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

The optimal solution has a time complexity of O(m*n*log(m*n)).

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

The space complexity is O(m*n).

What data structures are used to solve K Highest Ranked Items Within a Price Range?

This problem uses Array, Breadth-First Search, Sorting, Heap (Priority Queue), Matrix. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving K Highest Ranked Items Within a Price Range?

Insight 1: Breadth-First Search (BFS) is essential for finding the shortest distance from the starting position to all other cells in the grid. Insight 2: We need to maintain a list of items within the price range and sort them based on the specified ranking criteria (distance, price, row, column).

Which companies ask LeetCode 2146. K Highest Ranked Items Within a Price Range in interviews?

This problem is frequently asked at Booking.com.

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K Highest Ranked Items Within a Price Range - LeetCode 2146 Solution

K Highest Ranked Items Within a Price Range - Complete Solution Guide

K Highest Ranked Items Within a Price Range is LeetCode problem 2146, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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. 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 an

Detailed Explanation

The problem requires us to find the `k` highest-ranked items in a shop represented by a 2D grid, given a price range `[low, high]` and a starting position `[row, col]`. The rank of an item is determined by distance from the starting position (shortest path), price within the given range, row number, and column number, in that order of priority. We need to return the positions (row, column) of the top `k` items sorted by their rank. If there are fewer than `k` reachable items within the price range, return all of them.

Solution Approach

The solution utilizes Breadth-First Search (BFS) to traverse the grid from the starting position. During the BFS, we keep track of the distance from the starting point to each cell. When we encounter a cell with a price within the specified range, we add it to a list along with its distance. After each level of BFS, we sort the list of items found at that level based on the ranking criteria (distance, price, row, column) and append them to the result. If the size of the result becomes greater than or equal to k, we terminate the BFS and return the top k items.

Step-by-Step Algorithm

Step 1: Initialize a queue for BFS and a set to keep track of visited cells. Add the starting position to the queue and the visited set.
Step 2: Perform BFS. In each level, iterate through all the nodes in the current level of the queue.
Step 3: For each cell in the current level, check if its price falls within the given range `[low, high]`. If it does, store the cell's price, row, and column along with its distance from the start.
Step 4: Explore the neighbors of the current cell (up, down, left, right) that are within the grid boundaries, are not walls (value 0), and have not been visited yet. Add these neighbors to the queue and the visited set.
Step 5: After processing all cells in a level, sort the items found at that level based on the ranking criteria: distance (ascending), price (ascending), row (ascending), and column (ascending).
Step 6: Add the sorted items to the result list. If the size of the result list becomes greater than or equal to `k`, break the BFS loop.
Step 7: If fewer than `k` reachable items within the price range were found, return all of them. Otherwise, return the first `k` elements of the result list.

Key Insights

Insight 1: Breadth-First Search (BFS) is essential for finding the shortest distance from the starting position to all other cells in the grid.
Insight 2: We need to maintain a list of items within the price range and sort them based on the specified ranking criteria (distance, price, row, column).
Insight 3: Early termination of the BFS is possible once we have found `k` or more items within the price range to improve efficiency.

Complexity Analysis

Time Complexity: O(m*n*log(m*n))

Space Complexity: O(m*n)

Topics

This problem involves: Array, Breadth-First Search, Sorting, Heap (Priority Queue), Matrix.

Companies

Asked at: Booking.com.

K Highest Ranked Items Within a Price Range | LeetCode 2146

Medium

Array Breadth-First Search Sorting Heap (Priority Queue)Matrix

Time: O(m*n*log(m*n))Space: O(m*n)

Solution Code

Available in:

import java.util.*;

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 low = pricing[0];
        int high = pricing[1];
        int start_r = start[0];
        int start_c = start[1];

        Queue<int[]> queue = new LinkedList<>();
        Set<String> visited = new HashSet<>();

        queue.offer(new int[]{start_r, start_c});
        visited.add(start_r + "," + start_c);

        List<List<Integer>> result = new ArrayList<>();

        int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

        int distance = 0;
        while (!queue) {
            int level_size = queue.size();
            List<int[]> items_at_this_level = new ArrayList<>();

            for (int i = 0; i < level_size; i++) {
                int[] curr = queue.poll();
                int r = curr[0];
                int c = curr[1];

                int price = grid[r][c];
                if (low <= price && price <= high) {
                    items_at_this_level.add(new int[]{distance, price, r, c});
                }

                for (int[] dir : directions) {
                    int nr = r + dir[0];
                    int nc = c + dir[1];

                    if (0 <= nr && nr < m && 0 <= nc && nc < n && grid[nr][nc] != 0 && !visited.contains(nr + "," + nc)) {
                        visited.add(nr + "," + nc);
                        queue.offer(new int[]{nr, nc});
                    }
                }
            }
            distance++;
            
            Collections.sort(items_at_this_level, (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];
            });

            for (int[] item : items_at_this_level) {
                List<Integer> item_list = new ArrayList<>();
                item_list.add(item[2]);
                item_list.add(item[3]);
                result.add(item_list);
            }

            if (result.size() >= k) {
                break;
            }
        }

        return result.subList(0, Math.min(k, result.size()));
    }
}

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Problem Info

Description

Problem 2146: K Highest Ranked Items Within a Price Range - Detailed explanation and solution approaches.

Topics

Array Breadth-First Search Sorting Heap (Priority Queue)Matrix

Companies

Booking.com

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2146

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize a queue for BFS and a set to keep track of visited cells. Add the starting position to the queue and the visited set.
2
Step 2: Perform BFS. In each level, iterate through all the nodes in the current level of the queue.
3
Step 3: For each cell in the current level, check if its price falls within the given range `[low, high]`. If it does, store the cell's price, row, and column along with its distance from the start.
4
Step 4: Explore the neighbors of the current cell (up, down, left, right) that are within the grid boundaries, are not walls (value 0), and have not been visited yet. Add these neighbors to the queue and the visited set.
5
Step 5: After processing all cells in a level, sort the items found at that level based on the ranking criteria: distance (ascending), price (ascending), row (ascending), and column (ascending).
6
Step 6: Add the sorted items to the result list. If the size of the result list becomes greater than or equal to `k`, break the BFS loop.
7
Step 7: If fewer than `k` reachable items within the price range were found, return all of them. Otherwise, return the first `k` elements of the result list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n*log(m*n))

💾

Space Complexity

O(m*n)

📝

Why?

BFS explores the grid in a breadth-first manner, adding cells to the queue and marking them as visited. The worst-case scenario for BFS is visiting all cells in the grid, leading to O(m*n) time for traversal alone. The sorting operation in each level adds a factor of log(m*n) because the maximum number of items within the price range found at each level is O(m*n) so this step results in a time of O(m*n*log(m*n)). The space complexity is primarily determined by the queue, visited set, and the result list which can hold at most all the grid cells and at most `k` results with `k` <= `m*n`.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dijkstra's Algorithm

Dijkstra's algorithm could be used to find the shortest path to each cell. This approach is very similar to the BFS approach and is applicable because all edge weights are equal to 1. However, the BFS solution will be faster because it doesn't require maintaining a priority queue since all edges have the same cost.

Time: O(m*n*log(m*n)) with a priority queue or O((m*n)^2) with a simple array implementation. In this case, using a priority queue gives same complexity as current solution.Space: O(m*n)

Trade-offs: While Dijkstra's algorithm is a general-purpose shortest path algorithm, BFS is more efficient when all edge weights are equal, as it avoids the overhead of maintaining a priority queue.

✨

Key Insights

1Insight 1: Breadth-First Search (BFS) is essential for finding the shortest distance from the starting position to all other cells in the grid.
2Insight 2: We need to maintain a list of items within the price range and sort them based on the specified ranking criteria (distance, price, row, column).
3Insight 3: Early termination of the BFS is possible once we have found `k` or more items within the price range to improve efficiency.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Starting position is a wall (grid[row][col] == 0). The problem states that grid[row][col] > 0, so this is not an actual edge case to handle in the code itself, but an invalid input.
Edge case 2: No items within the price range are reachable from the starting position. The algorithm will return an empty list, which is the expected behavior.
Edge case 3: Fewer than `k` items are reachable within the price range. The algorithm will return all reachable items, which is the expected behavior.
Edge case 4: The start node itself fulfills the item criteria. The solution correctly processes the starting node and adds it to the result if its price falls within the specified range.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the ranking criteria, leading to incorrect sorting of the items. Ensure that the sorting is performed in the correct order of priority: distance, price, row, column.

✗Common mistake 2: Not checking for visited cells, leading to infinite loops and incorrect distance calculations. Always maintain a visited set to avoid revisiting cells.

✗Common mistake 3: Not handling the case where fewer than `k` items are reachable within the price range. The algorithm should return all reachable items in this case.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Recommendation systems where items need to be ranked based on distance, price, and other criteria from the user's current location.

🏦Real-world scenario 2: Navigation apps that suggest nearby points of interest (e.g., restaurants, shops) based on distance, rating, and price range.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 14, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "K Highest Ranked Items Within a Price Range"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?