**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

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

**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 <= 10`

^{5}`0 <= edges[i] < n`

`edges[i] != i`

```
int edgeScore(vector<int>& edges) {
vector<long long> score(edges.size());
for (int i = 0; i < edges.size(); ++i)
score[edges[i]] += i;
return max_element(begin(score), end(score)) - begin(score);
}
```

```
class Solution {
public int edgeScore(int[] edges) {
int n_length=edges.length;
//to avoid overflow we will take long size score array
long score_of_node[]=new long[n_length];
for(int i=0; i<n_length; i++){
score_of_node[edges[i]]+=i; //score of the node will be update to (prev score + incomming node) here every i denotes income incomming vertex
}
int max_ans_index=0;
for(int i=0; i<n_length; i++){
if(score_of_node[i]>score_of_node[max_ans_index]){
max_ans_index=i; // simply just find the node with maximum score and
}
}
return max_ans_index; // return the node
}
}
```

```
class Solution:
def edgeScore(self, edges: List[int]) -> int:
n = len(edges)
cnt = defaultdict(int)
ans = 0
for i in range(n):
cnt[edges[i]] += i
m = max(cnt.values())
for i in range(n):
if cnt[i] == m:
ans = i
break
return ans
```

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