2392. Build a Matrix With Conditions

You are given a positive integer k. You are also given:

• a 2D integer array rowConditions of size n where rowConditions[i] = [abovei, belowi], and
• a 2D integer array colConditions of size m where colConditions[i] = [lefti, righti].

The two arrays contain integers from 1 to k.

You have to build a k x k matrix that contains each of the numbers from 1 to k exactly once. The remaining cells should have the value 0.

The matrix should also satisfy the following conditions:

• The number abovei should appear in a row that is strictly above the row at which the number belowi appears for all i from 0 to n - 1.
• The number lefti should appear in a column that is strictly left of the column at which the number righti appears for all i from 0 to m - 1.

Return any matrix that satisfies the conditions. If no answer exists, return an empty matrix.

Example 1:

Input: k = 3, rowConditions = [[1,2],[3,2]], colConditions = [[2,1],[3,2]]
Output: [[3,0,0],[0,0,1],[0,2,0]]
Explanation: The diagram above shows a valid example of a matrix that satisfies all the conditions.
The row conditions are the following:
- Number 1 is in row 1, and number 2 is in row 2, so 1 is above 2 in the matrix.
- Number 3 is in row 0, and number 2 is in row 2, so 3 is above 2 in the matrix.
The column conditions are the following:
- Number 2 is in column 1, and number 1 is in column 2, so 2 is left of 1 in the matrix.
- Number 3 is in column 0, and number 2 is in column 1, so 3 is left of 2 in the matrix.
Note that there may be multiple correct answers.

Example 2:

Input: k = 3, rowConditions = [[1,2],[2,3],[3,1],[2,3]], colConditions = [[2,1]]
Output: []
Explanation: From the first two conditions, 3 has to be below 1 but the third conditions needs 3 to be above 1 to be satisfied.
No matrix can satisfy all the conditions, so we return the empty matrix.

Constraints:

• 2 <= k <= 400
• 1 <= rowConditions.length, colConditions.length <= 104
• rowConditions[i].length == colConditions[i].length == 2
• 1 <= abovei, belowi, lefti, righti <= k
• abovei != belowi
• lefti != righti

Approach 01:

class Solution {
public:
    vector<vector<int>> buildMatrix(int k, vector<vector<int>>& rowConditions, vector<vector<int>>& colConditions) {
        // Perform topological sorting for rows
        const vector<int> rowOrder = topologicalSort(rowConditions, k);
        if (rowOrder.empty())
            return {};

        // Perform topological sorting for columns
        const vector<int> colOrder = topologicalSort(colConditions, k);
        if (colOrder.empty())
            return {};

        // Initialize the result matrix
        vector<vector<int>> matrix(k, vector<int>(k));
        vector<int> nodeToRowIndex(k + 1);

        // Map nodes to their row indices based on rowOrder
        for (int row = 0; row < k; ++row)
            nodeToRowIndex[rowOrder[row]] = row;

        // Fill in the matrix based on the column order
        for (int col = 0; col < k; ++col) {
            const int node = colOrder[col];
            const int row = nodeToRowIndex[node];
            matrix[row][col] = node;
        }

        return matrix;
    }

private:
    // Perform topological sort on a directed graph defined by conditions
    vector<int> topologicalSort(const vector<vector<int>>& conditions, int n) {
        vector<int> sortedOrder;            // To store the topological order
        vector<vector<int>> adjList(n + 1); // Adjacency list for the graph
        vector<int> inDegrees(n + 1, 0);    // In-degree count for each node
        queue<int> zeroInDegreeNodes; // Queue for nodes with zero in-degree

        // Build the graph from conditions
        for (const vector<int>& condition : conditions) {
            const int from = condition[0];
            const int to = condition[1];
            adjList[from].push_back(to);
            ++inDegrees[to];
        }

        // Initialize the queue with nodes having zero in-degree
        for (int i = 1; i <= n; ++i)
            if (inDegrees[i] == 0)
                zeroInDegreeNodes.push(i);

        // Perform Kahn's algorithm for topological sorting
        while (!zeroInDegreeNodes.empty()) {
            const int node = zeroInDegreeNodes.front();
            zeroInDegreeNodes.pop();
            sortedOrder.push_back(node);
            for (const int neighbor : adjList[node])
                if (--inDegrees[neighbor] == 0)
                    zeroInDegreeNodes.push(neighbor);
        }

        // Check if a valid topological sort was found
        return sortedOrder.size() == n ? sortedOrder : vector<int>();
    }
};

from collections import deque, defaultdict
from typing import List

class Solution:
    def buildMatrix(self, k: int, rowConditions: List[List[int]], colConditions: List[List[int]]) -> List[List[int]]:
        # Perform topological sorting for rows
        rowOrder = self.topologicalSort(rowConditions, k)
        if not rowOrder:
            return []

        # Perform topological sorting for columns
        colOrder = self.topologicalSort(colConditions, k)
        if not colOrder:
            return []

        # Initialize the result matrix
        matrix = [[0] * k for _ in range(k)]
        nodeToRowIndex = [0] * (k + 1)

        # Map nodes to their row indices based on rowOrder
        for row, node in enumerate(rowOrder):
            nodeToRowIndex[node] = row

        # Fill in the matrix based on the column order
        for col, node in enumerate(colOrder):
            row = nodeToRowIndex[node]
            matrix[row][col] = node

        return matrix

    def topologicalSort(self, conditions: List[List[int]], n: int) -> List[int]:
        sortedOrder = []                     # To store the topological order
        adjList = defaultdict(list)          # Adjacency list for the graph
        inDegrees = [0] * (n + 1)            # In-degree count for each node
        zeroInDegreeNodes = deque()          # Queue for nodes with zero in-degree

        # Build the graph from conditions
        for from_node, to_node in conditions:
            adjList[from_node].append(to_node)
            inDegrees[to_node] += 1

        # Initialize the queue with nodes having zero in-degree
        for i in range(1, n + 1):
            if inDegrees[i] == 0:
                zeroInDegreeNodes.append(i)

        # Perform Kahn's algorithm for topological sorting
        while zeroInDegreeNodes:
            node = zeroInDegreeNodes.popleft()
            sortedOrder.append(node)
            for neighbor in adjList[node]:
                inDegrees[neighbor] -= 1
                if inDegrees[neighbor] == 0:
                    zeroInDegreeNodes.append(neighbor)

        # Check if a valid topological sort was found
        return sortedOrder if len(sortedOrder) == n else []

Time Complexity

• Topological Sorting:

  The time complexity for performing topological sorting using Kahn’s algorithm is \( O(V + E) \), where \( V \) is the number of nodes (which is \( k \) in this case) and \( E \) is the number of edges in the graph (which depends on the size of rowConditions and colConditions).

• Matrix Construction:

  After obtaining the topological order for rows and columns, constructing the matrix takes \( O(k^2) \) time, as it involves filling a \( k \times k \) matrix.

• Overall Time Complexity:

  The overall time complexity is \( O(k^2 + E) \), where \( E \) is the number of edges in the graph formed by rowConditions and colConditions.

Space Complexity

• Auxiliary Space:

  The space complexity for storing the adjacency list, in-degrees, and topological sort order is \( O(k + E) \). Additionally, the result matrix requires \( O(k^2) \) space.

• Overall Space Complexity:

  The overall space complexity is \( O(k^2 + E) \).