How do you solve Count All Possible Routes on LeetCode?

Count All Possible Routes (LeetCode #1575) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * fuel) time complexity and O(n * fuel) space complexity. Key insight: Using memoization can drastically reduce the number of calculations needed by storing results of subproblems.

What is the brute force solution for Count All Possible Routes?

The brute force approach for Count All Possible Routes is Brute Force, with O(n^2 * fuel) time complexity and O(n * fuel) space complexity. The brute-force approach explores all possible routes from the starting city to the finishing city by recursively visiting each city and checking if we have enough fuel to continue. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n * fuel).

What algorithm or data structure is used to solve Count All Possible Routes?

Count All Possible Routes uses the following concepts: Array, Dynamic Programming, Memoization. The recommended patterns to study are: Dynamic Programming, Memoization, Recursion.

What are the interview tips for Count All Possible Routes?

Always think about how to optimize your solution with memoization or dynamic programming. Clarify the constraints and edge cases before diving into coding. Practice explaining your thought process as you code, as this will help you articulate your approach in interviews.

What are common mistakes when solving Count All Possible Routes?

Failing to account for the fuel cost correctly when moving between cities. Not using memoization effectively, leading to excessive recomputation and timeouts.

#1575

Count All Possible Routes

Hard

ArrayDynamic ProgrammingMemoizationDynamic ProgrammingMemoizationRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n^2 * fuel)	O(n * fuel)
Space	O(n * fuel)	O(n * fuel)

💡

Intuition

Time O(n * fuel)Space O(n * fuel)

The optimal solution uses dynamic programming with memoization to store results of subproblems, avoiding redundant calculations. This significantly reduces the time complexity by reusing previously computed results.

⚙️

Algorithm

3 steps

1Step 1: Initialize a memoization table to store results for each city and remaining fuel.
2Step 2: Define a recursive function that checks all possible routes from the current city to the finish city, using the remaining fuel.
3Step 3: For each city, calculate the fuel cost to move to other cities and recursively call the function, storing results in the memoization table.

solution.py20 lines

1# Full working Python code
2class Solution:
3    def countRoutes(self, locations, start, finish, fuel):
4        mod = 10**9 + 7
5        n = len(locations)
6        memo = {}  # Memoization dictionary
7
8        def dfs(city, remaining_fuel):
9            if (city, remaining_fuel) in memo:
10                return memo[(city, remaining_fuel)]
11            count = 1 if city == finish else 0
12            for next_city in range(n):
13                if next_city != city:
14                    cost = abs(locations[city] - locations[next_city])
15                    if remaining_fuel >= cost:
16                        count += dfs(next_city, remaining_fuel - cost)
17            memo[(city, remaining_fuel)] = count % mod
18            return memo[(city, remaining_fuel)]
19
20        return dfs(start, fuel)

ℹ

Complexity note: The time complexity is O(n * fuel) because we explore each city and the remaining fuel can be up to 'fuel'. The space complexity is O(n * fuel) due to the memoization table.

1Using memoization can drastically reduce the number of calculations needed by storing results of subproblems.
2Recursive exploration of routes can be efficiently managed with a well-defined state (city and remaining fuel).

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.