Longest Path With Different Adjacent Characters

📋

Problem

Imagine you are designing a workflow system where each task depends on others, and you want to find the longest sequence of tasks where no two adjacent tasks share the same category.

Given a directed acyclic graph (DAG) with n nodes labeled from 0 to n-1, and a string s of length n where s[i] represents the character label of node i, find the length of the longest path in the graph such that no two adjacent nodes on the path have the same character. The graph is given as a list of edges, where each edge is a pair [u, v] indicating a directed edge from node u to node v. Return the length of the longest valid path. If the graph contains a cycle, return -1.

1 ≤ n ≤ 10^50 ≤ edges.length ≤ 10^5edges[i].length == 20 ≤ edges[i][0], edges[i][1] < ns consists of lowercase English lettersThe graph may be disconnected

Edge cases: Graph with no edges → longest path is 1 (single node)Graph with all nodes having the same character → longest path is 1Graph with multiple disconnected components → consider longest path in any component

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IDE

def longest_path_with_diff_adj_chars(n: int, edges: list[list[int]], s: str) -> int:public int longestPathWithDiffAdjChars(int n, int[][] edges, String s)int longestPathWithDiffAdjChars(int n, vector<vector<int>>& edges, string s)function longestPathWithDiffAdjChars(n, edges, s)

def longest_path_with_diff_adj_chars(n, edges, s):
    # Write your solution here
    pass

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

#include <vector>
using namespace std;

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

function longestPathWithDiffAdjChars(n, edges, s) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 0Returning 0 for single node or empty edge graph instead of 1.✅ Return 1 when n >= 1 even if edges are empty.

Wrong: 4Ignoring character difference constraint and counting path with same adjacent chars.✅ Add condition to skip edges where s[u] == s[v].

Wrong: Positive number instead of -1Missing cycle detection, causing infinite recursion or incorrect path length.✅ Implement cycle detection using DFS states or topological sort; return -1 if cycle found.

Wrong: Less than correct longest pathGreedy approach picking only one child or not considering all paths.✅ Use DP to consider all children and pick max path length.

Wrong: Incorrect due to revisiting nodes or cyclesConfusing 0/1 path with unbounded or revisiting nodes in DAG.✅ Use memoization or topological order to avoid revisiting nodes.

✓

Test Cases

Focus on handling base cases such as empty graphs and single nodes.

Implement cycle detection and avoid greedy approaches that miss longer paths.

Optimize your solution to linear time using topological sort and dynamic programming.

t1_01basic

Input{"n":5,"edges":[[0,1],[1,2],[2,3],[3,4]],"s":"abaca"}

Expected3

⏱ Performance - must finish in 2000ms

One longest path is 0 -> 1 -> 2 with characters 'a', 'b', 'a'. Adjacent characters differ, length is 3.

💡 Think about topological order and dynamic programming on DAGs.

💡 Use topological sort to process nodes and track longest paths with character constraints.

💡 For each node, update longest path length only if adjacent characters differ.

Why it failed: Incorrect output likely due to missing character difference check between adjacent nodes. Fix by adding condition to skip edges where s[u] == s[v]. This is the canonical example test.

✓ Correctly handles the canonical example with proper character difference checks.

t1_02basic

Input{"n":4,"edges":[[0,1],[1,2],[2,3]],"s":"abcd"}

Expected4

⏱ Performance - must finish in 2000ms

All characters differ along the path 0->1->2->3, so longest path length is 4.

💡 Check if your solution handles strictly increasing paths with all distinct characters.

💡 Ensure your DP accumulates path lengths correctly without skipping valid edges.

💡 Longest path length equals number of nodes in the path when all adjacent chars differ.

Why it failed: Output less than 4 indicates failure to accumulate path length correctly or skipping valid edges. Fix by correctly updating dp values for all edges with differing characters.

✓ Correctly computes longest path when all adjacent characters differ.

t2_01edge

Input{"n":1,"edges":[],"s":"a"}

Expected1

⏱ Performance - must finish in 2000ms

Single node graph with no edges; longest path is 1.

💡 Consider the base case of a graph with a single node.

💡 Longest path in a single node graph is always 1.

💡 Return 1 when no edges exist and n=1.

Why it failed: Returning 0 or other value indicates failure to handle single node base case. Fix by returning 1 when n=1 and no edges.

✓ Correctly handles single node graph.

t2_02edge

Input{"n":3,"edges":[[0,1],[1,2]],"s":"aaa"}

Expected1

⏱ Performance - must finish in 2000ms

All nodes have the same character; no two adjacent nodes can be on the path, so longest path is 1.

💡 Check if your solution handles cases where all characters are identical.

💡 Longest path length cannot exceed 1 if adjacent characters are the same.

💡 Return 1 when no edges can be used due to character constraints.

Why it failed: Output greater than 1 means ignoring character difference constraint. Fix by enforcing s[u] != s[v] before extending path.

✓ Correctly returns 1 when all characters are identical.

t2_03edge

Input{"n":4,"edges":[],"s":"abcd"}

Expected1

⏱ Performance - must finish in 2000ms

Graph with no edges; longest path is 1 (any single node).

💡 Consider graphs with no edges and multiple nodes.

💡 Longest path is the max length of any single node path, which is 1.

💡 Return 1 when edges list is empty regardless of n.

Why it failed: Returning 0 or other value indicates failure to handle empty edge list. Fix by returning 1 when edges are empty.

✓ Correctly handles graph with no edges.

t3_01corner

Input{"n":3,"edges":[[0,1],[1,2],[2,0]],"s":"abc"}

Expected-1

⏱ Performance - must finish in 2000ms

Graph contains a cycle (0->1->2->0), so return -1.

💡 Detect cycles in the graph to avoid infinite paths.

💡 Use DFS with visited states or topological sort to detect cycles.

💡 Return -1 immediately if a cycle is detected.

Why it failed: Returning a positive number means cycle detection is missing. Fix by adding cycle detection using DFS states or topological sort.

✓ Correctly detects cycles and returns -1.

t3_02corner

Input{"n":5,"edges":[[0,1],[0,2],[1,3],[2,3],[3,4]],"s":"ababa"}

Expected4

⏱ Performance - must finish in 2000ms

Longest path is 0->1->3->4 or 0->2->3->4 with alternating characters, length 4.

💡 Beware of greedy approaches that pick only one child per node.

💡 Use DP to consider all children and pick max path length.

💡 Update dp[node] = max(dp[node], 1 + dp[child]) only if characters differ.

Why it failed: Output less than 4 indicates greedy approach failing to consider all children. Fix by using DP to consider all outgoing edges.

✓ Correctly computes longest path considering all children.

t3_03corner

Input{"n":4,"edges":[[0,1],[1,2],[2,3]],"s":"abca"}

Expected4

⏱ Performance - must finish in 2000ms

Longest path 0->1->2->3 with characters 'a','b','c','a' all different adjacent pairs, length 4.

💡 Check if your solution mistakenly treats this as unbounded knapsack or allows revisiting nodes.

💡 Ensure each node is visited once in topological order or memoized DFS.

💡 Do not allow revisiting nodes or cycles; use memoization or topological sort.

Why it failed: Output incorrect due to revisiting nodes or ignoring DAG property. Fix by enforcing DAG traversal with memoization or topological sort.

✓ Correctly handles DAG traversal without revisiting nodes.

t4_01performance

Input{"_description":"n=100000 at constraint boundary - executor generates this input"}

Expectednull

⏱ Performance - must finish in 2000ms

Input size n=100000 with up to 100000 edges; solution must run in O(n + m) time within 2 seconds.

💡 Use topological sort and DP for O(n + m) complexity.

💡 Avoid DFS with repeated recomputation or brute force exponential approaches.

💡 Memoize results or use iterative DP after topological ordering.

Why it failed: TLE indicates use of brute force or exponential DFS. Fix by implementing topological sort + DP approach with O(n + m) complexity.

✓ Solution runs efficiently in O(n + m) time as expected.

Practice

1. Consider the following code snippet implementing the optimal solution for the largest color value in a directed graph. Given colors = "abaca" and edges = [[0,1],[0,2],[2,3],[3,4]], what is the value of max_color_value after processing all nodes?

easy

A. 3

B. 4

C. 2

D. 5

Solution

1. Step 1: Trace dp for node 0 (color 'a')

   dp[0]['a'] = 1, max_color_value = 1

2. Step 2: Propagate to nodes 1 and 2

   Node 1 (color 'b'): dp[1]['a'] = 1, dp[1]['b'] = 1; Node 2 (color 'a'): dp[2]['a'] = 2, max_color_value = 2

3. Step 3: Propagate to node 3 (color 'c')

   dp[3]['a'] = 2, dp[3]['c'] = 1, max_color_value = 2

4. Step 4: Propagate to node 4 (color 'a')

   dp[4]['a'] = 3, max_color_value = 3

5. Final Answer:

   Option A -> Option A

6. Quick Check:

   Maximum color count along path is 3 for 'a' [OK]

Hint: Track dp updates per node carefully [OK]

Common Mistakes:

• Off-by-one in dp increments or max updates

2. Consider the following Python code implementing the optimal leaf trimming approach for Minimum Height Trees. Given n = 4 and edges = [[1,0],[1,2],[1,3]], what is the final returned list?

easy

A. [3]

B. [0, 2, 3]

C. [2]

D. [1]

Solution

1. Step 1: Initialize graph and degrees

   Nodes: 0,1,2,3; edges connect 1 to 0,2,3. Degrees: 0->1, 1->3, 2->1, 3->1. Leaves initially: [0,2,3].

2. Step 2: Trim leaves until ≤ 2 nodes remain

   Remove leaves [0,2,3], decrement neighbors' degrees. Node 1 degree reduces from 3 to 0, becomes leaf. Remaining nodes = 4 - 3 = 1. Loop ends. Return [1].

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Only node 1 remains as centroid after trimming [OK]

Hint: Leaves trimmed until ≤2 nodes remain [OK]

Common Mistakes:

• Returning initial leaves
• Off-by-one in loop
• Not updating degrees correctly

3. 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

4. Suppose the problem is modified so that nodes can be revisited multiple times (cycles allowed). Which modification to the iterative DFS approach is necessary to correctly find all paths from source to target without infinite loops?

hard

A. Add a visited set per path to prevent revisiting nodes already in the current path

B. Remove path tracking and rely on global visited set to avoid cycles

C. Use BFS instead of DFS to guarantee shortest paths and avoid cycles

D. No modification needed; the current approach handles cycles correctly

Solution

1. Step 1: Identify cycle problem

   Allowing revisits means cycles can cause infinite loops if not tracked per path.

2. Step 2: Solution approach

   Maintain a visited set per path to avoid revisiting nodes already in the current path, preventing cycles.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Global visited set disallows legitimate different paths; per-path visited set is needed [OK]

Hint: Cycle detection requires per-path visited tracking, not global visited [OK]

Common Mistakes:

• Using global visited set incorrectly
• Assuming no cycles without checks

5. Suppose the problem is modified so that some courses can be taken multiple times (repeatedly) and prerequisites must be satisfied each time before retaking. Which of the following statements about adapting the BFS topological sort approach is correct?

hard

A. We should switch to a dynamic programming approach because BFS topological sort cannot handle repeated courses.

B. The original BFS topological sort can be used without changes since it handles cycles by returning -1.

C. We can use BFS topological sort by resetting indegrees after each semester and reprocessing courses as needed.

D. The problem becomes invalid because repeated courses create cycles; BFS topological sort cannot be applied directly.

Solution

1. Step 1: Understand repeated courses and prerequisites

   Allowing repeated courses with prerequisites means cycles can form because a course depends on itself indirectly.

2. Step 2: Analyze BFS topological sort applicability

   BFS topological sort requires a DAG; cycles break this assumption, so it cannot be applied directly.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Repeated courses induce cycles -> BFS topological sort invalid [OK]

Hint: Repeated courses create cycles, breaking DAG assumption [OK]

Common Mistakes:

• Assuming BFS topological sort handles cycles by returning -1
• Thinking resetting indegrees can simulate repeats correctly
• Believing DP is always needed without analyzing graph structure