Leetcode 2392 Build a Matrix With Conditions Solution in Java | Hindi Coding Community

 

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

Java Code :

public int[][] buildMatrix(int k, int[][] rowConditions, int[][] colConditions) {

List<Integer> order1 = GenerateTopologicalSort(rowConditions, k);

List<Integer> order2 = GenerateTopologicalSort(colConditions, k);

if (order1.size() < k || order2.size() < k) return new int[0][0];

Map<Integer, Integer> m = new HashMap();

for (int i = 0; i < k; i++) m.put(order2.get(i), i);

int[][] ans = new int[k][k];

for (int i = 0; i < k; i++)

ans[i][m.get(order1.get(i))] = order1.get(i);

return ans;

}

private List<Integer> GenerateTopologicalSort(int[][] A, int k) {

int[] deg = new int[k];

List<Integer> order = new ArrayList();

List<List<Integer>> graph = new ArrayList();

for (int i = 0; i < k; i++) graph.add(new ArrayList());

Queue<Integer> q = new LinkedList();

for (int[] c: A) {

graph.get(c[0] - 1).add(c[1] - 1);

deg[c[1] - 1]++;

}

for (int i = 0; i < k; i++)

if (deg[i] == 0) q.add(i);

while(!q.isEmpty()) {

int x = q.poll();

order.add(x + 1);

for (int y: graph.get(x))

if (--deg[y] == 0) q.add(y);

}

return order;

}