Leetcode #2392: Build a Matrix With Conditions
In this guide, we solve Leetcode #2392 Build a Matrix With Conditions in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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].
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Graph, Topological Sort, Array, Matrix
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
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.
Python Solution
class Solution:
def buildMatrix(
self, k: int, rowConditions: List[List[int]], colConditions: List[List[int]]
) -> List[List[int]]:
def f(cond):
g = defaultdict(list)
indeg = [0] * (k + 1)
for a, b in cond:
g[a].append(b)
indeg[b] += 1
q = deque([i for i, v in enumerate(indeg[1:], 1) if v == 0])
res = []
while q:
for _ in range(len(q)):
i = q.popleft()
res.append(i)
for j in g[i]:
indeg[j] -= 1
if indeg[j] == 0:
q.append(j)
return None if len(res) != k else res
row = f(rowConditions)
col = f(colConditions)
if row is None or col is None:
return []
ans = [[0] * k for _ in range(k)]
m = [0] * (k + 1)
for i, v in enumerate(col):
m[v] = i
for i, v in enumerate(row):
ans[i][m[v]] = v
return ans
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.