LeetCode in Kotlin

2503. Maximum Number of Points From Grid Queries

Hard

You are given an m x n integer matrix grid and an array queries of size k.

Find an array answer of size k such that for each integer queries[i] you start in the top left cell of the matrix and repeat the following process:

• If queries[i] is strictly greater than the value of the current cell that you are in, then you get one point if it is your first time visiting this cell, and you can move to any adjacent cell in all 4 directions: up, down, left, and right.
• Otherwise, you do not get any points, and you end this process.

After the process, answer[i] is the maximum number of points you can get. Note that for each query you are allowed to visit the same cell multiple times.

Return the resulting array answer.

Example 1:

Input: grid = [[1,2,3],[2,5,7],[3,5,1]], queries = [5,6,2]

Output: [5,8,1]

Explanation: The diagrams above show which cells we visit to get points for each query.

Example 2:

Input: grid = [[5,2,1],[1,1,2]], queries = [3]

Output: [0]

Explanation: We can not get any points because the value of the top left cell is already greater than or equal to 3.

Constraints:

• m == grid.length
• n == grid[i].length
• 2 <= m, n <= 1000
• 4 <= m * n <= 105
• k == queries.length
• 1 <= k <= 104
• 1 <= grid[i][j], queries[i] <= 106

Solution

import java.util.ArrayDeque
import java.util.Arrays
import java.util.PriorityQueue
import java.util.Queue

class Solution {
    private val dirs = arrayOf(intArrayOf(-1, 0), intArrayOf(1, 0), intArrayOf(0, -1), intArrayOf(0, 1))

    fun maxPoints(grid: Array<IntArray>, queries: IntArray): IntArray {
        val r = grid.size
        val c = grid[0].size
        val res = IntArray(queries.size)
        val index = arrayOfNulls<Int>(queries.size)
        for (i in queries.indices) {
            index[i] = i
        }
        Arrays.sort(index, { o: Int?, m: Int? -> queries[o!!].compareTo(queries[m!!]) })
        val q1: Queue<IntArray> = ArrayDeque()
        val q2 = PriorityQueue({ a: IntArray, b: IntArray -> a[2].compareTo(b[2]) })
        q2.offer(intArrayOf(0, 0, grid[0][0]))
        val visited = Array(r) { BooleanArray(c) }
        var count = 0
        visited[0][0] = true
        for (i in queries.indices) {
            val currLimit = queries[index[i]!!]
            while (q2.isNotEmpty() && q2.peek()[2] < currLimit) {
                q1.offer(q2.poll())
            }
            while (q1.isNotEmpty()) {
                val curr = q1.poll()
                count++
                for (dir in dirs) {
                    val x = dir[0] + curr[0]
                    val y = dir[1] + curr[1]
                    if (x < 0 || y < 0 || x >= r || y >= c || visited[x][y]) {
                        continue
                    }
                    if (!visited[x][y] && grid[x][y] < currLimit) {
                        q1.offer(intArrayOf(x, y, grid[x][y]))
                    } else {
                        q2.offer(intArrayOf(x, y, grid[x][y]))
                    }
                    visited[x][y] = true
                }
            }
            res[index[i]!!] = count
        }
        return res
    }
}

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