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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 nodes and 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.
The naive approach involves exploring every possible path in the graph to count color frequencies. This is typically implemented using Depth First Search (DFS).
recursion_stack set. If we encounter a node already in the current recursion stack, a cycle exists.python# Pseudo-code for Brute Force def dfs(node, path_counts): path_counts[color[node]] += 1 max_val = max(path_counts.values()) for neighbor in graph[node]: if neighbor in current_path: return -1 # Cycle max_val = max(max_val, dfs(neighbor, path_counts)) path_counts[color[node]] -= 1 # Backtrack 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 , an or even 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.
The key intuition is to treat the problem as a dynamic programming challenge on a Directed Acyclic Graph (DAG). If we know the maximum frequency of every color for all paths ending at a node , and there is an edge , we can use that information to update the potential maximum frequencies for paths ending at .
However, we cannot simply process nodes in arbitrary order. We must ensure that before we finalize the color counts for node , we have processed all nodes that have an edge pointing to . This is exactly what Kahn's Algorithm guarantees.
The Invariant:
We maintain a DP table dp[node][color], representing the maximum frequency of color in any path ending at node.
When traversing from , the transition is:
dp[v][color] = max(dp[v][color], dp[u][color]) for all 26 colors.
Once all parents of have been processed (indicated by 's in-degree dropping to 0), we increment the count for 's own color:
dp[v][color_of_v]++.
Visual Description: Imagine the graph as a flow of water. We start at the "source" nodes (in-degree 0). As we "flow" from node to , we carry the history of color counts with us. Node acts as a reservoir; it waits until all streams feeding into it (all incoming edges) have delivered their data. Only then does it add its own color contribution and release the flow to its neighbors. If we process nodes but the reservoir count is less than , water is trapped in a loop (cycle).

indegree array to store the number of incoming edges for each node.counts[n][26] where counts[i][j] stores the maximum count of the -th lowercase letter in any path ending at node .indegree of 0 into a queue. These are the starting points of our paths.counts[u].counts[u][color_of_u].counts[v][c] = max(counts[v][c], counts[u][c]).indegree[v].indegree[v] becomes 0, push into the queue.-1.Let's trace the algorithm with colors = "abaca", edges = [[0,1], [0,2], [2,3], [3,4]].
Setup:
0:0, 1:1, 2:1, 3:1, 4:1.[0] (only node 0 has 0 in-degree).counts table initialized to 0s.Process Node 0:
0. Color is 'a'.counts[0]['a'] becomes 1. Max answer = 1.1 and 2.1: counts[1] inherits counts[0]. indegree[1] becomes 0. Push 1.2: counts[2] inherits counts[0]. indegree[2] becomes 0. Push 2.Process Node 1:
1. Color is 'b'.counts[1]['b'] becomes 1 (inherited 'a' count is 1). Max answer = 1.Process Node 2:
2. Color is 'a'.counts[2]['a'] becomes 1 (inherited) + 1 (self) = 2. Max answer = 2.3.3: counts[3] inherits counts[2]. indegree[3] becomes 0. Push 3.Process Node 3:
3. Color is 'c'.counts[3]['c'] becomes 1. counts[3]['a'] is 2. Max answer = 2.4.4: counts[4] inherits counts[3]. indegree[4] becomes 0. Push 4.Process Node 4:
4. Color is 'a'.counts[4]['a'] becomes 2 (inherited) + 1 (self) = 3. Max answer = 3.Completion: Visited nodes (5) equals (5). No cycle. Return 3.
The correctness relies on the properties of Topological Sort and Dynamic Programming:
counts[v] has aggregated the maximums from all possible incoming paths before we finalize it.max over all incoming edges, we ensure optimality.cpp#include <vector> #include <string> #include <queue> #include <algorithm> using namespace std; class Solution { public: int largestPathValue(string colors, vector<vector<int>>& edges) { int n = colors.size(); vector<vector<int>> adj(n); vector<int> indegree(n, 0); // Build graph and calculate indegrees for (const auto& edge : edges) { adj[edge[0]].push_back(edge[1]); indegree[edge[1]]++; } // Initialize Queue with nodes having 0 indegree queue<int> q; for (int i = 0; i < n; i++) { if (indegree[i] == 0) { q.push(i); } } // dp[i][j] stores max count of color j in a path ending at node i vector<vector<int>> dp(n, vector<int>(26, 0)); int nodesSeen = 0; int maxColorValue = 0; while (!q.empty()) { int u = q.front(); q.pop(); nodesSeen++; // Process current node's color int colorIndex = colors[u] - 'a'; dp[u][colorIndex]++; maxColorValue = max(maxColorValue, dp[u][colorIndex]); // Propagate counts to neighbors for (int v : adj[u]) { for (int c = 0; c < 26; c++) { dp[v][c] = max(dp[v][c], dp[u][c]); } indegree[v]--; if (indegree[v] == 0) { q.push(v); } } } // Cycle detection if (nodesSeen < n) return -1; return maxColorValue; } };
javaimport java.util.*; class Solution { public int largestPathValue(String colors, int[][] edges) { int n = colors.length(); List<List<Integer>> adj = new ArrayList<>(); int[] indegree = new int[n]; for (int i = 0; i < n; i++) { adj.add(new ArrayList<>()); } // Build graph for (int[] edge : edges) { adj.get(edge[0]).add(edge[1]); indegree[edge[1]]++; } Queue<Integer> q = new LinkedList<>(); for (int i = 0; i < n; i++) { if (indegree[i] == 0) { q.offer(i); } } // dp[i][j] = max count of color j ending at node i int[][] dp = new int[n][26]; int nodesSeen = 0; int maxColorValue = 0; while (!q.isEmpty()) { int u = q.poll(); nodesSeen++; int uColor = colors.charAt(u) - 'a'; dp[u][uColor]++; maxColorValue = Math.max(maxColorValue, dp[u][uColor]); for (int v : adj.get(u)) { // Propagate max color counts to neighbor for (int c = 0; c < 26; c++) { dp[v][c] = Math.max(dp[v][c], dp[u][c]); } indegree[v]--; if (indegree[v] == 0) { q.offer(v); } } } if (nodesSeen < n) return -1; // Cycle detected return maxColorValue; } }
pythonfrom collections import deque class Solution: def largestPathValue(self, colors: str, edges: list[list[int]]) -> int: n = len(colors) adj = [[] for _ in range(n)] indegree = [0] * n # Build graph for u, v in edges: adj[u].append(v) indegree[v] += 1 # Initialize queue queue = deque([i for i in range(n) if indegree[i] == 0]) # dp[i][j] stores max count of color j ending at node i dp = [[0] * 26 for _ in range(n)] nodes_seen = 0 max_color_value = 0 while queue: u = queue.popleft() nodes_seen += 1 u_color = ord(colors[u]) - ord('a') dp[u][u_color] += 1 max_color_value = max(max_color_value, dp[u][u_color]) for v in adj[u]: # Propagate counts for c in range(26): dp[v][c] = max(dp[v][c], dp[u][c]) indegree[v] -= 1 if indegree[v] == 0: queue.append(v) if nodes_seen < n: return -1 return max_color_value
Let be the number of nodes and be the number of edges. Let be the size of the character set (26 for lowercase English letters).
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.
Why do I get TLE with standard DFS?
Why is my answer wrong for cyclic graphs?
nodes_seen == n at the end of Kahn's algorithm.-1.Why is my DP update logic failing?
dp[v] only when is pushed to the queue, or resetting it.counts DP table: .vdp[v] must accumulate the maximums from all parents . We must update dp[v] every time we process an incoming edge from , not just when we are ready to process .