# 2146. K Highest Ranked Items Within a Price Range

## Description

You are given a 0-indexed 2D integer array grid of size m x n that represents a map of the items in a shop. The integers in the grid represent the following:

• 0 represents a wall that you cannot pass through.
• 1 represents an empty cell that you can freely move to and from.
• All other positive integers represent the price of an item in that cell. You may also freely move to and from these item cells.

It takes 1 step to travel between adjacent grid cells.

You are also given integer arrays pricing and start where pricing = [low, high] and start = [row, col] indicates that you start at the position (row, col) and are interested only in items with a price in the range of [low, high] (inclusive). You are further given an integer k.

You are interested in the positions of the k highest-ranked items whose prices are within the given price range. The rank is determined by the first of these criteria that is different:

1. Distance, defined as the length of the shortest path from the start (shorter distance has a higher rank).
2. Price (lower price has a higher rank, but it must be in the price range).
3. The row number (smaller row number has a higher rank).
4. The column number (smaller column number has a higher rank).

Return the k highest-ranked items within the price range sorted by their rank (highest to lowest). If there are fewer than k reachable items within the price range, return all of them.

Example 1:

Input: grid = [[1,2,0,1],[1,3,0,1],[0,2,5,1]], pricing = [2,5], start = [0,0], k = 3
Output: [[0,1],[1,1],[2,1]]
Explanation: You start at (0,0).
With a price range of [2,5], we can take items from (0,1), (1,1), (2,1) and (2,2).
The ranks of these items are:
- (0,1) with distance 1
- (1,1) with distance 2
- (2,1) with distance 3
- (2,2) with distance 4
Thus, the 3 highest ranked items in the price range are (0,1), (1,1), and (2,1).


Example 2:

Input: grid = [[1,2,0,1],[1,3,3,1],[0,2,5,1]], pricing = [2,3], start = [2,3], k = 2
Output: [[2,1],[1,2]]
Explanation: You start at (2,3).
With a price range of [2,3], we can take items from (0,1), (1,1), (1,2) and (2,1).
The ranks of these items are:
- (2,1) with distance 2, price 2
- (1,2) with distance 2, price 3
- (1,1) with distance 3
- (0,1) with distance 4
Thus, the 2 highest ranked items in the price range are (2,1) and (1,2).


Example 3:

Input: grid = [[1,1,1],[0,0,1],[2,3,4]], pricing = [2,3], start = [0,0], k = 3
Output: [[2,1],[2,0]]
Explanation: You start at (0,0).
With a price range of [2,3], we can take items from (2,0) and (2,1).
The ranks of these items are:
- (2,1) with distance 5
- (2,0) with distance 6
Thus, the 2 highest ranked items in the price range are (2,1) and (2,0).
Note that k = 3 but there are only 2 reachable items within the price range.


Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 105
• 1 <= m * n <= 105
• 0 <= grid[i][j] <= 105
• pricing.length == 2
• 2 <= low <= high <= 105
• start.length == 2
• 0 <= row <= m - 1
• 0 <= col <= n - 1
• grid[row][col] > 0
• 1 <= k <= m * n

## Solutions

BFS.

• class Solution {
public List<List<Integer>> highestRankedKItems(
int[][] grid, int[] pricing, int[] start, int k) {
int m = grid.length, n = grid[0].length;
int row = start[0], col = start[1];
int low = pricing[0], high = pricing[1];
List<int[]> items = new ArrayList<>();
if (low <= grid[row][col] && grid[row][col] <= high) {
items.add(new int[] {0, grid[row][col], row, col});
}
grid[row][col] = 0;
Deque<int[]> q = new ArrayDeque<>();
q.offer(new int[] {row, col, 0});
int[] dirs = {-1, 0, 1, 0, -1};
while (!q.isEmpty()) {
int[] p = q.poll();
int i = p[0], j = p[1], d = p[2];
for (int l = 0; l < 4; ++l) {
int x = i + dirs[l], y = j + dirs[l + 1];
if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] > 0) {
if (low <= grid[x][y] && grid[x][y] <= high) {
items.add(new int[] {d + 1, grid[x][y], x, y});
}
grid[x][y] = 0;
q.offer(new int[] {x, y, d + 1});
}
}
}
items.sort((a, b) -> {
if (a[0] != b[0]) {
return a[0] - b[0];
}
if (a[1] != b[1]) {
return a[1] - b[1];
}
if (a[2] != b[2]) {
return a[2] - b[2];
}
return a[3] - b[3];
});
List<List<Integer>> ans = new ArrayList<>();
for (int i = 0; i < items.size() && i < k; ++i) {
int[] p = items.get(i);
}
return ans;
}
}

• class Solution {
public:
vector<vector<int>> highestRankedKItems(vector<vector<int>>& grid, vector<int>& pricing, vector<int>& start, int k) {
int m = grid.size(), n = grid[0].size();
int row = start[0], col = start[1];
int low = pricing[0], high = pricing[1];
vector<tuple<int, int, int, int>> items;
if (low <= grid[row][col] && grid[row][col] <= high)
items.emplace_back(0, grid[row][col], row, col);
queue<tuple<int, int, int>> q;
q.emplace(row, col, 0);
grid[row][col] = 0;
vector<int> dirs = {-1, 0, 1, 0, -1};
while (!q.empty()) {
auto [i, j, d] = q.front();
q.pop();
for (int l = 0; l < 4; ++l) {
int x = i + dirs[l], y = j + dirs[l + 1];
if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y]) {
if (low <= grid[x][y] && grid[x][y] <= high) items.emplace_back(d + 1, grid[x][y], x, y);
grid[x][y] = 0;
q.emplace(x, y, d + 1);
}
}
}
sort(items.begin(), items.end());
vector<vector<int>> ans;
for (int i = 0; i < items.size() && i < k; ++i) {
auto [d, p, x, y] = items[i];
ans.push_back({x, y});
}
return ans;
}
};

• class Solution:
def highestRankedKItems(
self, grid: List[List[int]], pricing: List[int], start: List[int], k: int
) -> List[List[int]]:
m, n = len(grid), len(grid[0])
row, col, low, high = start + pricing
items = []
if low <= grid[row][col] <= high:
items.append([0, grid[row][col], row, col])
q = deque([(row, col, 0)])
grid[row][col] = 0
while q:
i, j, d = q.popleft()
for a, b in [[0, 1], [0, -1], [1, 0], [-1, 0]]:
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n and grid[x][y]:
if low <= grid[x][y] <= high:
items.append([d + 1, grid[x][y], x, y])
q.append((x, y, d + 1))
grid[x][y] = 0
items.sort()
return [item[2:] for item in items][:k]


• func highestRankedKItems(grid [][]int, pricing []int, start []int, k int) [][]int {
m, n := len(grid), len(grid[0])
row, col := start[0], start[1]
low, high := pricing[0], pricing[1]
var items [][]int
if low <= grid[row][col] && grid[row][col] <= high {
items = append(items, []int{0, grid[row][col], row, col})
}
q := [][]int{ {row, col, 0} }
grid[row][col] = 0
dirs := []int{-1, 0, 1, 0, -1}
for len(q) > 0 {
p := q[0]
q = q[1:]
i, j, d := p[0], p[1], p[2]
for l := 0; l < 4; l++ {
x, y := i+dirs[l], j+dirs[l+1]
if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] > 0 {
if low <= grid[x][y] && grid[x][y] <= high {
items = append(items, []int{d + 1, grid[x][y], x, y})
}
grid[x][y] = 0
q = append(q, []int{x, y, d + 1})
}
}
}
sort.Slice(items, func(i, j int) bool {
a, b := items[i], items[j]
if a[0] != b[0] {
return a[0] < b[0]
}
if a[1] != b[1] {
return a[1] < b[1]
}
if a[2] != b[2] {
return a[2] < b[2]
}
return a[3] < b[3]
})
var ans [][]int
for i := 0; i < len(items) && i < k; i++ {
ans = append(ans, items[i][2:])
}
return ans
}

• function highestRankedKItems(
grid: number[][],
pricing: number[],
start: number[],
k: number,
): number[][] {
const [m, n] = [grid.length, grid[0].length];
const [row, col] = start;
const [low, high] = pricing;
let q: [number, number][] = [[row, col]];
const pq: [number, number, number, number][] = [];
if (low <= grid[row][col] && grid[row][col] <= high) {
pq.push([0, grid[row][col], row, col]);
}
grid[row][col] = 0;
const dirs = [-1, 0, 1, 0, -1];
for (let step = 1; q.length > 0; ++step) {
const nq: [number, number][] = [];
for (const [x, y] of q) {
for (let j = 0; j < 4; j++) {
const nx = x + dirs[j];
const ny = y + dirs[j + 1];
if (nx >= 0 && nx < m && ny >= 0 && ny < n && grid[nx][ny] > 0) {
if (low <= grid[nx][ny] && grid[nx][ny] <= high) {
pq.push([step, grid[nx][ny], nx, ny]);
}
grid[nx][ny] = 0;
nq.push([nx, ny]);
}
}
}
q = nq;
}
pq.sort((a, b) => {
if (a[0] !== b[0]) return a[0] - b[0];
if (a[1] !== b[1]) return a[1] - b[1];
if (a[2] !== b[2]) return a[2] - b[2];
return a[3] - b[3];
});
const ans: number[][] = [];
for (let i = 0; i < Math.min(k, pq.length); i++) {
ans.push([pq[i][2], pq[i][3]]);
}
return ans;
}