How do you solve K Highest Ranked Items Within a Price Range on LeetCode?

K Highest Ranked Items Within a Price Range (LeetCode #2146) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using BFS allows us to explore the grid efficiently while keeping track of distances.

What is the time complexity of K Highest Ranked Items Within a Price Range?

The optimal solution for K Highest Ranked Items Within a Price Range runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step after collecting items. The space complexity is O(n) because we store all reachable items in a list.

What is the brute force solution for K Highest Ranked Items Within a Price Range?

The brute force approach for K Highest Ranked Items Within a Price Range is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all reachable cells from the starting position and check their prices. This method is straightforward but inefficient as it involves checking each cell individually. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for K Highest Ranked Items Within a Price Range?

Clearly explain your thought process while coding. Discuss edge cases, such as when there are no reachable items. Practice BFS and sorting problems to become comfortable with these techniques.

What are common mistakes when solving K Highest Ranked Items Within a Price Range?

Not handling walls (0s) correctly during BFS traversal. Forgetting to check if the item price is within the specified range.

#2146

K Highest Ranked Items Within a Price Range

Medium

ArrayBreadth-First SearchSortingHeap (Priority Queue)MatrixBreadth-First SearchSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

Using BFS allows us to efficiently explore the grid while keeping track of distances and prices. We can prioritize items based on distance and price without needing to check every cell multiple times.

⚙️

Algorithm

4 steps

1Step 1: Perform a BFS from the starting position to find all reachable cells and their distances.
2Step 2: Filter the reachable items based on the price range [low, high].
3Step 3: Sort the items based on the ranking criteria: distance, price, row, and column.
4Step 4: Return the top k items.

solution.py22 lines

1from collections import deque
2
3def highestRankedItems(grid, pricing, start, k):
4    m, n = len(grid), len(grid[0])
5    low, high = pricing
6    directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
7    queue = deque([start])
8    visited = set([tuple(start)])
9    items = []
10
11    while queue:
12        x, y = queue.popleft()
13        if grid[x][y] != 0 and low <= grid[x][y] <= high:
14            items.append((grid[x][y], x, y))
15        for dx, dy in directions:
16            nx, ny = x + dx, y + dy
17            if 0 <= nx < m and 0 <= ny < n and (nx, ny) not in visited and grid[nx][ny] != 0:
18                visited.add((nx, ny))
19                queue.append((nx, ny))
20
21    items.sort(key=lambda item: (item[1], item[2], item[0]))
22    return items[:k]

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step after collecting items. The space complexity is O(n) because we store all reachable items in a list.

1Using BFS allows us to explore the grid efficiently while keeping track of distances.
2Sorting based on multiple criteria can be done using custom sort functions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.