In the Gas Station (Circular) problem, if you implement the optimal greedy solution by iterating through the stations once and updating the start position when the current tank becomes negative, what is the overall time complexity for n stations?

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Greedy Algorithms - Gas Station (Circular)

AO(n log n)

BO(n)

CO(n²)

DO(log n)

Step-by-Step Solution

Solution:

Step 1: Understand the greedy approach
The algorithm iterates through the gas stations once, maintaining a running tank balance.
Step 2: Reset start index when tank is negative
When the tank drops below zero, the start index is moved to the next station, and the tank is reset.
Step 3: Single pass through stations
Each station is visited at most twice (once in the main loop and once when resetting), resulting in linear time complexity.
Final Answer:
Option B -> Option B
Quick Check:
Single pass with resets means linear time [OK]

Quick Trick: Single pass with resets yields O(n) time complexity [OK]

Common Mistakes:

MISTAKES

Assuming nested loops cause O(n²) complexity
Confusing prefix sums approach with sorting
Thinking resetting start index causes multiple full passes

Trap Explanation:

PITFALL

Incorrectly assuming multiple resets cause quadratic time, but resets only move forward.

Interviewer Note:

CONTEXT

Tests understanding of greedy approach efficiency and time complexity analysis.

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More Greedy Algorithms Quizzes

In the Gas Station (Circular) problem, if you implement the optimal greedy solution by iterating through the stations once and updating the start position when the current tank becomes negative, what is the overall time complexity for n stations?

Step 1: Understand the greedy approach

Step 2: Reset start index when tank is negative

Step 3: Single pass through stations

Final Answer:

Quick Check:

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