Leetcode #2146: K Highest Ranked Items Within a Price Range
In this guide, we solve Leetcode #2146 K Highest Ranked Items Within a Price Range in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Breadth-First Search, Array, Matrix, Sorting, Heap (Priority Queue)
Intuition
We need to repeatedly access the smallest or largest element as the input changes.
A heap provides fast insertions and removals while keeping order.
Approach
Push candidates into the heap as you scan, and pop when you need the best element.
Keep the heap size bounded if the problem requires a top-k structure.
Steps:
- Push candidates into a heap.
- Pop the best candidate when needed.
- Maintain heap size or invariants.
Example
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).
Python Solution
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]]
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.