# Solution For Largest Color Value In A Directed Graph

The problem “Largest Color Value In A Directed Graph” on LeetCode can be solved using dynamic programming and topological sorting.

Problem Statement:

Given a directed graph, containing N nodes, and edges, each node i is assigned a color colors[i]. Find the largest number of colors in a longest path starting from the source(s) and ending at any node of the graph. (If a cycle is detected, return -1.)

Solution:

Step 1: Topological Sorting

The problem requires finding the longest path in the directed graph starting from the source(s). To achieve this, we can use topological sorting. If a directed graph has cycles, then it cannot be topologically sorted, which means that there is no longest path. Therefore, we need to check if the graph contains any cycle or not.

Step 2: Detecting Cycles

To check if a graph contains cycles, we can use Depth First Search (DFS) algorithm, to traverse the graph and maintain a visited array and a recursion stack. The visited array keeps track of all the nodes that have been visited during the DFS traversal, and the recursion stack is used to detect cycles.

We start by selecting a node as the source and calling DFS on it. If during the DFS traversal, we visit a node that is already present in the recursion stack, then it is a cycle, and we return -1.

Step 3: Dynamic Programming

Once we have checked for cycles and topologically sorted the graph, we can start optimizing the solution using dynamic programming. We can use an array, colorCount, to keep track of the maximum count of colors seen so far for each node in the topological order.

For each node, we iterate over its outgoing edges, and for each edge, we update the colorCount of the destination node by taking the maximum between the colorCount of the current node and the colorCount of the destination node plus 1, if the color of the destination node is not already visited.

Finally, we return the maximum value in the colorCount array as the answer.

Pseudo Code:

def largestPathColors(graph: List[List[int]], colors: str) -> int:

``````n = len(colors)

# Step 1: Topological Sorting
queue, indegree = [], [0] * n
for u in range(n):
for v in graph[u]:
indegree[v] += 1
for u in range(n):
if indegree[u] == 0:
queue.append(u)

# Step 2: Detecting Cycles
visited, stack = set(), set()
def hasCycle(u):
if u in visited:
return False             # Already visited node
for v in graph[u]:
if v in stack or hasCycle(v):
return True          # Cycle detected
stack.remove(u)
return False

while queue:
u = queue.pop(0)
if hasCycle(u):
return -1

# Step 3: Dynamic Programming
colorCount = [0] * n
colorCount[u] = 1
for v in graph[u]:
for i in range(26):
if ord(colors[v]) == (i + 97):
colorCount[v] = max(colorCount[v], colorCount[u] + (i == ord(colors[v]) - 97))
break
indegree[v] -= 1
if indegree[v] == 0:
queue.append(v)

return max(colorCount)
``````

Time Complexity:

The time complexity of the above solution is O(N + M), where N is the number of nodes and M is the number of edges in the graph. The cost of topological sorting is O(N + M), and the cost of detecting cycles using DFS is O(N + M). The cost of dynamic programming is also O(N + M) as we visit every edge exactly once.

Space Complexity:

The space complexity of the above solution is O(N) to store the visited set, recursion stack, queue, and colorCount array.

## Step by Step Implementation For Largest Color Value In A Directed Graph

```This problem can be solved using a breadth-first search algorithm. The idea is to start at the root node and explore all of the nodes at the same level before moving on to the next level. The node with the largest color value will be the one with the longest path from the root node.

Here is some pseudocode for the algorithm:

function largestColorValue(root) {

// create a queue for storing nodes to visit
var queue = [root];

// create a variable for storing the largest color value
var largest = root.color;

// while the queue is not empty
while (queue.length > 0) {

// take the first node from the queue
var node = queue.shift();

// update the largest color value if necessary
if (node.color > largest) {
largest = node.color;
}

// add all of the node's children to the queue
for (var i = 0; i < node.children.length; i++) {
queue.push(node.children[i]);
}
}

// return the largest color value
return largest;
}```
```This problem can be solved using a breadth-first search algorithm. The idea is to start at the root node and explore all of the nodes at the same level before moving on to the next level. We can keep track of the largest color value seen so far and return that value when we reach the end of the graph.

def largest_color_value(graph):

# keep track of the largest color value seen so far
max_color = 0

# start at the root node
root = graph[0]

# create a queue for storing nodes at the current level
queue = [root]

# create a set for storing visited nodes
visited = set()

# keep track of the current level
level = 0

while queue:

# get the next node
node = queue.pop(0)

# mark the node as visited

# update the max color value
max_color = max(max_color, node.color)

# get the neighbors of the node
neighbors = node.neighbors

# add the neighbors to the queue if they haven't been visited
for neighbor in neighbors:

if neighbor not in visited:

queue.append(neighbor)

# increment the level
level += 1

return max_color```
```var largestColorValueInGraph = function(graph) {
// create a visited array to keep track of which nodes have been visited
// create a max variable to keep track of the maximum color value
let visited = [], max = 0;

// create a helper function that takes in a node and a color value
let helper = function(node, color) {
// if the node is not in the visited array, add it
if (!visited.includes(node)) {
visited.push(node);
} else {
// otherwise, return
return;
}
// if the color value of the node is greater than the maximum, update the maximum
if (graph[node]['color'] > max) {
max = graph[node]['color'];
}
// iterate through the node's neighbors
for (let i = 0; i < graph[node]['neighbors'].length; i++) {
// call the helper function on the neighbor with the same color value
helper(graph[node]['neighbors'][i], graph[node]['color']);
}
};

// iterate through all the nodes in the graph
for (let node in graph) {
// call the helper function on the node with the node's color value
helper(node, graph[node]['color']);
}

// return the maximum color value
return max;
};```
`There are many possible solutions to this problem. One approach is to use a depth-first search algorithm to traverse the graph. For each node in the graph, we can keep track of the maximum color value that we have seen so far. When we visit a node, we update the maximum color value if the node's color is greater than the maximum color value that we have seen so far. Finally, we return the maximum color value that we have seen.`
```using System;
using System.Collections.Generic;
using System.Linq;

public class Solution {
public int[] FindOrder(int numCourses, int[][] prerequisites) {
// Create a list of edges
List edges = new List();
for (int i = 0; i < prerequisites.Length; i++) {
int[] edge = prerequisites[i];
}

// Create a list of vertices
List vertices = new List();
for (int i = 0; i < numCourses; i++) {
}

foreach (Edge edge in edges) {
}

// Initialize visited array
bool[] visited = new bool[numCourses];

// Initialize result list
List result = new List();

// Call DFS
foreach (Vertex vertex in vertices) {
if (!visited[vertex.label]) {
DFS(vertex, visited, result);
}
}

// Check if result contains all vertices
if (result.Count != numCourses) {
return new int[0];
}

// Reverse the result list
result.Reverse();

// Convert result list to array and return
return result.ToArray();
}

public void DFS(Vertex vertex, bool[] visited, List result) {
// Mark the current node as visited
visited[vertex.label] = true;

// Recurse for all the neighbors
foreach (Vertex neighbor in vertex.neighbors) {
if (!visited[neighbor.label]) {
DFS(neighbor, visited, result);
}
}

// Add the current node to the result list
}
}

public class Edge {
public int from;
public int to;

public Edge(int from, int to) {
this.from = from;
this.to = to;
}
}

public class Vertex {
public int label;
public List neighbors;

public Vertex(int label) {
this.label = label;
this.neighbors = new List();
}
}```

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