Largest Color Value in a Directed Graph - Solution & Explanation
Problem Statement
There is a directed graph of n colored nodes and m edges. The nodes are numbered from 0 to n - 1.
You are given a string colors where colors[i] is a lowercase English letter representing the color of the ith node in this graph (0-indexed). You are also given a 2D array edges where edges[j] = [aj, bj] indicates that there is a directed edge from node aj to node bj.
A valid path in the graph is a sequence of nodes x1 -> x2 -> x3 -> ... -> xk such that there is a directed edge from xi to xi+1 for every 1 <= i < k. The color value of the path is the number of nodes that are colored the most frequently occurring color along that path.
Return the largest color value of any valid path in the given graph, or -1 if the graph contains a cycle.
Example 1:
Input: colors = "abaca", edges = [[0,1],[0,2],[2,3],[3,4]]
Output: 3
Explanation: The path 0 -> 2 -> 3 -> 4 contains 3 nodes that are colored "a" (red in the above image).
Example 2:
Input: colors = "a", edges = [[0,0]] Output: -1 Explanation: There is a cycle from 0 to 0.
Constraints:
n == colors.lengthm == edges.length1 <= n <= 1050 <= m <= 105colorsconsists of lowercase English letters.0 <= aj, bj < n
Approach Overview
Problem Overview: You receive a directed graph where every node has a color. The goal is to find the largest number of occurrences of a single color along any valid path. If the graph contains a cycle, the answer must be -1 because paths can grow infinitely.
Approach 1: Topological Sort with Kahn's Algorithm and Color Frequency Tracking (O((V+E)*26) time, O(V*26) space)
This approach treats the problem as dynamic programming over a DAG. First build an adjacency list and compute the in-degree of each node. Using topological sort (Kahn's algorithm), process nodes with in-degree 0. For every node, maintain an array of size 26 representing the maximum count of each color seen along paths ending at that node. When visiting a node, increment the count for its own color, then propagate the updated counts to its neighbors using max updates. If the number of processed nodes is less than n, a cycle exists and the result is -1. This method efficiently combines graph traversal with dynamic programming to track color frequencies.
Approach 2: DFS with Cycle Detection and Memoization (O((V+E)*26) time, O(V*26) space)
This strategy uses depth‑first search to explore paths while detecting cycles using a recursion stack. For each node, maintain a memoized color-frequency vector describing the best counts achievable from that node downward. During DFS, mark nodes as visiting to detect back edges (cycles). If a cycle is found, return -1. Otherwise compute the best color counts by combining results from all neighbors and incrementing the current node’s color. Memoization prevents recomputation of subgraphs and ensures each node is processed once.
Recommended for interviews: The topological sort solution is the most common interview expectation because it explicitly handles cycle detection and processes nodes in dependency order. DFS with memoization demonstrates deeper graph reasoning and is also accepted, but candidates usually reach the Kahn's algorithm approach faster. Showing awareness of cycle detection and maintaining a 26‑length color frequency array is the key insight.
Approach 1: Topological Sort with Kahn's Algorithm and Color Frequency Tracking
This approach leverages a topological sort using Kahn's algorithm to process nodes in an order that respects the directed edges. We use an additional data structure to keep a running tally of the frequency of each color at every node, and update these tallies as we process nodes. This allows us to track the most frequently occurring color for all paths.
If we detect any cycles during this process, we can immediately return -1.
The implementation starts by constructing an adjacency list from the edges and calculates the in-degree for each node. We use Kahn's algorithm to perform a topological sort, starting with nodes with zero in-degree.
During the processing, we maintain a color frequency table for each node, updating the table as we process each node's neighbors. If the node count at the end doesn't match the number of nodes, a cycle exists, and we return -1. Otherwise, we return the maximum color frequency value found.
Complexity
Time Complexity: O(n + m), where n is the number of nodes and m is the number of edges. This accounts for the initial processing of nodes and edges and the BFS traversal of the graph.
Space Complexity: O(n), mainly due to the adjacency list, in-degree array, and color count table.
Approach 2: DFS with Cycle Detection and Memoization
This approach applies a Depth-First Search (DFS) on each node while using memoization to store and retrieve previously computed results. This aids in finding the most frequent color along paths derived from each node.
During DFS, we also check for cycles by marking nodes as currently being visited, immediately returning -1 upon detecting a cycle.
The C++ solution utilizes DFS to explore paths from all nodes. During exploration, if a node is detected as currently being visited, it indicates a cycle, and we return -1.
A memoization table stores the frequency of colors for paths from each node. After processing all nodes, the highest frequency from the table is the desired result.
Code
C++
JavaScript
Complexity
Time Complexity: O(n + m), for processing nodes, edges, and DFS traversal.
Space Complexity: O(n), driven by memoization and state tracking.
Approach 3: Topological Sort + Dynamic Programming
Calculate the in-degree of each node and perform a topological sort.
Define a 2D array dp, where dp[i][j] represents the number of nodes with color j on the path from the start node to node i.
From node i, traverse all outgoing edges i \to j, and update dp[j][k] = max(dp[j][k], dp[i][k] + (c == k)), where c is the color of node j.
The answer is the maximum value in the dp array.
If there is a cycle in the graph, it is impossible to visit all nodes, so return -1.
The time complexity is O((n + m) times |\Sigma|), and the space complexity is O(m + n times |\Sigma|). Here, |\Sigma| is the size of the alphabet (26 in this case), and n and m are the number of nodes and edges, respectively.
Code
Python
Java
C++
Go
TypeScript
Complexity Comparison
| Approach | Complexity |
|---|---|
| Topological Sort with Kahn's Algorithm and Color Frequency Tracking | Time Complexity: O(n + m), where n is the number of nodes and m is the number of edges. This accounts for the initial processing of nodes and edges and the BFS traversal of the graph. Space Complexity: O(n), mainly due to the adjacency list, in-degree array, and color count table. |
| DFS with Cycle Detection and Memoization | Time Complexity: O(n + m), for processing nodes, edges, and DFS traversal. Space Complexity: O(n), driven by memoization and state tracking. |
| Topological Sort + Dynamic Programming | — |
Detailed Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Topological Sort with Color Frequency DP | O((V+E)*26) | O(V*26) | Best general solution for directed graphs with cycle detection and path DP |
| DFS with Memoization and Cycle Detection | O((V+E)*26) | O(V*26) | Useful when solving with recursive graph traversal and caching results |
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