How do you solve Largest Color Value in a Directed Graph on LeetCode?

Largest Color Value in a Directed Graph (LeetCode #1857) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(n) space complexity. Key insight: Topological sorting is essential for processing nodes in a directed graph without cycles.

What is the time complexity of Largest Color Value in a Directed Graph?

The optimal solution for Largest Color Value in a Directed Graph runs in O(n + m) time and uses O(n) space. The complexity is O(n + m) because we traverse each node and edge once during the topological sort and color counting.

What is the brute force solution for Largest Color Value in a Directed Graph?

The brute force approach for Largest Color Value in a Directed Graph is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all possible paths in the directed graph to find the maximum color value. This involves checking each node and all its possible paths recursively, counting the occurrences of each color along the way. The optimal Optimal Solution approach improves this to O(n + m).

What are the interview tips for Largest Color Value in a Directed Graph?

Always clarify if the graph can have cycles and how to handle them. Discuss the importance of topological sorting in directed graphs. Practice explaining your thought process clearly while coding.

What are common mistakes when solving Largest Color Value in a Directed Graph?

Failing to check for cycles in the graph. Not properly managing color counts when traversing paths.

#1857

Largest Color Value in a Directed Graph

Hard

Hash TableStringDynamic ProgrammingGraph TheoryTopological SortMemoizationCountingGraph TraversalDynamic ProgrammingTopological Sort

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + m)
Space	O(1)	O(n)

💡

Intuition

Time O(n + m)Space O(n)

Using topological sorting allows us to process nodes in a way that ensures we handle all dependencies before a node is processed. This helps us avoid cycles and efficiently calculate the maximum color value along valid paths.

⚙️

Algorithm

4 steps

1Step 1: Build the graph and calculate in-degrees for each node.
2Step 2: Perform a topological sort using Kahn's algorithm, maintaining a queue of nodes with zero in-degrees.
3Step 3: For each node processed, update the color counts for its neighbors based on the current node's counts.
4Step 4: Track the maximum color value encountered during the updates.

solution.py27 lines

1def largestColorValue(colors, edges):
2    from collections import defaultdict, deque
3    n = len(colors)
4    graph = defaultdict(list)
5    in_degree = [0] * n
6
7    for u, v in edges:
8        graph[u].append(v)
9        in_degree[v] += 1
10
11    queue = deque([i for i in range(n) if in_degree[i] == 0])
12    color_count = [[0] * 26 for _ in range(n)]
13    max_color_value = 0
14
15    while queue:
16        node = queue.popleft()
17        color_count[node][ord(colors[node]) - ord('a')] += 1
18        max_color_value = max(max_color_value, max(color_count[node]))
19
20        for neighbor in graph[node]:
21            for c in range(26):
22                color_count[neighbor][c] = max(color_count[neighbor][c], color_count[node][c])
23            in_degree[neighbor] -= 1
24            if in_degree[neighbor] == 0:
25                queue.append(neighbor)
26
27    return max_color_value if all(d == 0 for d in in_degree) else -1

ℹ

Complexity note: The complexity is O(n + m) because we traverse each node and edge once during the topological sort and color counting.

1Topological sorting is essential for processing nodes in a directed graph without cycles.
2Dynamic programming can help track color counts efficiently across paths.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.