Minimum Fuel Cost to Report to the Capital

Tested Python solution for LeetCode 2477 with 20 pytest cases. Generate a practice environment with lcpy.

LeetCode 2477, Medium. Topics: Tree, Depth-First Search, Breadth-First Search, Graph Theory. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 20 parametrized pytest cases, and a playground notebook:

lcpy gen -n 2477   # by problem number
lcpy gen -s minimum_fuel_cost_to_report_to_the_capital   # by problem name

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Problem

There is a tree (i.e., a connected, undirected graph with no cycles) structure country network consisting of n cities numbered from 0 to n - 1 and exactly n - 1 roads. The capital city is city 0. You are given a 2D integer array roads where roads[i] = [a_i, b_i] denotes that there exists a bidirectional road connecting cities a_i and b_i.

There is a meeting for the representatives of each city. The meeting is in the capital city.

There is a car in each city. You are given an integer seats that indicates the number of seats in each car.

A representative can use the car in their city to travel or change the car and ride with another representative. The cost of traveling between two cities is one liter of fuel.

Return the minimum number of liters of fuel to reach the capital city.

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Examples

Input: roads = [[0,1],[0,2],[0,3]], seats = 5
Output: 3
Explanation:
- Representative1 goes directly to the capital with 1 liter of fuel.
- Representative2 goes directly to the capital with 1 liter of fuel.
- Representative3 goes directly to the capital with 1 liter of fuel.
It costs 3 liters of fuel at minimum.
It can be proven that 3 is the minimum number of liters of fuel needed.

Input: roads = [[3,1],[3,2],[1,0],[0,4],[0,5],[4,6]], seats = 2
Output: 7
Explanation:
- Representative2 goes directly to city 3 with 1 liter of fuel.
- Representative2 and representative3 go together to city 1 with 1 liter of fuel.
- Representative2 and representative3 go together to the capital with 1 liter of fuel.
- Representative1 goes directly to the capital with 1 liter of fuel.
- Representative5 goes directly to the capital with 1 liter of fuel.
- Representative6 goes directly to city 4 with 1 liter of fuel.
- Representative4 and representative6 go together to the capital with 1 liter of fuel.
It costs 7 liters of fuel at minimum.
It can be proven that 7 is the minimum number of liters of fuel needed.

Input: roads = [], seats = 1
Output: 0
Explanation: No representatives need to travel to the capital city.

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Constraints

• 1 <= n <= 10^5
• roads.length == n - 1
• roads[i].length == 2
• 0 <= a_i, b_i < n
• a_i != b_i
• roads represents a valid tree.
• 1 <= seats <= 10^5

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(n)
    # Space: O(n)
    def minimum_fuel_cost(self, roads: list[list[int]], seats: int) -> int:
        n = len(roads) + 1
        if n == 1:
            return 0

        adj: list[list[int]] = [[] for _ in range(n)]
        for a, b in roads:
            adj[a].append(b)
            adj[b].append(a)

        # Iterative BFS from the capital (n up to 1e5 rules out recursion).
        parent = [-1] * n
        order = [0]
        for u in order:
            for v in adj[u]:
                if v != parent[u]:
                    parent[v] = u
                    order.append(v)

        total = 0
        size = [1] * n
        for u in reversed(order):
            p = parent[u]
            if p != -1:
                size[p] += size[u]
                total += (size[u] + seats - 1) // seats
        return total

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Complexity

Time	Space
O(n)	O(n)

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Tags

NeetCode All.