Largest Color Value in a Directed Graph

▶

Steps

setup

Build graph and compute in-degree

We initialize the graph adjacency list and compute the in-degree for each node based on the edges. This prepares the data structures for topological sorting.

💡 Knowing each node's in-degree helps identify starting points (nodes with zero in-degree) for the topological sort.

Line:

for u, v in edges:
    graph[u].append(v)
    indegree[v] += 1

💡 The graph structure and in-degree array represent the dependencies between nodes.

setup

Initialize DP and queue with zero in-degree nodes

We create a DP array to track max color counts per node and color. Then, we enqueue all nodes with zero in-degree to start the topological sort.

💡 DP arrays will accumulate color counts along paths; starting with zero in-degree nodes ensures correct processing order.

Line:

dp = [[0]*26 for _ in range(n)]
queue = deque([i for i in range(n) if indegree[i] == 0])

💡 Nodes 0 has zero in-degree and is the first to be processed.

traverse

Process node 0: dequeue and update DP

We dequeue node 0 from the queue and increment the DP count for its color 'a'. This marks node 0 as processing.

💡 Incrementing DP for the node's color accounts for the node itself in the path color count.

Line:

node = queue.popleft()
c_index = ord(colors[node]) - ord('a')
dp[node][c_index] += 1

💡 Node 0's color count for 'a' is now 1, starting the accumulation for paths through node 0.

expand

Update neighbors of node 0 and enqueue if ready

For each neighbor of node 0 (nodes 1 and 2), we update their DP arrays by taking the max with node 0's DP. We decrement their in-degree and enqueue if it reaches zero.

💡 Propagating DP values ensures neighbors accumulate max color counts from all incoming paths.

Line:

for nei in graph[node]:
    for i in range(26):
        dp[nei][i] = max(dp[nei][i], dp[node][i])
    indegree[nei] -= 1
    if indegree[nei] == 0:
        queue.append(nei)

💡 Nodes 1 and 2 are now ready to process, inheriting DP values from node 0.

traverse

Process node 1: dequeue and update DP

Node 1 is dequeued and its DP count for its color 'b' is incremented by 1, accounting for itself.

💡 Each node increments its own color count in DP to reflect its presence on paths.

Line:

node = queue.popleft()
c_index = ord(colors[node]) - ord('a')
dp[node][c_index] += 1

💡 Node 1's DP now reflects one occurrence of color 'b'.

done

Node 1 has no neighbors, mark done

Node 1 has no outgoing edges, so it is marked done and removed from processing.

💡 Nodes without neighbors finalize their DP counts immediately.

Line:

count += 1  # implicit in loop
# no neighbors to update

💡 Node 1's DP is final and contributes to the max color value if applicable.

traverse

Process node 2: dequeue and update DP

Node 2 is dequeued and increments its DP count for color 'a' by 1, reflecting its own color.

💡 Incrementing DP for the node's color is essential to count occurrences along paths.

Line:

node = queue.popleft()
c_index = ord(colors[node]) - ord('a')
dp[node][c_index] += 1

💡 Node 2's DP now shows color 'a' count including itself and inherited from node 0.

expand

Update neighbor 3 of node 2 and enqueue if ready

We update DP for node 3 by taking max with node 2's DP, decrement in-degree of node 3, and enqueue node 3 since its in-degree becomes zero.

💡 DP propagation and in-degree updates ensure correct processing order and color count accumulation.

Line:

for nei in graph[node]:
    for i in range(26):
        dp[nei][i] = max(dp[nei][i], dp[node][i])
    indegree[nei] -= 1
    if indegree[nei] == 0:
        queue.append(nei)

💡 Node 3 is now ready to process with DP values inherited from node 2.

traverse

Process node 3: dequeue and update DP

Node 3 is dequeued and increments its DP count for color 'c' by 1, reflecting its own color.

💡 Each node adds itself to the DP counts to represent its color on paths.

Line:

node = queue.popleft()
c_index = ord(colors[node]) - ord('a')
dp[node][c_index] += 1

💡 Node 3's DP now includes one occurrence of color 'c' plus inherited counts.

expand

Update neighbor 4 of node 3 and enqueue if ready

We update DP for node 4 by taking max with node 3's DP, decrement in-degree of node 4, and enqueue node 4 since its in-degree becomes zero.

💡 This step propagates accumulated color counts further along the path.

Line:

for nei in graph[node]:
    for i in range(26):
        dp[nei][i] = max(dp[nei][i], dp[node][i])
    indegree[nei] -= 1
    if indegree[nei] == 0:
        queue.append(nei)

💡 Node 4 is ready to process with DP values inherited from node 3.

traverse

Process node 4: dequeue and update DP

Node 4 is dequeued and increments its DP count for color 'a' by 1, reflecting its own color.

💡 Final node increments its color count to complete the path color accumulation.

Line:

node = queue.popleft()
c_index = ord(colors[node]) - ord('a')
dp[node][c_index] += 1

💡 Node 4's DP now reflects the total count of color 'a' along the path 0->2->3->4.

done

Mark node 4 done and finalize processing

Node 4 has no neighbors, so it is marked done. We have processed all nodes, so the topological sort is complete.

💡 Completing processing of all nodes confirms no cycles and full DP computation.

Line:

count += 1  # implicit in loop
# no neighbors to update

💡 All nodes processed, so the graph is acyclic and DP arrays are complete.

reconstruct

Compute and return the maximum color value

Since all nodes were processed, we find the maximum color count in the DP arrays, which is 3 for color 'a' along path 0->2->3->4.

💡 The maximum DP value across all nodes and colors is the answer to the problem.

Line:

if count < n:
    return -1
return max_color_value

💡 The largest color value is the highest count of any color along any path in the graph.

from collections import defaultdict, deque

def largestPathValue(colors, edges):
    n = len(colors)  # STEP 1
    graph = defaultdict(list)  # STEP 1
    indegree = [0] * n  # STEP 1
    for u, v in edges:  # STEP 1
        graph[u].append(v)  # STEP 1
        indegree[v] += 1  # STEP 1

    dp = [[0]*26 for _ in range(n)]  # STEP 2
    queue = deque([i for i in range(n) if indegree[i] == 0])  # STEP 2
    count = 0
    max_color_value = 0

    while queue:  # STEP 3-12
        node = queue.popleft()  # STEP 3,5,7,9,11
        count += 1
        c_index = ord(colors[node]) - ord('a')  # STEP 3,5,7,9,11
        dp[node][c_index] += 1  # STEP 3,5,7,9,11
        max_color_value = max(max_color_value, dp[node][c_index])

        for nei in graph[node]:  # STEP 4,8,10
            for i in range(26):
                dp[nei][i] = max(dp[nei][i], dp[node][i])
            indegree[nei] -= 1
            if indegree[nei] == 0:
                queue.append(nei)

    if count < n:  # STEP 13
        return -1
    return max_color_value  # STEP 13

if __name__ == '__main__':
    print(largestPathValue("abaca", [[0,1],[0,2],[2,3],[3,4]]))  # 3
    print(largestPathValue("a", [[0,0]]))  # -1

📊

Largest Color Value in a Directed Graph - Watch the Algorithm Execute, Step by Step

Watching each step of the topological sort combined with DP updates reveals how the algorithm accumulates color counts along paths and detects cycles, which is hard to grasp from code alone.

Visiting: none

0

1

2

3

4

Visiting: none

0

1

2

3

4

Queue / Stack

0

Visiting: 0

0

1

2

3

4

Visiting: 0

Edge: 0 → 2

0

1

2

3

4

Queue / Stack

12

Visiting: 1

0

1

2

3

4

Queue / Stack

2

Visiting: none

0

1

2

3

4

Queue / Stack

2

Visiting: 2

0

1

2

3

4

Visiting: 2

Edge: 2 → 3

0

1

2

3

4

Queue / Stack

3

Visiting: 3

0

1

2

3

4

Visiting: 3

Edge: 3 → 4

0

1

2

3

4

Queue / Stack

4

Visiting: 4

0

1

2

3

4

Visiting: none

0

1

2

3

4

Key Takeaways

✓ Topological sorting ensures nodes are processed only after all dependencies are resolved.

This ordering is crucial to correctly propagate color counts along valid paths without cycles.

✓ Dynamic programming arrays accumulate maximum color counts per node and color along all incoming paths.

Watching DP updates visually clarifies how color counts combine from multiple predecessors.

✓ Cycle detection is implicit by comparing processed node count to total nodes; if fewer processed, a cycle exists.

This insight is subtle in code but clear when seeing the queue empty before all nodes are processed.

Practice

1. Consider the following code snippet implementing the optimal solution for the longest path with different adjacent characters problem. Given n = 3, edges = [[0,1],[1,2]], and s = "aba", what is the final returned value?

easy

A. -1

B. 3

C. 1

D. 2

Solution

1. Step 1: Trace topological order and dp array

   Topological order is [0,1,2]. Initialize dp = [1,1,1]. - For node 0: neighbor 1, s[1] = 'b' != s[0] = 'a', dp[1] = max(1, 1+1) = 2. - For node 1: neighbor 2, s[2] = 'a' != s[1] = 'b', dp[2] = max(1, 2+1) = 3.

2. Step 2: Compute final answer

   Maximum dp value is 3, which is returned.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   dp array ends as [1, 2, 3], max is 3 [OK]

Hint: Check dp updates along topo order carefully [OK]

Common Mistakes:

• Returning dp[node] instead of max(dp)
• Ignoring character difference condition
• Miscomputing topological order

2. You are given a list of sequences, each representing partial orderings of elements. Your task is to determine if there exists exactly one sequence that can be constructed which is consistent with all these partial orderings and matches a given original sequence. Which algorithmic approach best guarantees an efficient and correct solution for this problem?

easy

A. Use brute force to generate all permutations of the original sequence and check which ones are supersequences of all given sequences.

B. Apply a greedy approach by always choosing the smallest available element not yet placed, hoping to build the unique sequence.

C. Use dynamic programming to count the number of ways to interleave the sequences and verify if only one matches the original.

D. Construct a directed graph from the sequences and perform a DFS postorder topological sort with cycle detection and uniqueness verification.

Solution

1. Step 1: Understand problem constraints

   The problem requires verifying if exactly one sequence matches all partial orderings, which is a classic unique topological order problem.

2. Step 2: Identify suitable algorithm

   Brute force is inefficient; greedy and DP approaches do not guarantee correctness or uniqueness checks. DFS postorder topological sort with cycle detection and uniqueness verification efficiently solves the problem.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   DFS topological sort with uniqueness check is standard for unique sequence reconstruction [OK]

Hint: Unique topo order requires graph + DFS with uniqueness check [OK]

Common Mistakes:

• Assuming greedy or DP can guarantee uniqueness
• Trying brute force despite factorial complexity

3. What is the worst-case time complexity of the iterative DFS approach that enumerates all paths from source to target in a DAG with n nodes, where the number of paths can be exponential?

medium

A. O(n * 2^n) accounting for path copying and stack operations

B. O(2^n) due to exponential number of paths in worst case

C. O(n!) because permutations of nodes are explored

D. O(n^2) due to visiting each edge once

Solution

1. Step 1: Identify number of paths

   In worst case, number of paths can be exponential, roughly 2^n for DAGs with branching.

2. Step 2: Analyze per-path cost

   Each path is stored and copied when pushed to stack, costing O(n) per path. Total time is O(n * 2^n).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Exponential paths times linear path length copying [OK]

Hint: Exponential paths x linear path copying = O(n * 2^n) [OK]

Common Mistakes:

• Ignoring path copying cost
• Confusing edge visits with path enumeration

4. What is the time complexity of the optimal DFS postorder topological sort approach for sequence reconstruction, given n as the length of the original sequence and m as the total length of all sequences in seqs?

medium

A. O(n + m)

B. O(n! * m)

C. O(n * m)

D. O(m^2)

Solution

1. Step 1: Identify graph construction cost

   Building graph from seqs takes O(m) time, where m is total length of all sequences.

2. Step 2: Analyze DFS traversal cost

   DFS visits each node once and edges once, total O(n + m). No extra factorial or quadratic factors.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Linear time in nodes plus edges [OK]

Hint: DFS topo sort is O(V+E) -> O(n+m) here [OK]

Common Mistakes:

• Confusing with brute force factorial complexity
• Assuming quadratic due to nested loops

5. If the problem is extended so that the same element can appear multiple times in the matrix (i.e., reuse allowed), which modification to the topological sort approach is necessary to correctly build the matrix?

hard

A. No change needed; reuse does not affect topological ordering

B. Replace topological sort with a longest path algorithm to handle repeated elements

C. Track counts of each element and assign positions based on frequency, not just order

D. Modify graph to allow self-loops and update cycle detection accordingly

Solution

1. Step 1: Understand reuse impact on matrix construction

   Allowing reuse means elements can appear multiple times, so positions must reflect frequency, not just order.

2. Step 2: Adjust algorithm to track counts and assign positions accordingly

   Topological order alone does not handle multiple occurrences; tracking counts and assigning positions based on frequency is necessary.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Frequency tracking ensures correct placement of repeated elements [OK]

Hint: Reuse implies tracking element counts and positions [OK]

Common Mistakes:

• Assuming reuse needs no algorithm change
• Ignoring frequency of elements
• Replacing topological sort unnecessarily