Question

2285. Maximum Total Importance of Roads

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Description

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:

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 <= ai, bi <= n - 1
• ai != bi
• There are no duplicate roads.

Solutions

Solution 1

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))

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;
  }
}

class Solution {
public:
  long long maximumImportance(int n, vector<vector<int>>& roads) {
    vector<int> deg(n);
    for (auto& r : roads) {
      ++deg[r[0]];
      ++deg[r[1]];
    }
    sort(deg.begin(), deg.end());
    long long ans = 0;
    for (int i = 0; i < n; ++i) ans += 1ll * (i + 1) * deg[i];
    return ans;
  }
};

func maximumImportance(n int, roads [][]int) int64 {
  deg := make([]int, n)
  for _, r := range roads {
    deg[r[0]]++
    deg[r[1]]++
  }
  sort.Ints(deg)
  var ans int64
  for i := 0; i < n; i++ {
    ans += int64((i + 1) * deg[i])
  }
  return ans
}