days
The bottom-up DP iterates over all n travel days.
Step 2: Binary search per iteration costs O(log n)
For each day, binary search is used to find the next day not covered by the current ticket.
Step 3: Combine complexities for total time
Total time is O(n) iterations * O(log n) binary search = O(n log n).
Final Answer:
Option A -> Option A
Quick Check:
Binary search inside a loop over n elements yields O(n log n) [OK]

Quick Trick: Binary search inside DP loop leads to n log n time [OK]

Common Mistakes:

MISTAKES

Assuming O(n) ignoring binary search cost
Confusing with O(n^2) by assuming linear search
Mistaking binary search cost as O(log log n)

Trap Explanation:

PITFALL

Ignoring binary search cost leads to underestimating complexity

Interviewer Note:

CONTEXT

Tests understanding of time complexity in DP with binary search optimization

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