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2392. Build a Matrix With Conditions

Difficulty Topics

Description

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

Solution

build-a-matrix-with-conditions.py
class Solution:
    def buildMatrix(self, k: int, rowConditions: List[List[int]], colConditions: List[List[int]]) -> List[List[int]]:
        res = [[0] * k for _ in range(k)]

        def topoSort(A):
            adj = [[] for _ in range(k + 1)] 

            for a, b in A:
                adj[a].append(b)

            seen = [False] * (k + 1)
            order = []

            def dfs(x):
                if seen[x]: return

                seen[x] = True

                for nei in adj[x]:
                    if not seen[nei]:
                        dfs(nei)

                order.append(x)

            for x in range(1, k + 1):
                dfs(x)

            order.reverse()
            M = {j: i for i, j in enumerate(order)}

            if all(M[i] < M[j] for i, j in A):
                return order

            return None

        R = topoSort(rowConditions)
        C = topoSort(colConditions)

        if R is None or C is None:
            return []

        RM = {j : i for i, j in enumerate(R)}
        CM = {j : i for i, j in enumerate(C)}

        for x in range(1, k + 1):
            res[RM[x]][CM[x]] = x

        return res