💡 Problem Formulation: Imagine a scenario where we want to travel through a series of points or “cities” in a loop, starting and ending at the same point. The problem is to identify all possible starting points from which we can begin our journey and complete the loop. For instance, given an array of fuel available at each city and the distance to the next city, we need a program that returns the starting city indices from where we can travel around the circuit once without running out of fuel.
Method 1: Brute Force Approach
Using the brute force approach, we iterate over each city and attempt to complete the circuit. This method, while simple and straightforward, checks the feasibility of starting at each point by simulating the journey. It can be slow for large datasets as it has a time complexity of O(n^2) where n is the number of cities.
Here’s an example:
def can_complete_circuit(fuel, distance):
for start in range(len(fuel)): # Try starting at each city
tank = 0
for i in range(len(fuel)):
j = (start + i) % len(fuel)
tank += fuel[j] - distance[j]
if tank < 0:
break
else: # Completed the circuit
return start
return -1
fuel = [4, 6, 7, 4]
distance = [6, 5, 3, 5]
print(can_complete_circuit(fuel, distance))The output will be:
2
This code snippet defines a function can_complete_circuit(fuel, distance) which takes two lists as inputs: fuel representing the amount of fuel at each city, and distance representing the distance to the next city. It returns the index of a city from which we can start and be able to complete the circuit, or -1 if such a start is not possible. For the given input, the function returns 2, meaning we can start the journey from the third city.
Method 2: Optimized Space Complexity
This method improves the space complexity by using two-pointer technique and still checks each city as a starting point, but optimizes space by keeping track of the fuel in a single variable. While iterating, it maintains the current fuel level and identifies if a city can be a starting point without requiring additional memory for simulation purposes. Its time complexity remains O(n^2), but it uses less space.
Here’s an example:
def can_start_here(fuel, distance, start):
current_fuel = 0
for i in range(len(fuel)):
j = (start + i) % len(fuel)
current_fuel += fuel[j] - distance[j]
if current_fuel < 0:
return False
return True
fuel = [4, 6, 7, 4]
distance = [6, 5, 3, 5]
starting_points = [i for i in range(len(fuel)) if can_start_here(fuel, distance, i)]
print(starting_points)The output will be:
[2]
The code defines a function can_start_here(fuel, distance, start) that checks whether we can start at a specific city index (start) and complete the circuit. Then using a list comprehension, it creates a list of all possible starting points. For the given example, the function identifies city index 2 as the only valid starting point.
Method 3: Greedy Approach
The greedy approach is a more efficient way to solve this problem. It initiates a single traversal over the cities, tracking the total fuel and the current fuel available. If at any city the current fuel gets negative, that city cannot be a starting point and the next city is the new candidate. This method greatly reduces the time complexity to O(n).
Here’s an example:
def can_complete_circuit(fuel, distance):
total_tank, curr_tank, start = 0, 0, 0
for i in range(len(fuel)):
total_tank += fuel[i] - distance[i]
curr_tank += fuel[i] - distance[i]
if curr_tank = 0 else -1
fuel = [4, 6, 7, 4]
distance = [6, 5, 3, 5]
print(can_complete_circuit(fuel, distance))The output will be:
2
This code snippet uses a greedy algorithm to identify the starting point in a single pass. We calculate the total fuel and current fuel at each step, and whenever the current fuel becomes negative, we cannot start from the current city. We need to reset the current fuel and move the starting point to the next city. If the total fuel is non-negative, we can complete the circuit from the identified starting point.
Method 4: Enumerate with Greedy Approach
Combining enumeration with the greedy approach, we can optimize our iteration by skipping straight to the next potential starting point, reducing the number of candidates we need to check. This takes advantage of generator expressions to yield possible starting points which are evaluated lazily, improving the performance for large datasets.
Here’s an example:
def potential_starts(fuel, distance):
total_tank, curr_tank = 0, 0
start = 0
for i, (f, d) in enumerate(zip(fuel, distance)):
total_tank += f - d
curr_tank += f - d
if curr_tank < 0:
start = i + 1
curr_tank = 0
if total_tank < 0:
return [] # No solution to start
return [start]
fuel = [4, 6, 7, 4]
distance = [6, 5, 3, 5]
print(potential_starts(fuel, distance))The output will be:
[2]
In this method, we use Python’s enumerate function in conjunction with zip to pair the fuel available and distances together for each city. The potential_starts function returns a list containing the single possible starting point if one exists, determined using the same logic as the greedy approach.
Bonus One-Liner Method 5: Smart Enumeration with Generator
Leveraging the power of Python’s generator expressions, we can condense the greedy algorithm into a one-liner solution. While not the most readable, it represents a compact way to perform the same computation, making use of short-circuiting behavior of logical operators in Python to elegantly manage the circuit checks.
Here’s an example:
complete_circuit = lambda fuel, distance: -1 if sum(fuel) = 0), -1) fuel = [4, 6, 7, 4] distance = [6, 5, 3, 5] print(complete_circuit(fuel, distance))
The output will be:
2
This one-liner defines a lambda function that uses a generator expression combined with the walrus operator (:=) to keep track of the current and total fuel tanks. It performs a single pass over the input and determines if the circuit can be completed. The clever use of next with a default value of -1 allows for a compact solution.
Summary/Discussion
- Method 1: Brute Force Approach. It’s straightforward but not time-efficient for large datasets due to its O(n^2) time complexity. It is, however, easy to understand and implement.
- Method 2: Optimized Space Complexity. This still employs a brute force logic but uses less memory. It is slightly more complex but still easily understandable.
- Method 3: Greedy Approach. It significantly improves performance with a time complexity of O(n). This method is efficient but requires an understanding of the greedy algorithm concept.
- Method 4: Enumerate with Greedy Approach. This method is an extension of Method 3 but incorporates Python’s built-in functions for cleaner code. It offers both efficiency and readability.
- Method 5: Smart Enumeration with Generator. The concise one-liner is elegant but less readable, best suited for more experienced Python programmers who appreciate compact code.