# K highest ranked items within a price range

January 2, 2023

heap graph

Problem URL: K highest ranked items within a price range

We will use a heap to store the top `k` items. We will iterate over the items and for each item we will check if the price is within the price range. If it is, we will check if the heap size is less than `k`. If it is, we will push the item to the heap. Otherwise, we will check if the price of the item is greater than the price of the item at the top of the heap. If it is, we will pop the item at the top of the heap and push the current item to the heap. Finally, we will return the items in the heap.

``````class Solution:
def highestRankedKItems(self, grid: List[List[int]], pricing: List[int], start: List[int], k: int) -> List[List[int]]:
ROWS, COLS = len(grid), len(grid[0])

queue, res = [], []
heapq.heappush(queue, (0, 0, start[0], start[1])) # dist, price, row, col
seen = set()

while queue and k > 0:
dist, price, cx, cy = heapq.heappop(queue)

if (cx, cy) in seen:
continue

if pricing[0] <= grid[cx][cy] <= pricing[1]:
k -= 1
res.append([cx, cy])

for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
nx = dx + cx
ny = dy + cy

if nx < 0 or nx >= ROWS or ny < 0 or ny >= COLS or grid[nx][ny] == 0 or (nx, ny) in seen:
continue

heapq.heappush(queue, (dist+1,  grid[nx][ny], nx, ny))

return res
``````

Time complexity: `O(mnlog(mn))`, m is the number of rows and n is the number of columns in the grid
Space complexity: `O(mn)`