Approach
Breadth-first search
For K Highest Ranked Items Within a Price Range, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.
- Model each valid configuration as a state and each legal move as an edge.
- Seed the queue with the starting state and mark it immediately.
- Expand each state once, recording distance or reachability for unseen neighbours.
Code notes
- 47 lines of Python from the credited upstream file 2146.py.
- The implementation visibly relies on sequence storage, work queue.
- No explicit loop blocks detected.
Complexity
Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.
Check the problem constraints before deciding whether this complexity will pass.
Use this to learn the idea, then write your own version.
1class Solution:2 def highestRankedKItems(3 self,4 grid: list[list[int]],5 pricing: list[int],6 start: list[int],7 k: int8 ) -> list[list[int]]:9 DIRS = ((0, 1), (1, 0), (0, -1), (-1, 0))10 m = len(grid)11 n = len(grid[0])12 low, high = pricing13 row, col = start14 ans = []15 16 if low <= grid[row][col] <= high:17 ans.append([row, col])18 if k == 1:19 return ans20 21 q = collections.deque([(row, col)])22 seen = {(row, col)} 23 24 while q:25 neighbors = []26 for _ in range(len(q)):27 i, j = q.popleft()28 for t in range(4):29 x = i + DIRS[t][0]30 y = j + DIRS[t][1]31 if x < 0 or x == m or y < 0 or y == n:32 continue33 if not grid[x][y] or (x, y) in seen:34 continue35 if low <= grid[x][y] <= high:36 neighbors.append([x, y])37 q.append((x, y))38 seen.add((x, y))39 neighbors.sort(key=lambda x: (grid[x[0]][x[1]], x[0], x[1]))40 for neighbor in neighbors:41 if len(ans) < k:42 ans.append(neighbor)43 if len(ans) == k:44 return ans45 46 return ans47