LeetCode in Kotlin

2201. Count Artifacts That Can Be Extracted

Medium

There is an n x n 0-indexed grid with some artifacts buried in it. You are given the integer n and a 0-indexed 2D integer array artifacts describing the positions of the rectangular artifacts where artifacts[i] = [r1i, c1i, r2i, c2i] denotes that the ith artifact is buried in the subgrid where:

You will excavate some cells of the grid and remove all the mud from them. If the cell has a part of an artifact buried underneath, it will be uncovered. If all the parts of an artifact are uncovered, you can extract it.

Given a 0-indexed 2D integer array dig where dig[i] = [ri, ci] indicates that you will excavate the cell (ri, ci), return the number of artifacts that you can extract.

The test cases are generated such that:

Example 1:

Input: n = 2, artifacts = [[0,0,0,0],[0,1,1,1]], dig = [[0,0],[0,1]]

Output: 1

Explanation: The different colors represent different artifacts. Excavated cells are labeled with a ā€˜Dā€™ in the grid. There is 1 artifact that can be extracted, namely the red artifact. The blue artifact has one part in cell (1,1) which remains uncovered, so we cannot extract it. Thus, we return 1.

Example 2:

Input: n = 2, artifacts = [[0,0,0,0],[0,1,1,1]], dig = [[0,0],[0,1],[1,1]]

Output: 2

Explanation: Both the red and blue artifacts have all parts uncovered (labeled with a ā€˜Dā€™) and can be extracted, so we return 2.

Constraints:

Solution

class Solution {
    fun digArtifacts(n: Int, artifacts: Array<IntArray>, dig: Array<IntArray>): Int {
        val ar = Array(n) { IntArray(n) }
        for (ints in dig) {
            ar[ints[0]][ints[1]] = 1
        }
        var ans = 0
        for (artifact in artifacts) {
            val x1 = artifact[0]
            val y1 = artifact[1]
            val x2 = artifact[2]
            val y2 = artifact[3]
            var flag = 0
            var a = x1
            var b = y1
            while (a <= x2) {
                b = y1
                while (b <= y2) {
                    if (ar[a][b] != 1) {
                        flag = 1
                        break
                    }
                    b++
                }
                if (flag == 1) {
                    break
                }
                a++
            }
            if (a == x2 + 1 && b == y2 + 1) {
                ans++
            }
        }
        return ans
    }
}