Largest Color Value in a Directed Graph

LeetCode 1857: Largest Color Value in a Directed Graph Solution & Explanation

Problem Overview

TL;DR: The optimal solution uses Kahn's Algorithm (Topological Sort) combined with dynamic programming to propagate the maximum frequency of each color through the graph while simultaneously detecting cycles.

You are provided with a directed graph containing $n$ nodes and $m$ edges, where each node is assigned a specific color (represented by a lowercase English letter). Your task is to find a valid path through the graph such that the frequency of the most common color on that path is maximized. If the graph contains a cycle, no finite path exists, and you must return -1.

This problem, "LeetCode 1857: Largest Color Value in a Directed Graph," is a challenging interview question because it requires combining graph traversal for cycle detection with state propagation to track color counts efficiently.

Brute Force Approach for Largest Color Value in a Directed Graph

The naive approach involves exploring every possible path in the graph to count color frequencies. This is typically implemented using Depth First Search (DFS).

1. Start a DFS from every node in the graph.
2. Maintain a frequency map for the current path.
3. Upon extending the path to a neighbor, update the counts and track the maximum value seen so far.
4. Backtrack to explore other branches.
5. To detect cycles, maintain a recursion_stack set. If we encounter a node already in the current recursion stack, a cycle exists.

1# Pseudo-code for Brute Force
2def dfs(node, path_counts):
3    path_counts[color[node]] += 1
4    max_val = max(path_counts.values())
5    
6    for neighbor in graph[node]:
7        if neighbor in current_path: return -1 # Cycle
8        max_val = max(max_val, dfs(neighbor, path_counts))
9        
10    path_counts[color[node]] -= 1 # Backtrack
11    return max_val

Why this fails: The time complexity of this approach is exponential in the worst case. A graph can have an exponential number of paths. For constraints where $N, M \le 10^5$, an $O(N!)$ or even $O(2^N)$ approach will immediately trigger a Time Limit Exceeded (TLE). Furthermore, without memoization, we re-calculate the max color values for the same sub-paths repeatedly.

Algorithm Strategy: Graph BFS - Topological Sort (Kahn's Algorithm)

1. Graph Construction: Build an adjacency list to represent the graph and an indegree array to store the number of incoming edges for each node.
2. DP State Initialization: Create a 2D array counts[n][26] where counts[i][j] stores the maximum count of the $j$-th lowercase letter in any path ending at node $i$.
3. Queue Initialization: Push all nodes with an indegree of 0 into a queue. These are the starting points of our paths.
4. Process Queue (Kahn's Loop):
   • Dequeue a node $u$.
   • Increment the count corresponding to node $u$'s specific color in counts[u].
   • Update the global maximum answer with counts[u][color_of_u].
   • Iterate through all neighbors $v$ of $u$:
     • For every color $c$ (0 to 25), update $v$'s history: counts[v][c] = max(counts[v][c], counts[u][c]).
     • Decrement indegree[v].
     • If indegree[v] becomes 0, push $v$ into the queue.
5. Cycle Detection: Maintain a counter of visited nodes. If the number of visited nodes is less than $n$ after the queue is empty, a cycle exists. Return -1.
6. Result: If no cycle is found, return the global maximum.

Execution Flow

Let's trace the algorithm with colors = "abaca", edges = [[0,1], [0,2], [2,3], [3,4]].

1. Setup:

   • Indegrees: 0:0, 1:1, 2:1, 3:1, 4:1.
   • Queue: [0] (only node 0 has 0 in-degree).
   • counts table initialized to 0s.

2. Process Node 0:

   • Pop 0. Color is 'a'.
   • counts[0]['a'] becomes 1. Max answer = 1.
   • Neighbors: 1 and 2.
   • Update 1: counts[1] inherits counts[0]. indegree[1] becomes 0. Push 1.
   • Update 2: counts[2] inherits counts[0]. indegree[2] becomes 0. Push 2.
   • Nodes visited: 1.

3. Process Node 1:

   • Pop 1. Color is 'b'.
   • counts[1]['b'] becomes 1 (inherited 'a' count is 1). Max answer = 1.
   • No neighbors.
   • Nodes visited: 2.

4. Process Node 2:

   • Pop 2. Color is 'a'.
   • counts[2]['a'] becomes 1 (inherited) + 1 (self) = 2. Max answer = 2.
   • Neighbor: 3.
   • Update 3: counts[3] inherits counts[2]. indegree[3] becomes 0. Push 3.
   • Nodes visited: 3.

5. Process Node 3:

   • Pop 3. Color is 'c'.
   • counts[3]['c'] becomes 1. counts[3]['a'] is 2. Max answer = 2.
   • Neighbor: 4.
   • Update 4: counts[4] inherits counts[3]. indegree[4] becomes 0. Push 4.
   • Nodes visited: 4.

6. Process Node 4:

   • Pop 4. Color is 'a'.
   • counts[4]['a'] becomes 2 (inherited) + 1 (self) = 3. Max answer = 3.
   • No neighbors.
   • Nodes visited: 5.

7. Completion: Visited nodes (5) equals $n$ (5). No cycle. Return 3.

Proof of Correctness

The correctness relies on the properties of Topological Sort and Dynamic Programming:

1. Dependency Resolution: Kahn's algorithm ensures that a node $v$ is processed only after all nodes $u$ pointing to it have been processed. This guarantees that counts[v] has aggregated the maximums from all possible incoming paths before we finalize it.
2. Optimal Substructure: The maximum color value of a path ending at $v$ is determined by the maximum color values of paths ending at its predecessors. By taking the max over all incoming edges, we ensure optimality.
3. Cycle Detection: In a DAG, a topological sort will always visit every node. If the graph has a cycle, the nodes involved in the cycle (and those reachable only from the cycle) will never reach an in-degree of 0 and will never be added to the queue. Comparing the count of processed nodes to $n$ correctly identifies this state.

Pattern Reuse Notes

The Graph BFS - Topological Sort (Kahn's Algorithm) pattern is highly versatile. Understanding how to attach additional state (like the color counts here) to the topological order is key for many advanced graph problems.

• LeetCode 207: Course Schedule: The pure form of this pattern. Use Kahn's algorithm strictly for cycle detection.
• LeetCode 210: Course Schedule II: Same as above, but requires returning the order of nodes processed.
• LeetCode 269: Alien Dictionary: Constructs a graph from string comparisons and uses topological sort to derive character order.
• LeetCode 310: Minimum Height Trees: Uses a variation of topological sort (trimming leaves/onions peeling) to find the centroid of a graph.

Reference Implementation

C++ Solution for LeetCode 1857

1#include <vector>
2#include <string>
3#include <queue>
4#include <algorithm>
5
6using namespace std;
7
8class Solution {
9public:
10    int largestPathValue(string colors, vector<vector<int>>& edges) {
11        int n = colors.size();
12        vector<vector<int>> adj(n);
13        vector<int> indegree(n, 0);
14        
15        // Build graph and calculate indegrees
16        for (const auto& edge : edges) {
17            adj[edge[0]].push_back(edge[1]);
18            indegree[edge[1]]++;
19        }
20        
21        // Initialize Queue with nodes having 0 indegree
22        queue<int> q;
23        for (int i = 0; i < n; i++) {
24            if (indegree[i] == 0) {
25                q.push(i);
26            }
27        }
28        
29        // dp[i][j] stores max count of color j in a path ending at node i
30        vector<vector<int>> dp(n, vector<int>(26, 0));
31        
32        int nodesSeen = 0;
33        int maxColorValue = 0;
34        
35        while (!q.empty()) {
36            int u = q.front();
37            q.pop();
38            nodesSeen++;
39            
40            // Process current node's color
41            int colorIndex = colors[u] - 'a';
42            dp[u][colorIndex]++;
43            maxColorValue = max(maxColorValue, dp[u][colorIndex]);
44            
45            // Propagate counts to neighbors
46            for (int v : adj[u]) {
47                for (int c = 0; c < 26; c++) {
48                    dp[v][c] = max(dp[v][c], dp[u][c]);
49                }
50                
51                indegree[v]--;
52                if (indegree[v] == 0) {
53                    q.push(v);
54                }
55            }
56        }
57        
58        // Cycle detection
59        if (nodesSeen < n) return -1;
60        
61        return maxColorValue;
62    }
63};

Java Solution for LeetCode 1857

1import java.util.*;
2
3class Solution {
4    public int largestPathValue(String colors, int[][] edges) {
5        int n = colors.length();
6        List<List<Integer>> adj = new ArrayList<>();
7        int[] indegree = new int[n];
8        
9        for (int i = 0; i < n; i++) {
10            adj.add(new ArrayList<>());
11        }
12        
13        // Build graph
14        for (int[] edge : edges) {
15            adj.get(edge[0]).add(edge[1]);
16            indegree[edge[1]]++;
17        }
18        
19        Queue<Integer> q = new LinkedList<>();
20        for (int i = 0; i < n; i++) {
21            if (indegree[i] == 0) {
22                q.offer(i);
23            }
24        }
25        
26        // dp[i][j] = max count of color j ending at node i
27        int[][] dp = new int[n][26];
28        int nodesSeen = 0;
29        int maxColorValue = 0;
30        
31        while (!q.isEmpty()) {
32            int u = q.poll();
33            nodesSeen++;
34            
35            int uColor = colors.charAt(u) - 'a';
36            dp[u][uColor]++;
37            maxColorValue = Math.max(maxColorValue, dp[u][uColor]);
38            
39            for (int v : adj.get(u)) {
40                // Propagate max color counts to neighbor
41                for (int c = 0; c < 26; c++) {
42                    dp[v][c] = Math.max(dp[v][c], dp[u][c]);
43                }
44                
45                indegree[v]--;
46                if (indegree[v] == 0) {
47                    q.offer(v);
48                }
49            }
50        }
51        
52        if (nodesSeen < n) return -1; // Cycle detected
53        
54        return maxColorValue;
55    }
56}

Python Solution for LeetCode 1857

1from collections import deque
2
3class Solution:
4    def largestPathValue(self, colors: str, edges: list[list[int]]) -> int:
5        n = len(colors)
6        adj = [[] for _ in range(n)]
7        indegree = [0] * n
8        
9        # Build graph
10        for u, v in edges:
11            adj[u].append(v)
12            indegree[v] += 1
13            
14        # Initialize queue
15        queue = deque([i for i in range(n) if indegree[i] == 0])
16        
17        # dp[i][j] stores max count of color j ending at node i
18        dp = [[0] * 26 for _ in range(n)]
19        
20        nodes_seen = 0
21        max_color_value = 0
22        
23        while queue:
24            u = queue.popleft()
25            nodes_seen += 1
26            
27            u_color = ord(colors[u]) - ord('a')
28            dp[u][u_color] += 1
29            max_color_value = max(max_color_value, dp[u][u_color])
30            
31            for v in adj[u]:
32                # Propagate counts
33                for c in range(26):
34                    dp[v][c] = max(dp[v][c], dp[u][c])
35                
36                indegree[v] -= 1
37                if indegree[v] == 0:
38                    queue.append(v)
39                    
40        if nodes_seen < n:
41            return -1
42            
43        return max_color_value