Largest Color Value in a Directed Graph

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Largest Color Value in a Directed Graph

Edge cases:

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IDE

def largestPathValue(colors: str, edges: list[list[int]]) -> int:public int largestPathValue(String colors, int[][] edges)int largestPathValue(string colors, vector<vector<int>>& edges)function largestPathValue(colors, edges)

def largestPathValue(colors, edges):
    # Write your solution here
    pass

class Solution {
    public int largestPathValue(String colors, int[][] edges) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int largestPathValue(string colors, vector<vector<int>>& edges) {
    // Write your solution here
    return 0;
}

function largestPathValue(colors, edges) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: -1 on acyclic graphCycle detection incorrectly flags acyclic graphs as cyclic.✅ Use a 3-state visited array (0=unvisited,1=visiting,2=visited) to detect cycles precisely.

Wrong: 0 for single node graphBase case for single node not handled; DP arrays not initialized properly.✅ Initialize DP counts for each node's color to 1 at start.

Wrong: Value less than actual max color frequencyDP does not aggregate max counts correctly across all paths.✅ Update DP for each node by taking max over all predecessors for each color.

Wrong: Value greater than max color frequencyDouble counting nodes or unbounded counting bug.✅ Ensure each node counted once per path; no repeated counting in DP.

Wrong: TLE on large inputsExponential DFS or inefficient approach used.✅ Implement topological sort with DP for O(n + e) complexity.

✓

Test Cases

Focus on handling base cases like empty graphs and single nodes correctly.

Watch out for cycles and greedy traps; ensure DP aggregates color counts properly.

Optimize your solution to run in linear time using topological sort and DP.

t1_01basic

Input{"colors":"abaca","edges":[[0,1],[0,2],[2,3],[3,4]]}

Expected3

⏱ Performance - must finish in 2000ms

The path 0 -> 2 -> 3 -> 4 has colors 'a', 'a', 'c', 'a'. The color 'a' appears 3 times, which is the largest color value.

💡 Consider using topological sort to process nodes in dependency order.

💡 Track the frequency of each color along paths using dynamic programming.

💡 Use a DP table with dimensions n x 26 to store max counts of each color per node.

Why it failed: Incorrect output indicates failure to detect the longest path color frequency or cycle detection. Fix by implementing topological sort with DP to track color counts.

✓ Correctly computes max color frequency along any valid path using topological sort and DP.

t1_02basic

Input{"colors":"abcde","edges":[[0,1],[1,2],[2,3],[3,4]]}

Expected1

⏱ Performance - must finish in 2000ms

All nodes have distinct colors along the path 0->1->2->3->4, so max color frequency is 1.

💡 Check if your solution handles paths where all colors are unique.

💡 Ensure DP updates max frequency per color correctly even if counts remain 1.

💡 Verify that no cycles cause incorrect -1 output in acyclic graphs.

Why it failed: Output greater than 1 means color frequency aggregation is incorrect. Fix by updating DP arrays correctly for each color per node.

✓ Correctly handles paths with all unique colors, returning max frequency 1.

t2_01edge

Input{"colors":"a","edges":[]}

Expected1

⏱ Performance - must finish in 2000ms

Single node with no edges; max color frequency is 1.

💡 Consider the base case of a graph with one node and no edges.

💡 Your solution should handle graphs with no edges gracefully.

💡 Return 1 since the single node's color count is 1.

Why it failed: Returning 0 or -1 means failure to handle single node base case. Fix by initializing DP for nodes without edges.

✓ Correctly returns 1 for single node graph.

t2_02edge

Input{"colors":"","edges":[]}

Expected0

⏱ Performance - must finish in 2000ms

Empty graph (n=0); no nodes means max color value is 0.

💡 Check how your code handles empty input with zero nodes.

💡 Return 0 when there are no nodes to process.

💡 Avoid index errors by checking for empty colors string.

Why it failed: Code crashes or returns non-zero for empty graph. Fix by adding guard for n=0 case returning 0.

✓ Correctly returns 0 for empty graph.

t2_03edge

Input{"colors":"aaaaa","edges":[[0,1],[1,2],[2,3],[3,4]]}

Expected5

⏱ Performance - must finish in 2000ms

All nodes have the same color 'a' along a linear path of length 5; max color frequency is 5.

💡 Test if your DP accumulates counts correctly when all colors are identical.

💡 Ensure your solution does not reset counts incorrectly between nodes.

💡 The max frequency should equal the length of the longest path in this case.

Why it failed: Output less than 5 means DP does not accumulate color counts properly. Fix by correctly propagating max counts along edges.

✓ Correctly computes max frequency equal to path length for identical colors.

t2_04edge

Input{"colors":"abc","edges":[]}

Expected1

⏱ Performance - must finish in 2000ms

Multiple disconnected nodes with no edges; max color frequency is 1 for any single node.

💡 Handle multiple disconnected components by considering each node independently.

💡 DP initialization should cover all nodes even if isolated.

💡 Return max frequency among all nodes.

Why it failed: Returning 0 or less means isolated nodes not handled. Fix by initializing DP for all nodes.

✓ Correctly returns 1 for disconnected nodes.

t3_01corner

Input{"colors":"ab","edges":[[0,1],[1,0]]}

Expected-1

⏱ Performance - must finish in 2000ms

Graph contains a cycle between nodes 0 and 1; output must be -1.

💡 Detect cycles using DFS with a visited state array.

💡 Return -1 immediately if a cycle is found.

💡 Use a 3-state visited array: unvisited, visiting, visited.

Why it failed: Returning a non -1 value means cycle detection is missing or incorrect. Fix by adding cycle detection with 3-state DFS.

✓ Correctly detects cycle and returns -1.

t3_02corner

Input{"colors":"abcabc","edges":[[0,1],[1,2],[2,3],[3,4],[4,5]]}

Expected2

⏱ Performance - must finish in 2000ms

Longest path is 0->1->2->3->4->5 with colors 'a','b','c','a','b','c'. Max frequency is 2 for colors 'a','b', or 'c'.

💡 Beware of greedy approaches that pick only local max color counts.

💡 DP must consider all paths and aggregate max counts per color.

💡 Update DP arrays by taking max over all predecessors for each color.

Why it failed: Output 1 means greedy approach failed to aggregate max counts globally. Fix by using DP to track max color counts over all paths.

✓ Correctly computes max color frequency 2 by aggregating counts.

t3_03corner

Input{"colors":"aabbcc","edges":[[0,1],[1,2],[2,3],[3,4],[4,5]]}

Expected2

⏱ Performance - must finish in 2000ms

Longest path 0->1->2->3->4->5 with colors 'a','a','b','b','c','c'. Max frequency is 2 for colors 'a','b', or 'c'.

💡 Check that your solution does not confuse 0/1 knapsack style counting with this problem.

💡 Each node can be counted once per path; no unbounded counting.

💡 DP must propagate max counts without double counting nodes.

Why it failed: Output greater than 2 means double counting nodes or unbounded counting bug. Fix by ensuring each node counted once per path.

✓ Correctly handles counting each node once, returning max frequency 2.

t4_01performance

Input{"colors":"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa","edges":[]}

Expectednull

⏱ Performance - must finish in 2000ms

n=100 nodes with no edges; solution must run in O(n) time within 2 seconds.

💡 Use topological sort and DP with O(n + e) complexity.

💡 Avoid exponential DFS exploring all paths.

💡 Pre-allocate DP arrays and use iterative approach if possible.

Why it failed: TLE indicates exponential or inefficient approach. Fix by implementing topological sort with DP in O(n + e) time.

✓ Solution runs efficiently within time limits using topological sort and DP.

Practice

1. Given the following code snippet implementing topological sort positions, what is the final value of the pos array after processing conditions = [[1,2],[1,3],[3,4]] with n=4?

easy

A. [0, 0, 1, 2, 3]

B. [0, 0, 2, 1, 3]

C. [0, 0, 1, 3, 2]

D. [0, 1, 2, 3, 0]

Solution

1. Step 1: Initialize graph and in-degree arrays

   Edges: 1->2, 1->3, 3->4; in_degree: [0,0,1,1,1]

2. Step 2: Process queue and assign positions

   Queue starts with node 1 (in_degree 0). Assign pos[1]=0. Decrement in_degree of 2 and 3 to 0, enqueue both. Next, dequeue 2 (pos[2]=1), then dequeue 3 (pos[3]=2), decrement in_degree of 4 to 0, enqueue 4, assign pos[4]=3.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Positions reflect valid topological order [OK]

Hint: Track in-degree and queue order carefully [OK]

Common Mistakes:

• Swapping positions of nodes 2 and 3
• Off-by-one indexing in pos array

2. Consider the following Python code implementing the optimal approach for minimum semesters. Given n = 3 and relations = [[1,3],[2,3]], what is the value of semester returned by the function?

from collections import deque, defaultdict

def min_semesters_optimal(n, relations):
    graph = defaultdict(list)
    indegree = [0] * (n + 1)
    for u, v in relations:
        graph[u].append(v)
        indegree[v] += 1

    queue = deque([i for i in range(1, n + 1) if indegree[i] == 0])
    semester = 0
    processed = 0

    while queue:
        semester += 1
        for _ in range(len(queue)):
            course = queue.popleft()
            processed += 1
            for nei in graph[course]:
                indegree[nei] -= 1
                if indegree[nei] == 0:
                    queue.append(nei)

    return semester if processed == n else -1

print(min_semesters_optimal(3, [[1,3],[2,3]]))

easy

A. 2

B. 1

C. 3

D. -1

Solution

1. Step 1: Initialize graph and indegree

   Courses 1 and 2 have indegree 0, course 3 has indegree 2.

2. Step 2: Process BFS levels

   Semester 1: process courses 1 and 2 (both indegree 0), decrement indegree of 3 twice -> indegree[3] becomes 0, add 3 to queue. Semester 2: process course 3.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Two BFS levels -> 2 semesters [OK]

Hint: Courses with zero indegree processed per semester [OK]

Common Mistakes:

• Counting each course as a separate semester
• Off-by-one error in semester increment
• Returning -1 incorrectly when all courses processed

3. The following code attempts to find eventual safe states using the reversed graph and Kahn's algorithm. Identify the line containing the subtle bug that causes incorrect results on some inputs.

from collections import deque

def eventualSafeNodes(graph):
    n = len(graph)
    reverse_graph = [[] for _ in range(n)]
    in_degree = [0] * n

    for u in range(n):
        for v in graph[u]:
            reverse_graph[v].append(u)
        in_degree[u] = len(graph[u])  # Bug fixed here

    queue = deque([i for i in range(n) if in_degree[i] == 0])
    safe = []

    while queue:
        node = queue.popleft()
        safe.append(node)
        for prev in reverse_graph[node]:
            in_degree[prev] -= 1
            if in_degree[prev] == 0:
                queue.append(prev)

    return sorted(safe)

medium

A. Line initializing queue with nodes having in_degree zero

B. Line appending u to reverse_graph[v]

C. Line decrementing in_degree[prev] inside the while loop

D. Line setting in_degree[v] = len(graph[u]) inside the outer loop

Solution

1. Step 1: Identify in-degree assignment error

   In-degree should be assigned per node u, but the original code assigned in_degree[v] inside inner loop, overwriting multiple times incorrectly.

2. Step 2: Correct assignment

   It should be in_degree[u] = len(graph[u]) after inner loop completes, not in_degree[v].

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Incorrect in-degree causes wrong queue initialization and wrong safe nodes [OK]

Hint: In-degree must be assigned per node, not per neighbor [OK]

Common Mistakes:

• Confusing in-degree and out-degree assignment
• Assigning in-degree inside inner loop incorrectly
• Misusing loop variables for in-degree

4. What is the time complexity of the optimal BFS-based topological sort algorithm for the Parallel Courses problem, given n courses and m prerequisite relations?

medium

A. O(n + m) because each node and edge is processed once in BFS

B. O(m^2) because updating indegrees requires scanning all edges repeatedly

C. O(n * m) because for each course you process all edges

D. O(n^2) because in the worst case all courses depend on all others

Solution

1. Step 1: Analyze graph traversal

   Each course (node) is enqueued and dequeued at most once; each edge is visited once when decrementing indegree.

2. Step 2: Combine node and edge processing

   Total operations proportional to number of nodes plus edges -> O(n + m).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Linear time in nodes and edges [OK]

Hint: Each edge and node processed once in BFS [OK]

Common Mistakes:

• Assuming nested loops cause O(n*m)
• Confusing indegree updates as repeated scans
• Assuming dense graph means O(n^2) always

5. Suppose the input graph is no longer guaranteed to be a tree but can contain cycles. Which modification to the leaf trimming algorithm is necessary to correctly find minimum height roots?

hard

A. Use BFS from every node to find minimum height roots since leaf trimming fails on cycles.

B. Run the same leaf trimming algorithm; cycles do not affect the result.

C. Detect and remove cycles first by pruning edges until the graph becomes a tree, then apply leaf trimming.

D. Apply a topological sort to the graph and pick nodes with zero in-degree as roots.

Solution

1. Step 1: Understand impact of cycles

   Leaf trimming relies on tree properties (acyclic). Cycles cause infinite loops or incorrect leaf identification.

2. Step 2: Modify algorithm to handle cycles

   Detect cycles and remove edges to convert graph into a tree or forest, then apply leaf trimming to find centroids.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Cycle removal is prerequisite for leaf trimming correctness [OK]

Hint: Leaf trimming requires acyclic graphs [OK]

Common Mistakes:

• Assuming leaf trimming works on cyclic graphs
• Using BFS from every node inefficiently
• Applying topological sort on undirected graph