Problem
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 * 10^4
1 <= roads.length <= 5 * 10^4
roads[i].length == 2
0 <= ai, bi <= n - 1
ai != bi
There are no duplicate roads.
Solution (Java)
class Solution {
public long maximumImportance(int n, int[][] roads) {
int[] degree = new int[n];
int maxdegree = 0;
for (int[] r : roads) {
degree[r[0]]++;
degree[r[1]]++;
maxdegree = Math.max(maxdegree, Math.max(degree[r[0]], degree[r[1]]));
}
Map<Integer, Integer> rank = new HashMap<>();
int i = n;
while (i > 0) {
for (int j = 0; j < n; j++) {
if (degree[j] == maxdegree) {
rank.put(j, i--);
degree[j] = Integer.MIN_VALUE;
}
}
maxdegree = 0;
for (int d : degree) {
maxdegree = Math.max(maxdegree, d);
}
}
long res = 0;
for (int[] r : roads) {
res += rank.get(r[0]) + rank.get(r[1]);
}
return res;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).