LeetCode #2285 — MEDIUM

Maximum Total Importance of Roads

Move from brute-force thinking to an efficient approach using greedy strategy.

The Problem

Problem Statement

You are given an integer n denoting the number of cities in a country. The cities are numbered from 0 to n - 1.

You are also 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.

You need to assign each city with an integer value from 1 to n, where each value can only be used once. The importance of a road is then defined as the sum of the values of the two cities it connects.

Return the maximum total importance of all roads possible after assigning the values optimally.

Example 1:

Input: n = 5, roads = [[0,1],[1,2],[2,3],[0,2],[1,3],[2,4]]
Output: 43
Explanation: The figure above shows the country and the assigned values of [2,4,5,3,1].
- The road (0,1) has an importance of 2 + 4 = 6.
- The road (1,2) has an importance of 4 + 5 = 9.
- The road (2,3) has an importance of 5 + 3 = 8.
- The road (0,2) has an importance of 2 + 5 = 7.
- The road (1,3) has an importance of 4 + 3 = 7.
- The road (2,4) has an importance of 5 + 1 = 6.
The total importance of all roads is 6 + 9 + 8 + 7 + 7 + 6 = 43.
It can be shown that we cannot obtain a greater total importance than 43.

Example 2:

Input: n = 5, roads = [[0,3],[2,4],[1,3]]
Output: 20
Explanation: The figure above shows the country and the assigned values of [4,3,2,5,1].
- The road (0,3) has an importance of 4 + 5 = 9.
- The road (2,4) has an importance of 2 + 1 = 3.
- The road (1,3) has an importance of 3 + 5 = 8.
The total importance of all roads is 9 + 3 + 8 = 20.
It can be shown that we cannot obtain a greater total importance than 20.

Constraints:

2 <= n <= 5 * 10⁴
1 <= roads.length <= 5 * 10⁴
roads[i].length == 2
0 <= a_i, b_i <= n - 1
a_i != b_i
There are no duplicate roads.

Step 01

Brute Force Baseline

Problem summary: You are given an integer n denoting the number of cities in a country. The cities are numbered from 0 to n - 1. You are also given a 2D integer array roads where roads[i] = [ai, bi] denotes that there exists a bidirectional road connecting cities ai and bi. You need to assign each city with an integer value from 1 to n, where each value can only be used once. The importance of a road is then defined as the sum of the values of the two cities it connects. Return the maximum total importance of all roads possible after assigning the values optimally.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Greedy

Example 1

5
[[0,1],[1,2],[2,3],[0,2],[1,3],[2,4]]

Example 2

5
[[0,3],[2,4],[1,3]]

Step 02

Core Insight

What unlocks the optimal approach

Consider what each city contributes to the total importance of all roads.
Based on that, how can you sort the cities such that assigning them values in that order will yield the maximum total importance?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2285: Maximum Total Importance of Roads
class Solution {
    public long maximumImportance(int n, int[][] roads) {
        int[] deg = new int[n];
        for (int[] r : roads) {
            ++deg[r[0]];
            ++deg[r[1]];
        }
        Arrays.sort(deg);
        long ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += (long) (i + 1) * deg[i];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2285: Maximum Total Importance of Roads
func maximumImportance(n int, roads [][]int) (ans int64) {
	deg := make([]int, n)
	for _, r := range roads {
		deg[r[0]]++
		deg[r[1]]++
	}
	sort.Ints(deg)
	for i, x := range deg {
		ans += int64(x) * int64(i+1)
	}
	return
}

# Accepted solution for LeetCode #2285: Maximum Total Importance of Roads
class Solution:
    def maximumImportance(self, n: int, roads: List[List[int]]) -> int:
        deg = [0] * n
        for a, b in roads:
            deg[a] += 1
            deg[b] += 1
        deg.sort()
        return sum(i * v for i, v in enumerate(deg, 1))

// Accepted solution for LeetCode #2285: Maximum Total Importance of Roads
/**
 * [2285] Maximum Total Importance of Roads
 *
 * You are given an integer n denoting the number of cities in a country. The cities are numbered from 0 to n - 1.
 * You are also given a 2D integer array roads where roads[i] = [ai, bi] denotes that there exists a bidirectional road connecting cities ai and bi.
 * You need to assign each city with an integer value from 1 to n, where each value can only be used once. The importance of a road is then defined as the sum of the values of the two cities it connects.
 * Return the maximum total importance of all roads possible after assigning the values optimally.
 *  
 * Example 1:
 * <img alt="" src="https://assets.leetcode.com/uploads/2022/04/07/ex1drawio.png" style="width: 290px; height: 215px;" />
 * Input: n = 5, roads = [[0,1],[1,2],[2,3],[0,2],[1,3],[2,4]]
 * Output: 43
 * Explanation: The figure above shows the country and the assigned values of [2,4,5,3,1].
 * - The road (0,1) has an importance of 2 + 4 = 6.
 * - The road (1,2) has an importance of 4 + 5 = 9.
 * - The road (2,3) has an importance of 5 + 3 = 8.
 * - The road (0,2) has an importance of 2 + 5 = 7.
 * - The road (1,3) has an importance of 4 + 3 = 7.
 * - The road (2,4) has an importance of 5 + 1 = 6.
 * The total importance of all roads is 6 + 9 + 8 + 7 + 7 + 6 = 43.
 * It can be shown that we cannot obtain a greater total importance than 43.
 *
 * Example 2:
 * <img alt="" src="https://assets.leetcode.com/uploads/2022/04/07/ex2drawio.png" style="width: 281px; height: 151px;" />
 * Input: n = 5, roads = [[0,3],[2,4],[1,3]]
 * Output: 20
 * Explanation: The figure above shows the country and the assigned values of [4,3,2,5,1].
 * - The road (0,3) has an importance of 4 + 5 = 9.
 * - The road (2,4) has an importance of 2 + 1 = 3.
 * - The road (1,3) has an importance of 3 + 5 = 8.
 * The total importance of all roads is 9 + 3 + 8 = 20.
 * It can be shown that we cannot obtain a greater total importance than 20.
 *
 *  
 * Constraints:
 *
 * 	2 <= n <= 5 * 10^4
 * 	1 <= roads.length <= 5 * 10^4
 * 	roads[i].length == 2
 * 	0 <= ai, bi <= n - 1
 * 	ai != bi
 * 	There are no duplicate roads.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/maximum-total-importance-of-roads/
// discuss: https://leetcode.com/problems/maximum-total-importance-of-roads/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    // Credit: https://leetcode.com/problems/maximum-total-importance-of-roads/solutions/5380938/rust-32ms-555mb-one-liner-solution-by-ro-xpyj/
    pub fn maximum_importance(n: i32, roads: Vec<Vec<i32>>) -> i64 {
        roads
            .into_iter()
            .fold(vec![0; n as usize], |mut v, x| {
                v[x[0] as usize] += 1;
                v[x[1] as usize] += 1;
                v
            })
            .into_iter()
            .collect::<std::collections::BinaryHeap<i32>>()
            .into_sorted_vec()
            .into_iter()
            .enumerate()
            .map(|(i, x)| (i + 1) as i64 * x as i64)
            .sum::<i64>()
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2285_example_1() {
        let n = 5;
        let roads = vec![
            vec![0, 1],
            vec![1, 2],
            vec![2, 3],
            vec![0, 2],
            vec![1, 3],
            vec![2, 4],
        ];

        let result = 43;

        assert_eq!(Solution::maximum_importance(n, roads), result);
    }

    #[test]
    fn test_2285_example_2() {
        let n = 5;
        let roads = vec![vec![0, 3], vec![2, 4], vec![1, 3]];

        let result = 20;

        assert_eq!(Solution::maximum_importance(n, roads), result);
    }
}

// Accepted solution for LeetCode #2285: Maximum Total Importance of Roads
function maximumImportance(n: number, roads: number[][]): number {
    const deg: number[] = Array(n).fill(0);
    for (const [a, b] of roads) {
        ++deg[a];
        ++deg[b];
    }
    deg.sort((a, b) => a - b);
    return deg.reduce((acc, cur, idx) => acc + (idx + 1) * cur, 0);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.