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2392. Build a Matrix With Conditions

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

Solutions

Solution 1

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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

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class Solution {
    private int k;

    public int[][] buildMatrix(int k, int[][] rowConditions, int[][] colConditions) {
        this.k = k;
        List<Integer> row = f(rowConditions);
        List<Integer> col = f(colConditions);
        if (row == null || col == null) {
            return new int[0][0];
        }
        int[][] ans = new int[k][k];
        int[] m = new int[k + 1];
        for (int i = 0; i < k; ++i) {
            m[col.get(i)] = i;
        }
        for (int i = 0; i < k; ++i) {
            ans[i][m[row.get(i)]] = row.get(i);
        }
        return ans;
    }

    private List<Integer> f(int[][] cond) {
        List<Integer>[] g = new List[k + 1];
        Arrays.setAll(g, key -> new ArrayList<>());
        int[] indeg = new int[k + 1];
        for (var e : cond) {
            int a = e[0], b = e[1];
            g[a].add(b);
            ++indeg[b];
        }
        Deque<Integer> q = new ArrayDeque<>();
        for (int i = 1; i < indeg.length; ++i) {
            if (indeg[i] == 0) {
                q.offer(i);
            }
        }