How do you solve Build a Matrix With Conditions on LeetCode?

Build a Matrix With Conditions (LeetCode #2392) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(n + m) space complexity. Key insight: Understanding the problem as a graph allows for efficient ordering of elements.

What is the brute force solution for Build a Matrix With Conditions?

The brute force approach for Build a Matrix With Conditions is Brute Force, with O(n!) time complexity and O(n²) space complexity. The brute-force approach involves generating all possible k x k matrices and checking if they satisfy the given row and column conditions. This is straightforward but inefficient due to the high number of combinations. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve Build a Matrix With Conditions?

Build a Matrix With Conditions uses the following concepts: Array, Graph Theory, Topological Sort, Matrix. The recommended patterns to study are: Graph Theory, Topological Sort, Dependency Resolution.

What are the interview tips for Build a Matrix With Conditions?

Clearly explain your thought process and how you visualize the problem. Be prepared to discuss the implications of your chosen algorithm on performance. Practice implementing graph algorithms, as they often come up in similar problems.

What are common mistakes when solving Build a Matrix With Conditions?

Failing to recognize the need for a topological sort. Not checking for cycles in the graph, which can lead to incorrect assumptions about the order.

#2392

Build a Matrix With Conditions

Hard

ArrayGraph TheoryTopological SortMatrixGraph TheoryTopological SortDependency Resolution

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n + m)
Space	O(n²)	O(n + m)

💡

Intuition

Time O(n + m)Space O(n + m)

The optimal solution utilizes graph theory and topological sorting to determine the order of numbers based on the given conditions. This approach is efficient and directly addresses the constraints of the problem.

⚙️

Algorithm

3 steps

1Step 1: Construct a directed graph where each number points to the numbers that must be below or to the right of it.
2Step 2: Perform a topological sort on the graph to determine a valid order of numbers.
3Step 3: Place the sorted numbers in the matrix according to their positions derived from the topological sort.

solution.py41 lines

1from collections import defaultdict, deque
2
3def buildMatrix(k, rowConditions, colConditions):
4    row_graph = defaultdict(list)
5    row_indegree = [0] * (k + 1)
6    col_graph = defaultdict(list)
7    col_indegree = [0] * (k + 1)
8
9    for above, below in rowConditions:
10        row_graph[above].append(below)
11        row_indegree[below] += 1
12    for left, right in colConditions:
13        col_graph[left].append(right)
14        col_indegree[right] += 1
15
16    row_order = topologicalSort(row_graph, row_indegree, k)
17    col_order = topologicalSort(col_graph, col_indegree, k)
18
19    if not row_order or not col_order:
20        return []
21
22    matrix = [[0] * k for _ in range(k)]
23    for i in range(k):
24        matrix[row_order[i]][col_order[i]] = i + 1
25    return matrix
26
27
28def topologicalSort(graph, indegree, k):
29    queue = deque()
30    for i in range(1, k + 1):
31        if indegree[i] == 0:
32            queue.append(i)
33    order = []
34    while queue:
35        node = queue.popleft()
36        order.append(node)
37        for neighbor in graph[node]:
38            indegree[neighbor] -= 1
39            if indegree[neighbor] == 0:
40                queue.append(neighbor)
41    return order if len(order) == k else []

ℹ

Complexity note: The time complexity is O(n + m) where n is the number of row conditions and m is the number of column conditions. The space complexity is O(n + m) for storing the graph and indegree arrays.

1Understanding the problem as a graph allows for efficient ordering of elements.
2Topological sorting is crucial for resolving dependencies in directed graphs.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.