LeetCode #2374 — MEDIUM

Node With Highest Edge Score

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

The Problem

Problem Statement

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.

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

[1,0,0,0,0,7,7,5]

Example 2

[2,0,0,2]

Core Insight

What unlocks the optimal approach

Create an array arr where arr[i] is the edge score for node i.
How does the edge score for node edges[i] change? It increases by i.
The edge score may not fit within a standard 32-bit integer.

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 #2374: Node With Highest Edge Score
class Solution {
    public int edgeScore(int[] edges) {
        int n = edges.length;
        long[] cnt = new long[n];
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int j = edges[i];
            cnt[j] += i;
            if (cnt[ans] < cnt[j] || (cnt[ans] == cnt[j] && j < ans)) {
                ans = j;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2374: Node With Highest Edge Score
func edgeScore(edges []int) (ans int) {
	cnt := make([]int, len(edges))
	for i, j := range edges {
		cnt[j] += i
		if cnt[ans] < cnt[j] || (cnt[ans] == cnt[j] && j < ans) {
			ans = j
		}
	}
	return
}

# Accepted solution for LeetCode #2374: Node With Highest Edge Score
class Solution:
    def edgeScore(self, edges: List[int]) -> int:
        ans = 0
        cnt = [0] * len(edges)
        for i, j in enumerate(edges):
            cnt[j] += i
            if cnt[ans] < cnt[j] or (cnt[ans] == cnt[j] and j < ans):
                ans = j
        return ans

// Accepted solution for LeetCode #2374: Node With Highest Edge Score
impl Solution {
    pub fn edge_score(edges: Vec<i32>) -> i32 {
        let n = edges.len();
        let mut cnt = vec![0_i64; n];
        let mut ans = 0;

        for (i, &j) in edges.iter().enumerate() {
            let j = j as usize;
            cnt[j] += i as i64;
            if cnt[ans] < cnt[j] || (cnt[ans] == cnt[j] && j < ans) {
                ans = j;
            }
        }

        ans as i32
    }
}

// Accepted solution for LeetCode #2374: Node With Highest Edge Score
function edgeScore(edges: number[]): number {
    const n = edges.length;
    const cnt: number[] = Array(n).fill(0);
    let ans: number = 0;
    for (let i = 0; i < n; ++i) {
        const j = edges[i];
        cnt[j] += i;
        if (cnt[ans] < cnt[j] || (cnt[ans] === cnt[j] && j < ans)) {
            ans = j;
        }
    }
    return ans;
}

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)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.