2392. Build a Matrix With Conditions

# Time:  O(k^2 + r + c), r = len(rowConditions), c = len(colConditions)
# Space: O(k + r + c)

# topological sort
class Solution(object):
    def buildMatrix(self, k, rowConditions, colConditions):
        """
        :type k: int
        :type rowConditions: List[List[int]]
        :type colConditions: List[List[int]]
        :rtype: List[List[int]]
        """
        def topological_sort(conditions):
            adj = [[] for _ in xrange(k)]
            in_degree = [0]*k
            for u, v in conditions:
                u -= 1
                v -= 1
                adj[u].append(v)
                in_degree[v] += 1
            result = []
            q = [u for u in xrange(k) if not in_degree[u]]
            while q:
                new_q = []
                for u in q:
                    result.append(u)
                    for v in adj[u]:
                        in_degree[v] -= 1
                        if in_degree[v]:
                            continue
                        new_q.append(v)
                q = new_q
            return result

        row_order = topological_sort(rowConditions)
        if len(row_order) != k:
            return []
        col_order = topological_sort(colConditions)
        if len(col_order) != k:
            return []
        row_idx = {x:i for i, x in enumerate(row_order)}
        col_idx = {x:i for i, x in enumerate(col_order)}
        result = [[0]*k for _ in xrange(k)]
        for i in xrange(k):
            result[row_idx[i]][col_idx[i]] = i+1
        return result

Solution from kamyu104/LeetCode-Solutions · MIT