You are given a directed graph with n
nodes labeled from 0
to n - 1
, where each node has exactly one outgoing edge.
The graph is represented by a given 0-indexed integer array edges
of length n
, where edges[i]
indicates that there is a directed edge from node i
to node edges[i]
.
The edge score of a node i
is defined as the sum of the labels of all the nodes that have an edge pointing to i
.
Return the node with the highest edge score. If multiple nodes have the same edge score, return the node with the smallest index.
Node With Highest Edge Score LeetCode Contest
Example 1:
Input: edges = [1,0,0,0,0,7,7,5] Output: 7 Explanation: - The nodes 1, 2, 3 and 4 have an edge pointing to node 0. The edge score of node 0 is 1 + 2 + 3 + 4 = 10. - The node 0 has an edge pointing to node 1. The edge score of node 1 is 0. - The node 7 has an edge pointing to node 5. The edge score of node 5 is 7. - The nodes 5 and 6 have an edge pointing to node 7. The edge score of node 7 is 5 + 6 = 11. Node 7 has the highest edge score so return 7.
Example 2:
Input: edges = [2,0,0,2] Output: 0 Explanation: - The nodes 1 and 2 have an edge pointing to node 0. The edge score of node 0 is 1 + 2 = 3. - The nodes 0 and 3 have an edge pointing to node 2. The edge score of node 2 is 0 + 3 = 3. Nodes 0 and 2 both have an edge score of 3. Since node 0 has a smaller index, we return 0.
Constraints:
n == edges.length
2 <= n <= 105
0 <= edges[i] < n
edges[i] != i
Solution
class Solution:
def edgeScore(self, edges: List[int]) -> int:
res = []
dic = collections.defaultdict(int)
for i, n in enumerate(edges):
dic[n] = i + dic.get(n, 0)
for i, n in dic.items():
if len(res) == 0:
res = [n, i]
else:
temp = res
if n > temp[0]:
res = [n, i]
elif n == temp[0]:
if i < temp[1]:
res = [n, i]
return res[1]
class Solution {
public:
int edgeScore(vector<int>& edges) {
int n = edges.size();
vector<long long> score(n);
//score[i] = sum of all the nodes that have an edge pointing to i.
for(int i = 0; i < n; i++){
score[edges[i]] += i;
}
long long mx = INT_MIN;
int idx = 0;
//Finding the maximum score and it's index
for(int i = 0; i < n; i++){
if(score[i] > mx){
mx = score[i];
idx = i;
}
}
return idx;
}
};
class Solution {
public int edgeScore(int[] edges) {
final int n = edges.length;
// avoid overflow!!!
long[] score = new long[n];
// i == label
for (int i = 0; i < n; i++) {
// add each label to pointing node
score[edges[i]] += i;
}
// get smallest index with highest edge score
int result = 0;
for (int i = 0; i < n; i++) {
if (score[i] > score[result]) {
result = i;
}
}
return result;
}
}
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