Problem Description

In this problem, we are tasked with constructing a k x k matrix where each number from 1 to k appears exactly once and the rest of the cells are filled with 0. The placement of numbers must follow certain conditions given by two 2D arrays: rowConditions and colConditions. rowConditions[i] = [above_i, below_i] indicates that the number above_i must be placed in a row above the number below_i. Similarly, colConditions[i] = [left_i, right_i] indicates that the number left_i must be placed in a column to the left of the number right_i.

The challenge lies in placing all numbers from 1 to k such that these above and left-below and right-side conditions are met simultaneously. If it's impossible to satisfy all conditions, we must return an empty matrix.

Intuition

The given problem has a natural correspondence to topological sorting, which is a linear ordering of vertices of a directed graph such that for every directed edge (u, v), vertex u comes before v in the ordering. Here, each vertex can represent a number from 1 to k, and the directed edges can represent the rows and columns ordering constraints provided by rowConditions and colConditions.

To arrive at the solution, we apply topological sort separately for the row and column conditions. This will give us a valid order for the rows and columns if a valid solution exists. If topological sort isn’t possible (which would indicate that there is a cycle in the graph of conditions), that implies there's no way to satisfy the conditions, and hence we return an empty matrix.

For the implementation of topological sort, we define a function f that does the following:

1. Build a directed graph from the conditions given. We also maintain an in-degree count array to keep track of how many edges are coming into each node.
2. Use a queue to maintain the nodes with an in-degree of 0. These nodes can immediately be added to the sorted result as there are no preceding nodes according to the conditions.
3. When adding a node to the result, we also decrease the in-degrees of its connected nodes. If a node’s in-degree reaches 0, we add it to the queue as it’s ready to be part of the sorted result.
4. If the resulting sorted array doesn't contain exactly k elements, we return None, signaling that a topological sort is not possible.

After carrying out the topological sort for both rowConditions and colConditions, we receive two ordered arrays, one for the rows and one for the columns. If either sort fails, we return an empty matrix. Otherwise, we use a mapping to translate column orders and populate the final answer matrix according to the sorted rows and columns.

Learn more about Graph and Topological Sort patterns.

Solution Approach

The solution for building the matrix adheres to a two-step approach using topological sorting. Let’s break down each step with its corresponding algorithms, data structures, and patterns applied:

1. Topological Sorting via BFS (Breadth-First Search):

   Topological sort requires us to arrange nodes (in this case, numbers from 1 to k) such that for every directed edge (u, v), u is before v. We implement topological sorting through a modified BFS algorithm.

   • Graph Representation: We use a defaultdict of lists to create an adjacency list representation of a graph. The keys are our nodes (numbers) and the values are lists of adjacent nodes that must appear after the key node.

   • In-Degree Calculation: An array indeg is utilized to count how many edges point to each node. It essentially tells us if there are any conditions that need to be met before placing this node into our matrix.

   • Queue for Nodes with Zero In-Degree: A deque is utilized to keep track of the nodes that have no incoming edges (in-degree of 0). These are the nodes that can be placed immediately since there are no constraints on them.

   • BFS and Updating In-Degree: From the nodes in the queue, we perform a BFS. When a node is placed into the result, we iterate over its adjacent nodes in the graph, decrease their in-degree by one, and if any of these adjacent nodes now have an in-degree of 0, they are added to the queue.

   • Checking for a Successful Sort: If we cannot visit all k nodes during the topological sort (i.e., the resulting list is shorter than k), this indicates that there are cyclic dependencies and we cannot build a valid matrix. Therefore, we return None to indicate failure.

2. Building the Result Matrix:

   Once we have a valid topological sort for both rows (rowConditions) and columns (colConditions), we proceed to populate our k x k matrix.

   • Mapping Columns: We create an array m that maps each number to its location in the column order. This allows us to know where to place each number in the matrix columns.

   • Populating the Matrix: With row and m giving us the row and column indices respectively, we iterate through the sorted row indices, and for each number, we use the column index from the mapping m. This is where we place the value v using ans[i][m[v]] = v.

   • Resulting Matrix: The result ans is the final k x k matrix that satisfies all the given conditions, filled with numbers from 1 to k, respecting the topological order, and with all remaining unfilled cells populated with zeroes.

In the case where either the row topological sort or column topological sort fails, we know that there is no valid solution and we return an empty matrix []. Otherwise, the filled ans matrix is returned as the solution.

Example Walkthrough

Let's go through an example to illustrate the solution approach described above. Assume we have a 3x3 matrix (k=3) and we need to place numbers 1, 2, and 3 in the matrix according to the given conditions.

Suppose rowConditions and colConditions are given as follows:

1rowConditions = [[2, 3], [1, 2]]
2colConditions = [[1, 3], [2, 3]]

Step 1: Topological Sorting

We start with topological sorting for the rows and columns separately. We'll use a Breadth-First Search (BFS) algorithm for this sorting.

• For rows:

1. Build a graph from the rowConditions and calculate the in-degrees:

   • Node 2 should come before 3, and Node 1 should come before 2.
   • Graph: 1 -> 2 -> 3 (1 points to 2, and 2 points to 3)
   • In-degree: [0, 1, 1] since Node 1 has 0 in-degree (no conditions), Node 2 has 1 in-degree (from 1), and Node 3 has 1 in-degree (from 2).

2. Perform BFS:

   • Queue: Start with Node 1 (in-degree 0).
   • Add Node 1 to the result, queue next is Node 2 (decrement in-degrees and check).
   • Add Node 2 to the result, queue next is Node 3 (decrement in-degrees and check).

• For columns:

1. Build a graph from the colConditions and calculate the in-degrees:

   • Graph similar to row graph: 1 -> 3, 2 -> 3
   • In-degree: [0, 0, 2] (Node 3 has 2 in-degrees, Node 1 and Node 2 have 0).

2. Perform BFS:

   • Queue: Start with Node 1 and Node 2 (both have in-degree 0).
   • Add Node 1 and Node 2 to the result, queue next is Node 3.

Step 2: Building the Result Matrix

Now, we use the obtained topological sorts to fill in our matrix.

• Row order from topological sort: [1, 2, 3]
• Column order from topological sort: [1, 2, 3]

Create a mapping of column indices: m = [0, 1, 2]

We now populate the matrix ans, using the order from row and column topological sorts and the mapping m:

• Place 1 at ans[0][m[0]], which is ans[0][0]
• Place 2 at ans[1][m[1]], which is ans[1][1]
• Place 3 at ans[2][m[2]], which is ans[2][2]

The remaining cells are filled with zeros. Our final matrix looks like this:

1ans = [
2    [1, 0, 0],
3    [0, 2, 0],
4    [0, 0, 3]
5]

This matrix satisfies the given row and column conditions:

• Number 2 is below number 1 (row condition).
• Number 3 is below number 2 (row condition).
• Number 3 is to the right of number 1 (column condition).
• Number 3 is to the right of number 2 (column condition).

In conclusion, the matrix ans is the valid configuration we were looking to construct based on the given rowConditions and colConditions. If at any point during the topological sorting step we found a cycle or could not visit all nodes, that would have indicated there is no valid matrix configuration, and the result would have been an empty matrix. But in this example, everything checks out and we successfully built the matrix.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code consists primarily of two parts: constructing the order of elements according to the input conditions (rowConditions and colConditions) and then building the matrix based on these orders.

1. For constructing the order with the function f(cond):

   • It creates a graph where each element i points to the elements it must precede in sequence. The graph creation takes O(E) time where E is the total number of conditions in rowConditions or colConditions.
   • It initializes an indegree array indeg which takes O(k) time.
   • It processes each node exactly once using a queue to perform topological sort, which will be O(V + E), where V is the number of vertices (which is k) and E is the number of edges.
   • As f(cond) is called twice (once for rows and once for columns), the time complexity for this part is 2 * O(V + E) = O(2k + 2E) = O(k + E).

2. For the construction of the answer matrix:

   • The code uses a map m to keep track of the index at which each value should be placed in the matrix, which takes O(k) time.
   • It then iterates over each element of row and places it in the correct position in the ans matrix, which takes O(k) time.

Overall, the time complexity of the code is determined by the topological sort and the construction of the result matrix, which would be O(k + E) + O(k) = O(k + E).

Space Complexity

The space complexity of the code is determined by:

• The graph g and indegree array indeg which collectively take O(k + E) space.
• The queue q used for storing intermediate values while doing topological sort, which in the worst case could store all k vertices, taking O(k) space.
• The result array res that will contain k elements when the order is successfully constructed, taking O(k) space.
• The answer matrix ans which is a k x k matrix, so it takes O(k^2) space.
• Auxiliary space for the map m and variables used to iterate, which take O(k).

When added together, the space complexity is O(k + E) + O(k) + O(k) + O(k^2) + O(k) = O(k^2 + k + E).

Learn more about how to find time and space complexity quickly using problem constraints.