Gas Station


1. Clarify the problem: Before diving into solving the problem, let's clarify the requirements:

  • We are given two arrays: gas and cost, both of size n.
  • The gas array represents the amount of gas available at each gas station.
  • The cost array represents the amount of gas required to travel from one station to the next.
  • We need to find the starting gas station index that allows us to complete a circular trip around all the gas stations.
  • If it is impossible to complete the trip, we should return -1.

2. Analyze the problem: To solve this problem, we can use the concept of a "circular tour". We need to find a starting point such that we can make a complete circular tour without running out of gas.

3. Design an algorithm: Here is the algorithm to find the starting gas station:

  1. Initialize start and total to 0.
  2. Initialize tank to 0.
  3. Iterate through the gas stations from i = 0 to n-1.
    • Increment total by gas[i] - cost[i].
    • Increment tank by gas[i] - cost[i].
    • If tank is negative, update start to i+1 and reset tank to 0.
  4. If total is negative, return -1 (no solution).
  5. Otherwise, return start.

4. Explain your approach: The approach involves iterating through the gas stations while keeping track of the total gas and the current tank. We increment the total gas by gas[i] - cost[i] at each station, representing the net amount of gas we gain or lose by reaching that station. We also increment the current tank by the same amount. If the tank becomes negative at any station, it means we cannot start the tour from the current station and need to update the starting point to the next station. We reset the tank to 0 and continue the iteration. If the total gas is negative at the end, it means we cannot complete the circular tour, so we return -1. Otherwise, we return the starting point.

5. Write clean and readable code:

def canCompleteCircuit(gas, cost): n = len(gas) start = 0 total = 0 tank = 0 for i in range(n): total += gas[i] - cost[i] tank += gas[i] - cost[i] if tank < 0: start = i + 1 tank = 0 if total < 0: return -1 else: return start

6. Test your code:

# Test case 1 gas1 = [1, 2, 3, 4, 5] cost1 = [3, 4, 5, 1, 2] print(canCompleteCircuit(gas1, cost1)) # Output: 3 # Test case 2 gas2 = [2, 3, 4] cost2 = [3, 4, 3] print(canCompleteCircuit(gas2, cost2)) # Output: -1 # Test case 3 gas3 = [5, 1, 2, 3, 4] cost3 = [4, 4, 1, 5, 1] print(canCompleteCircuit(gas3, cost3)) # Output: 4

7. Optimize if necessary: The provided solution already has an optimal time complexity of O(n), where n is the length of the gas and cost arrays. We only traverse the arrays once to find the starting point. There is no need for further optimization.

8. Handle error cases: The code assumes that the inputs are valid arrays of the same length. However, it does not explicitly handle invalid input or error cases. If the input arrays are empty or have different lengths, the behavior of the code may be unpredictable.

9. Discuss complexity analysis: The time complexity of the solution is O(n), where n is the length of the gas and cost arrays. This is because we traverse the arrays once to find the starting point. The space complexity is O(1) as we only use a constant amount of additional space to store the start, total, and tank variables.

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