Build a Matrix With Conditions

📋

Build a Matrix With Conditions

Edge cases:

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IDE

def buildMatrix(n: int, rowConditions: list[list[int]], colConditions: list[list[int]]) -> list[list[int]]:public int[][] buildMatrix(int n, int[][] rowConditions, int[][] colConditions)vector<vector<int>> buildMatrix(int n, vector<vector<int>>& rowConditions, vector<vector<int>>& colConditions)function buildMatrix(n, rowConditions, colConditions)

def buildMatrix(n, rowConditions, colConditions):
    # Write your solution here
    pass

class Solution {
    public int[][] buildMatrix(int n, int[][] rowConditions, int[][] colConditions) {
        // Write your solution here
        return new int[0][0];
    }
}

#include <vector>
using namespace std;

vector<vector<int>> buildMatrix(int n, vector<vector<int>>& rowConditions, vector<vector<int>>& colConditions) {
    // Write your solution here
    return {};
}

function buildMatrix(n, rowConditions, colConditions) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: Non-empty matrix despite cycles in conditionsMissing cycle detection in topological sort, producing invalid orderings.✅ Implement cycle detection using visited states in DFS or in-degree checks in Kahn's algorithm; return empty matrix if cycle detected.

Wrong: Incorrect position mapping of elements in matrixTopological order computed but positions assigned incorrectly or swapped rows and columns.✅ Map each element's position by its index in row order and column order arrays consistently.

Wrong: Empty matrix returned for n=1 with no conditionsNot handling base case n=1 properly, returning empty instead of [[1]].✅ Add explicit check for n=1 and no conditions to return [[1]].

Wrong: TLE on large inputsUsing exponential DFS or inefficient graph traversal instead of O(n+m) topological sort.✅ Use Kahn's algorithm or optimized DFS with memoization for topological sort.

✓

Test Cases

Focus on handling empty and minimal inputs correctly.

Watch out for cycles and greedy ordering traps.

Optimize your topological sort to linear time complexity.

t1_01basic

Input{"n":3,"rowConditions":[[1,2],[2,3]],"colConditions":[[2,1],[3,2]]}

Expected[[0,0,1],[0,2,0],[3,0,0]]

⏱ Performance - must finish in 2000ms

Row order is 1->2->3, column order is 3->2->1, so 1 is at row 0 col 2, 2 at row 1 col 1, 3 at row 2 col 0.

💡 Think about topological sorting rows and columns separately.

💡 Use Kahn's algorithm or DFS to get valid orderings for rows and columns.

💡 Map each number to its row and column index from the topological orders.

Why it failed: Incorrect matrix output: likely failed to produce valid topological order or mismapped positions. Fix by ensuring cycle detection and correct mapping of nodes to positions.

✓ Correctly computed topological orders and constructed matrix.

t1_02basic

Input{"n":4,"rowConditions":[[1,3],[2,3],[3,4]],"colConditions":[[4,2],[2,1],[1,3]]}

Expected[[0,0,1,0],[0,0,0,2],[0,3,0,0],[4,0,0,0]]

⏱ Performance - must finish in 2000ms

Row order: 1->2->3->4; column order: 4->2->1->3; positions mapped accordingly.

💡 Separate topological sorts for rows and columns are needed.

💡 Check for cycles in both rowConditions and colConditions.

💡 Assign positions based on topological order indices.

Why it failed: Output matrix incorrect due to ignoring cycle detection or incorrect position mapping. Fix by implementing cycle detection and consistent mapping.

✓ Successfully handled multiple conditions and constructed correct matrix.

t2_01edge

Input{"n":0,"rowConditions":[],"colConditions":[]}

Expected[]

⏱ Performance - must finish in 2000ms

Empty input n=0 means empty matrix.

💡 Consider the base case when n=0.

💡 Return an empty matrix immediately if no elements.

💡 Check for empty input before processing.

Why it failed: Failed to handle n=0 case, possibly causing index errors or returning non-empty matrix. Fix by adding base case check for n=0.

✓ Correctly returned empty matrix for n=0.

t2_02edge

Input{"n":1,"rowConditions":[],"colConditions":[]}

Expected[[1]]

⏱ Performance - must finish in 2000ms

Single element with no conditions results in 1x1 matrix with element 1.

💡 For single element, no ordering constraints exist.

💡 Matrix should contain the single element at position (0,0).

💡 Return [[1]] when n=1 and no conditions.

Why it failed: Incorrect output for single element case, possibly empty or wrong matrix. Fix by returning [[1]] when n=1 and no conditions.

✓ Correctly handled single element with no conditions.

t2_03edge

Input{"n":3,"rowConditions":[[1,2],[2,1]],"colConditions":[]}

Expected[]

⏱ Performance - must finish in 2000ms

Conflicting rowConditions create a cycle, so no valid matrix exists.

💡 Detect cycles in rowConditions using DFS or Kahn's algorithm.

💡 If a cycle exists, return empty matrix immediately.

💡 Check both row and column conditions for cycles separately.

Why it failed: Failed to detect cycle in rowConditions, returning invalid matrix. Fix by implementing cycle detection and returning empty matrix on cycle.

✓ Cycle detected correctly, returned empty matrix.

t2_04edge

Input{"n":3,"rowConditions":[],"colConditions":[[1,2],[2,1]]}

Expected[]

⏱ Performance - must finish in 2000ms

Conflicting colConditions create a cycle, so no valid matrix exists.

💡 Detect cycles in colConditions using DFS or Kahn's algorithm.

💡 Return empty matrix if cycle detected.

💡 Check both row and column conditions for cycles.

Why it failed: Failed to detect cycle in colConditions, returning invalid matrix. Fix by implementing cycle detection and returning empty matrix on cycle.

✓ Cycle detected correctly, returned empty matrix.

t3_01corner

Input{"n":4,"rowConditions":[[1,2],[2,3],[3,4]],"colConditions":[[1,2],[2,3],[3,4]]}

Expected[[1,0,0,0],[0,2,0,0],[0,0,3,0],[0,0,0,4]]

⏱ Performance - must finish in 2000ms

Both row and column conditions are identical linear orders, so matrix is diagonal 1 to 4.

💡 Beware of greedy approaches that assign positions without full topological sort.

💡 Ensure topological sort respects all conditions, not just partial.

💡 Use full topological sort and map positions accordingly.

Why it failed: Greedy approach assigning positions without full topological sort leads to incorrect matrix. Fix by performing complete topological sort for both row and column conditions.

✓ Correctly handled linear orders with full topological sort.

t3_02corner

Input{"n":3,"rowConditions":[[1,2],[2,3],[3,1]],"colConditions":[[1,3],[3,2]]}

Expected[]

⏱ Performance - must finish in 2000ms

Cycle in rowConditions (1->2->3->1) invalidates matrix.

💡 Check for cycles separately in row and column conditions.

💡 Return empty matrix if any cycle detected.

💡 Use DFS or Kahn's algorithm for cycle detection.

Why it failed: Missed cycle detection in rowConditions, returning invalid matrix. Fix by adding cycle detection and returning empty matrix on cycle.

✓ Cycle detected correctly, returned empty matrix.

t3_03corner

Input{"n":3,"rowConditions":[[1,2],[2,3]],"colConditions":[[1,2],[2,3],[3,1]]}

Expected[]

⏱ Performance - must finish in 2000ms

Cycle in colConditions (1->2->3->1) invalidates matrix.

💡 Check cycles in both row and column conditions independently.

💡 Return empty matrix if cycle found in either.

💡 Use topological sort with cycle detection.

Why it failed: Failed to detect cycle in colConditions, producing invalid matrix. Fix by implementing cycle detection and returning empty matrix on cycle.

✓ Cycle detected in colConditions, returned empty matrix.

t4_01performance

Input{"n":100,"rowConditions":[[1,2],[2,3],[3,4],[4,5],[5,6],[6,7],[7,8],[8,9],[9,10],[10,11],[11,12],[12,13],[13,14],[14,15],[15,16],[16,17],[17,18],[18,19],[19,20],[20,21],[21,22],[22,23],[23,24],[24,25],[25,26],[26,27],[27,28],[28,29],[29,30],[30,31],[31,32],[32,33],[33,34],[34,35],[35,36],[36,37],[37,38],[38,39],[39,40],[40,41],[41,42],[42,43],[43,44],[44,45],[45,46],[46,47],[47,48],[48,49],[49,50],[50,51],[51,52],[52,53],[53,54],[54,55],[55,56],[56,57],[57,58],[58,59],[59,60],[60,61],[61,62],[62,63],[63,64],[64,65],[65,66],[66,67],[67,68],[68,69],[69,70],[70,71],[71,72],[72,73],[73,74],[74,75],[75,76],[76,77],[77,78],[78,79],[79,80],[80,81],[81,82],[82,83],[83,84],[84,85],[85,86],[86,87],[87,88],[88,89],[89,90],[90,91],[91,92],[92,93],[93,94],[94,95],[95,96],[96,97],[97,98],[98,99],[99,100]],"colConditions":[[1,2],[2,3],[3,4],[4,5],[5,6],[6,7],[7,8],[8,9],[9,10],[10,11],[11,12],[12,13],[13,14],[14,15],[15,16],[16,17],[17,18],[18,19],[19,20],[20,21],[21,22],[22,23],[23,24],[24,25],[25,26],[26,27],[27,28],[28,29],[29,30],[30,31],[31,32],[32,33],[33,34],[34,35],[35,36],[36,37],[37,38],[38,39],[39,40],[40,41],[41,42],[42,43],[43,44],[44,45],[45,46],[46,47],[47,48],[48,49],[49,50],[50,51],[51,52],[52,53],[53,54],[54,55],[55,56],[56,57],[57,58],[58,59],[59,60],[60,61],[61,62],[62,63],[63,64],[64,65],[65,66],[66,67],[67,68],[68,69],[69,70],[70,71],[71,72],[72,73],[73,74],[74,75],[75,76],[76,77],[77,78],[78,79],[79,80],[80,81],[81,82],[82,83],[83,84],[84,85],[85,86],[86,87],[87,88],[88,89],[89,90],[90,91],[91,92],[92,93],[93,94],[94,95],[95,96],[96,97],[97,98],[98,99],[99,100]]}

Expectednull

⏱ Performance - must finish in 2000ms

n=100 with linear chain conditions for rows and columns; O(n+m) topological sort must complete within 2s.

💡 Use Kahn's algorithm for topological sort to achieve O(n+m) time.

💡 Avoid DFS recursion stack overflow by iterative approach if needed.

💡 Preallocate data structures and avoid unnecessary overhead.

Why it failed: TLE due to using exponential or inefficient approach like naive DFS without pruning. Fix by implementing O(n+m) Kahn's algorithm for topological sort.

✓ Efficient O(n+m) topological sort confirmed by passing performance test.

Practice

1. You are given a directed acyclic graph (DAG) represented as an adjacency list. The task is to find all possible paths from the source node (0) to the target node (n-1). Which algorithmic approach guarantees finding all such paths efficiently without missing any, considering the graph's acyclic nature?

easy

A. Breadth-first search (BFS) that stops at the first found path to the target

B. Greedy algorithm that picks the next node with the fewest outgoing edges at each step

C. Dynamic programming with bottom-up tabulation to count the number of paths

D. Depth-first search (DFS) with backtracking to explore all paths from source to target

Solution

1. Step 1: Understand the problem constraints

   The graph is a DAG, and we need to find all paths from source to target, not just one or count them.

2. Step 2: Analyze algorithm suitability

   Greedy and BFS stopping at first path won't find all paths. DP counting paths doesn't return actual paths. DFS with backtracking explores all paths exhaustively, ensuring completeness.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   DFS with backtracking enumerates all paths [OK]

Hint: All paths require exhaustive DFS, not greedy or BFS [OK]

Common Mistakes:

• Assuming BFS finds all paths
• Confusing counting paths with enumerating paths

2. 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

3. 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

4. What is the time complexity of Kahn's Algorithm for finding a topological order in a graph with V vertices and E edges?

medium

A. O(V * E) because each vertex processes all edges repeatedly.

B. O(V^2) because in-degree updates require scanning all vertices.

C. O(V + E) because each vertex and edge is processed exactly once.

D. O(E log V) due to priority queue operations for ordering.

Solution

1. Step 1: Identify operations on vertices and edges

   Each vertex is enqueued and dequeued once; each edge is visited once to decrement in-degree.

2. Step 2: Sum total work

   Processing all vertices and edges once leads to O(V + E) time complexity.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Linear time in graph size is standard for BFS topological sort [OK]

Hint: Each edge and vertex processed once -> O(V + E) [OK]

Common Mistakes:

• Assuming nested loops cause O(V*E)
• Confusing in-degree updates as scanning all vertices
• Thinking priority queues are used by default

5. The following code attempts to find the longest path with different adjacent characters in a DAG. Identify the line containing the subtle bug that causes incorrect results on graphs with cycles or repeated characters.

medium

A. Line with 'for node in topo_order:' loop

B. Line with 'dp = [1] * n' initialization

C. Line with 'if indegree[nei] <= 0:' condition

D. Line with 'return max(dp)' statement

Solution

1. Step 1: Analyze indegree decrement and queue append

   The condition 'if indegree[nei] <= 0:' incorrectly allows nodes with negative indegree to be enqueued multiple times, breaking topological order.

2. Step 2: Consequence of bug

   This causes incorrect topo_order and potentially infinite loops or wrong dp updates.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Correct condition should be 'indegree[nei] == 0' to enqueue once [OK]

Hint: Check indegree condition carefully for cycle detection [OK]

Common Mistakes:

• Using <= 0 instead of == 0 for indegree check
• Ignoring character difference in dp update
• Incorrect dp initialization