2392. Build a Matrix With Conditions

Problem

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

/**
 * @param {number} k
 * @param {number[][]} rowConditions
 * @param {number[][]} colConditions
 * @return {number[][]}
 */
var buildMatrix = function(k, rowConditions, colConditions) {
    var rowOrder = topologicalSort(k, rowConditions);
    var colOrder = topologicalSort(k, colConditions);
    if (!rowOrder || !colOrder) return [];
    var colOrderMap = colOrder.reduce((map, n, i) => {
        map[n] = i;
        return map;
    }, {});
    var matrix = Array(k).fill(0).map(() => Array(k).fill(0));
    rowOrder.forEach((n, i) => matrix[i][colOrderMap[n]] = n);
    return matrix;
};

var topologicalSort = function(k, arr) {
    var beforeMap = Array(k).fill(0);
    var afterMap = Array(k).fill(0).map(() => []);
    for (var i = 0; i < arr.length; i++) {
        beforeMap[arr[i][1] - 1] += 1;
        afterMap[arr[i][0] - 1].push(arr[i][1]);
    }
    var queue = [];
    for (var j = 0; j < k; j++) {
        if (beforeMap[j] === 0) {
            queue.push(j + 1);
        }
    }
    var res = [];
    while (queue.length) {
        var num = queue.shift();
        afterMap[num - 1].forEach(n => {
            if (--beforeMap[n - 1] === 0) {
                queue.push(n);
            }
        });
        res.push(num);
    }
    if (res.length === k) {
        return res;
    }
    return null;
};

Explain:

nope.

Complexity:

• Time complexity : O(k).
• Space complexity : O(k ^ 2).