LeetCode-in-Java

2392. Build a Matrix With Conditions

Hard

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

import java.util.ArrayList;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Queue;

public class Solution {
    // Using topological sort to solve this problem
    public int[][] buildMatrix(int k, int[][] rowC, int[][] colC) {
        // First, get the topo-sorted of row and col
        List<Integer> row = toposort(k, rowC);
        List<Integer> col = toposort(k, colC);
        // base case: when the length of row or col is less than k, return empty.
        // That is: there is a loop in established graph
        if (row.size() < k || col.size() < k) {
            return new int[0][0];
        }
        int[][] res = new int[k][k];
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < k; i++) {
            // we record the number corresbonding to each column:
            // [number, column index]
            map.put(col.get(i), i);
        }
        // col: 3 2 1
        // row: 1 3 2
        for (int i = 0; i < k; i++) {
            // For each row: we have number row.get(i). And we need to know
            // which column we need to assign, which is from map.get(row.get(i))
            // known by map.get()
            res[i][map.get(row.get(i))] = row.get(i);
        }
        return res;
    }

    private List<Integer> toposort(int k, int[][] matrix) {
        // need a int[] to record the indegree of each number [1, k]
        int[] deg = new int[k + 1];
        // need a list to record the order of each number, then return this list
        List<Integer> res = new ArrayList<>();
        // need a 2-D list to be the graph, and fill the graph
        List<List<Integer>> graph = new ArrayList<>();
        for (int i = 0; i < k; i++) {
            graph.add(new ArrayList<>());
        }
        // need a queue to do the BFS
        Queue<Integer> queue = new LinkedList<>();
        // First, we need to establish the graph, following the given matrix
        for (int[] a : matrix) {
            int from = a[0];
            int to = a[1];
            graph.get(from - 1).add(to);
            deg[to]++;
        }
        // Second, after building a graph, we start the bfs,
        // that is, traverse the node with 0 degree
        for (int i = 1; i <= k; i++) {
            if (deg[i] == 0) {
                queue.offer(i);
                res.add(i);
            }
        }
        // Third, start the topo sort
        while (!queue.isEmpty()) {
            int node = queue.poll();
            List<Integer> list = graph.get(node - 1);
            for (int i : list) {
                if (--deg[i] == 0) {
                    queue.offer(i);
                    res.add(i);
                }
            }
        }
        return res;
    }
}

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