What is the time complexity of Most Beautiful Item for Each Query?

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

What data structures are used to solve Most Beautiful Item for Each Query?

This problem uses Array, Binary Search, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Most Beautiful Item for Each Query?

Insight 1: Sorting the items by price allows efficient searching for items within the price limit of each query using binary search. Insight 2: Precomputing the maximum beauty 'so far' for the sorted items optimizes the beauty lookup by ensuring that `items[i][1]` always stores the maximum beauty among items with prices up to `items[i][0]`.

Which companies ask LeetCode 2070. Most Beautiful Item for Each Query in interviews?

This problem is frequently asked at Postmates, razorpay.

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Most Beautiful Item for Each Query - LeetCode 2070 Solution

Most Beautiful Item for Each Query - Complete Solution Guide

Most Beautiful Item for Each Query is LeetCode problem 2070, 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 2D integer array items where items[i] = [price i , beauty i ] denotes the price and beauty of an item respectively. You are also given a 0-indexed integer array queries . For each queries[j] , you want to determine the maximum beauty of an item whose price is less than or equal to queries[j] . If no such item exists, then the answer to this query is 0 . Return an array answer of the same length as queries where answer[j] is the answer to the j th query . Example 1: Input: items =

Detailed Explanation

The problem requires finding the maximum beauty of an item for each query, given a list of items with their prices and beauties. For each query (representing a maximum price), we need to find the item with a price less than or equal to the query and has the maximum beauty among all such items. If no item satisfies the price constraint, the beauty is 0.

Solution Approach

The solution involves sorting the items by their price, calculating a running maximum of the beauty values, and then using binary search to efficiently find the maximum beauty for each query. First, the `items` array is sorted in ascending order based on the `price` of each item. Subsequently, the `beauty` of each item is updated to store the maximum beauty seen so far, up to that item's price. Finally, for each query, binary search is used to find the index of the last item with a price less than or equal to the query. The beauty of that item is then returned as the answer to the query. If no item satisfies the price condition, 0 is returned.

Step-by-Step Algorithm

Step 1: Sort the `items` array in ascending order based on their price (items[i][0]).
Step 2: Iterate through the sorted `items` array and calculate the running maximum of the beauty values. Update each item's beauty (items[i][1]) to be the maximum beauty seen so far, up to that item's price.
Step 3: For each query in the `queries` array:
Step 4: Perform a binary search (using bisect_right) on the sorted list of prices to find the index `idx` of the first item with a price greater than the query. This effectively finds the last item with a price less than or equal to the query.
Step 5: If `idx` is 0, it means no item has a price less than or equal to the query, so append 0 to the result.
Step 6: Otherwise, the maximum beauty for the current query is `items[idx - 1][1]`, so append it to the result.
Step 7: Return the list of results (the `answer` array).

Key Insights

Insight 1: Sorting the items by price allows efficient searching for items within the price limit of each query using binary search.
Insight 2: Precomputing the maximum beauty 'so far' for the sorted items optimizes the beauty lookup by ensuring that `items[i][1]` always stores the maximum beauty among items with prices up to `items[i][0]`.
Insight 3: Using binary search (specifically `bisect_right`) to find the insertion point efficiently determines the last item that satisfies the price condition.

Complexity Analysis

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

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Sorting.

Companies

Asked at: Postmates, razorpay.

Most Beautiful Item for Each Query | LeetCode 2070

Medium

Array Binary Search Sorting

Time: O(n log n + m log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;
import java.util.List;
import java.util.ArrayList;

class Solution {
    public int[] maximumBeauty(int[][] items, int[] queries) {
        Arrays.sort(items, (a, b) -> a[0] - b[0]);
        
        int max_beauty_so_far = 0;
        for (int[] item : items) {
            max_beauty_so_far = Math.max(max_beauty_so_far, item[1]);
            item[1] = max_beauty_so_far;
        }
        
        int[] prices = new int[items.length];
        for (int i = 0; i < items.length; i++) {
            prices[i] = items[i][0];
        }
        
        int[] answer = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int query = queries[i];
            int idx = bisect_right(prices, query);
            if (idx == 0) {
                answer[i] = 0;
            } else {
                answer[i] = items[idx - 1][1];
            }
        }
        
        return answer;
    }
    
    private int bisect_right(int[] arr, int target) {
        int left = 0;
        int right = arr.length;
        
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (arr[mid] <= target) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        
        return left;
    }
}

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Problem Info

Description

Problem 2070: Most Beautiful Item for Each Query - Detailed explanation and solution approaches.

Topics

Array Binary Search Sorting

Companies

Postmatesrazorpay

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2070

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: Sort the `items` array in ascending order based on their price (items[i][0]).
2
Step 2: Iterate through the sorted `items` array and calculate the running maximum of the beauty values. Update each item's beauty (items[i][1]) to be the maximum beauty seen so far, up to that item's price.
3
Step 3: For each query in the `queries` array:
4
Step 4: Perform a binary search (using bisect_right) on the sorted list of prices to find the index `idx` of the first item with a price greater than the query. This effectively finds the last item with a price less than or equal to the query.
5
Step 5: If `idx` is 0, it means no item has a price less than or equal to the query, so append 0 to the result.
6
Step 6: Otherwise, the maximum beauty for the current query is `items[idx - 1][1]`, so append it to the result.
7
Step 7: Return the list of results (the `answer` array).

💡

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(n log n + m log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the most time-consuming operations: sorting and the binary searches. The space complexity is determined by the additional data structures created during the execution of the algorithm, excluding the space needed for the output.

📖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

Brute-Force Approach

Iterate through the items for each query and find the maximum beauty among items with price less than or equal to the query. This involves nested loops.

Time: O(n*m), where n is the number of items and m is the number of queries.Space: O(1)

Trade-offs: Simple to implement but inefficient for large inputs due to the nested loops. The main solution provides a much better time complexity. Brute force is only suitable when n and m are small.

✨

Key Insights

1Insight 1: Sorting the items by price allows efficient searching for items within the price limit of each query using binary search.
2Insight 2: Precomputing the maximum beauty 'so far' for the sorted items optimizes the beauty lookup by ensuring that `items[i][1]` always stores the maximum beauty among items with prices up to `items[i][0]`.
3Insight 3: Using binary search (specifically `bisect_right`) to find the insertion point efficiently determines the last item that satisfies the price condition.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sorting

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

Edge Cases

Edge case 1: If a query is smaller than the smallest price in the `items` array, the solution correctly returns 0 because the binary search `bisect_right` will return 0.
Edge case 2: If all items have the same price, the precomputed maximum beauty ensures that the correct maximum beauty among these items is returned for queries greater than or equal to this price.
Edge case 3: If there are duplicate items with the same price, the running maximum beauty ensures that the item with the highest beauty among these duplicates is considered when processing queries greater than or equal to that price.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the items by price. Binary search relies on sorted data to function correctly, so omitting this step will lead to incorrect results.

✗Common mistake 2: Not calculating the running maximum of beauty values. This leads to returning the beauty of a specific item rather than the maximum beauty among all items with a price less than or equal to the query.

✗Common mistake 3: Using `bisect_left` instead of `bisect_right`. `bisect_right` returns the index of the first element *greater* than the target, while `bisect_left` returns the index of the first element *not less than* the target. Using `bisect_left` would cause incorrect results in cases where there exist items with prices equal to the query value, potentially missing such items when returning maximum beauty.

✅

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: E-commerce product search. Given a list of products with prices and quality ratings (beauty), allow users to specify a maximum price and return the product with the highest quality rating within that price range.

🏦Real-world scenario 2: Financial analysis. Given a list of assets with their cost and return, find the asset with the highest return for a given investment limit.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: August 30, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

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

@AlgoMasterJul 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.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

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 "Most Beautiful Item for Each Query"?

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 binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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