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Kotlin-based LeetCode algorithm problem solutions, regularly updated
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:**
![](https://assets.leetcode.com/uploads/2022/07/06/gridosdrawio.png)
**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