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

Solution 1: BFS + Sorting

We can start from $(\textit{row}, \textit{col})$ and use breadth-first search to find all items with prices in the range $[\textit{low}, \textit{high}]$. Store the distance, price, row coordinate, and column coordinate of these items in the array $\textit{pq}$.

Finally, sort $\textit{pq}$ by distance, price, row coordinate, and column coordinate, and return the coordinates of the first $k$ items.

The time complexity is $O(m \times n \times \log (m \times n))$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the number of rows and columns of the 2D array $\textit{grid}$, respectively.

Python3JavaC++GoTypeScript

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 = start
        low, high = pricing
        q = deque([(row, col)])
        pq = []
        if low <= grid[row][col] <= high:
            pq.append((0, grid[row][col], row, col))
        grid[row][col] = 0
        dirs = (-1, 0, 1, 0, -1)
        step = 0
        while q:
            step += 1
            for _ in range(len(q)):
                x, y = q.popleft()
                for a, b in pairwise(dirs):
                    nx, ny = x + a, y + b
                    if 0 <= nx < m and 0 <= ny < n and grid[nx][ny] > 0:
                        if low <= grid[nx][ny] <= high:
                            pq.append((step, grid[nx][ny], nx, ny))
                        grid[nx][ny] = 0
                        q.append((nx, ny))
        pq.sort()
        return [list(x[2:]) for x in pq[:k]]

class Solution {
    public List<List<Integer>> highestRankedKItems(
        int[][] grid, int[] pricing, int[] start, int k) {
        int m = grid.length;
        int n = grid[0].length;
        int row = start[0], col = start[1];
        int low = pricing[0], high = pricing[1];
        Deque<int[]> q = new ArrayDeque<>();
        q.offer(new int[] {row, col});
        List<int[]> pq = new ArrayList<>();
        if (low <= grid[row][col] && grid[row][col] <= high) {
            pq.add(new int[] {0, grid[row][col], row, col});
        }
        grid[row][col] = 0;
        final int[] dirs = {-1, 0, 1, 0, -1};
        for (int step = 1; !q.isEmpty(); ++step) {
            for (int size = q.size(); size > 0; --size) {
                int[] curr = q.poll();
                int x = curr[0], y = curr[1];
                for (int j = 0; j < 4; j++) {
                    int nx = x + dirs[j];
                    int ny = y + dirs[j + 1];
                    if (0 <= nx && nx < m && 0 <= ny && ny < n && grid[nx][ny] > 0) {
                        if (low <= grid[nx][ny] && grid[nx][ny] <= high) {
                            pq.add(new int[] {step, grid[nx][ny], nx, ny});
                        }
                        grid[nx][ny] = 0;
                        q.offer(new int[] {nx, ny});
                    }
                }
            }
        }

        pq.sort((a, b) -> {
            if (a[0] != b[0]) return Integer.compare(a[0], b[0]);
            if (a[1] != b[1]) return Integer.compare(a[1], b[1]);
            if (a[2] != b[2]) return Integer.compare(a[2], b[2]);
            return Integer.compare(a[3], b[3]);
        });

        List<List<Integer>> ans = new ArrayList<>();
        for (int i = 0; i < Math.min(k, pq.size()); i++) {
            ans.add(List.of(pq.get(i)[2], pq.get(i)[3]));
        }
        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];
        queue<pair<int, int>> q;
        q.push({row, col});
        vector<tuple<int, int, int, int>> pq;
        if (low <= grid[row][col] && grid[row][col] <= high) {
            pq.push_back({0, grid[row][col], row, col});
        }
        grid[row][col] = 0;
        vector<int> dirs = {-1, 0, 1, 0, -1};
        for (int step = 1; q.size(); ++step) {
            int sz = q.size();
            for (int i = 0; i < sz; ++i) {
                auto [x, y] = q.front();
                q.pop();
                for (int j = 0; j < 4; ++j) {
                    int nx = x + dirs[j];
                    int ny = y + dirs[j + 1];
                    if (0 <= nx && nx < m && 0 <= ny && ny < n && grid[nx][ny] > 0) {
                        if (low <= grid[nx][ny] && grid[nx][ny] <= high) {
                            pq.push_back({step, grid[nx][ny], nx, ny});
                        }
                        grid[nx][ny] = 0;
                        q.push({nx, ny});
                    }
                }
            }
        }
        sort(pq.begin(), pq.end());
        vector<vector<int>> ans;
        for (int i = 0; i < min(k, (int) pq.size()); ++i) {
            ans.push_back({get<2>(pq[i]), get<3>(pq[i])});
        }
        return ans;
    }
};

func highestRankedKItems(grid [][]int, pricing []int, start []int, k int) (ans [][]int) {
    m, n := len(grid), len(grid[0])
    row, col := start[0], start[1]
    low, high := pricing[0], pricing[1]
    q := [][2]int{{row, col}}
    pq := [][]int{}
    if low <= grid[row][col] && grid[row][col] <= high {
        pq = append(pq, []int{0, grid[row][col], row, col})
    }
    grid[row][col] = 0
    dirs := [5]int{-1, 0, 1, 0, -1}
    for step := 1; len(q) > 0; step++ {
        for sz := len(q); sz > 0; sz-- {
            x, y := q[0][0], q[0][1]
            q = q[1:]
            for j := 0; j < 4; j++ {
                nx, ny := x+dirs[j], 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 = append(pq, []int{step, grid[nx][ny], nx, ny})
                    }
                    grid[nx][ny] = 0
                    q = append(q, [2]int{nx, ny})
                }
            }
        }
    }
    sort.Slice(pq, func(i, j int) bool {
        a, b := pq[i], pq[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]
    })
    for i := 0; i < len(pq) && i < k; i++ {
        ans = append(ans, pq[i][2:])
    }
    return
}

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;
}

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