##### Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/2477.html

# 2477. Minimum Fuel Cost to Report to the Capital

• Difficulty: Medium.
• Related Topics: .
• Similar Questions: Binary Tree Postorder Traversal.

## 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] = [ai, bi] denotes that there exists a bidirectional road connecting cities ai and bi.

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.

Example 1:

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.


Example 2:

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.


Example 3:

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


Constraints:

• 1 <= n <= 105

• roads.length == n - 1

• roads[i].length == 2

• 0 <= ai, bi < n

• ai != bi

• roads represents a valid tree.

• 1 <= seats <= 105

## Solution (Java, C++, Python)

• class Solution {
private List<Integer>[] g;
private long ans;
private int seats;

public long minimumFuelCost(int[][] roads, int seats) {
int n = roads.length + 1;
g = new List[n];
Arrays.setAll(g, k -> new ArrayList<>());
this.seats = seats;
for (var e : roads) {
int a = e[0], b = e[1];
}
dfs(0, -1);
return ans;
}

private int dfs(int a, int fa) {
int size = 1;
for (int b : g[a]) {
if (b != fa) {
int t = dfs(b, a);
ans += (t + seats - 1) / seats;
size += t;
}
}
return size;
}
}

• class Solution {
public:
long long minimumFuelCost(vector<vector<int>>& roads, int seats) {
int n = roads.size() + 1;
vector<vector<int>> g(n);
for (auto& e : roads) {
int a = e[0], b = e[1];
g[a].emplace_back(b);
g[b].emplace_back(a);
}
long long ans = 0;
function<int(int, int)> dfs = [&](int a, int fa) -> int {
int size = 1;
for (int b : g[a]) {
if (b != fa) {
int t = dfs(b, a);
ans += (t + seats - 1) / seats;
size += t;
}
}
return size;
};
dfs(0, -1);
return ans;
}
};

• class Solution:
def minimumFuelCost(self, roads: List[List[int]], seats: int) -> int:
def dfs(a, fa):
size = 1
for b in g[a]:
if b != fa:
t = dfs(b, a)
nonlocal ans
ans += (t + seats - 1) // seats
size += t
return size

g = defaultdict(list)
g[a].append(b)
g[b].append(a)
ans = 0
dfs(0, -1)
return ans


• func minimumFuelCost(roads [][]int, seats int) int64 {
g := make([][]int, n)
for _, e := range roads {
a, b := e[0], e[1]
g[a] = append(g[a], b)
g[b] = append(g[b], a)
}
ans := 0
var dfs func(int, int) int
dfs = func(a, fa int) int {
size := 1
for _, b := range g[a] {
if b != fa {
t := dfs(b, a)
ans += (t + seats - 1) / seats
size += t
}
}
return size
}
dfs(0, -1)
return int64(ans)
}

• function minimumFuelCost(roads: number[][], seats: number): number {
const n = roads.length + 1;
const g: number[][] = Array.from({ length: n }, () => []);
for (const [a, b] of roads) {
g[a].push(b);
g[b].push(a);
}
let ans = 0;
const dfs = (a: number, fa: number): number => {
let sz = 1;
for (const b of g[a]) {
if (b !== fa) {
const t = dfs(b, a);
ans += Math.ceil(t / seats);
sz += t;
}
}
return sz;
};
dfs(0, -1);
return ans;
}


• impl Solution {
pub fn minimum_fuel_cost(roads: Vec<Vec<i32>>, seats: i32) -> i64 {
let n = roads.len() + 1;
let mut g: Vec<Vec<usize>> = vec![vec![]; n];
let a = road[0] as usize;
let b = road[1] as usize;
g[a].push(b);
g[b].push(a);
}
let mut ans = 0;
fn dfs(a: usize, fa: i32, g: &Vec<Vec<usize>>, ans: &mut i64, seats: i32) -> i32 {
let mut sz = 1;
for &b in g[a].iter() {
if (b as i32) != fa {
let t = dfs(b, a as i32, g, ans, seats);
*ans += ((t + seats - 1) / seats) as i64;
sz += t;
}
}
sz
}
dfs(0, -1, &g, &mut ans, seats);
ans
}
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).