What is the time complexity of Minimized Maximum of Products Distributed to Any Store?

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

What data structures are used to solve Minimized Maximum of Products Distributed to Any Store?

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

What are the key insights for solving Minimized Maximum of Products Distributed to Any Store?

Insight 1: The problem asks to minimize the maximum value, which hints towards using binary search. Insight 2: The key to the binary search condition is checking if a given 'mid' value (potential minimized maximum) allows us to distribute all products among the 'n' stores.

Which companies ask LeetCode 2064. Minimized Maximum of Products Distributed to Any Store in interviews?

This problem is frequently asked at Siemens.

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Minimized Maximum of Products Distributed to Any Store - LeetCode 2064 Solution

Minimized Maximum of Products Distributed to Any Store - Complete Solution Guide

Minimized Maximum of Products Distributed to Any Store is LeetCode problem 2064, 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 an integer n indicating there are n specialty retail stores. There are m product types of varying amounts, which are given as a 0-indexed integer array quantities , where quantities[i] represents the number of products of the i th product type. You need to distribute all products to the retail stores following these rules: A store can only be given at most one product type but can be given any amount of it. After distribution, each store will have been given some number of products

Detailed Explanation

The problem requires us to distribute 'm' different product types (given by the 'quantities' array) among 'n' retail stores. Each store can only receive products of one type, but any number of them. The goal is to minimize the maximum number of products any single store receives. Essentially, we want to find the smallest possible value 'x' such that we can distribute all products, and no store receives more than 'x' products.

Solution Approach

The solution utilizes binary search to find the minimized maximum number of products a store can receive. We initialize the search range from 1 (minimum possible products) to the maximum quantity in the 'quantities' array (maximum possible products). In each iteration, we calculate the 'mid' value and determine if it's a feasible solution. We iterate through the 'quantities' array, calculating the number of stores needed for each product type given the current 'mid' value. We sum the stores needed and compare it with the total number of stores 'n'. If the stores needed are less than or equal to 'n', it means 'mid' is a possible solution (or can be improved), and we search in the lower half. Otherwise, 'mid' is too small, and we search in the upper half.

Step-by-Step Algorithm

Step 1: Initialize 'low' to 1 (minimum possible maximized value) and 'high' to the maximum value in the 'quantities' array (maximum possible maximized value).
Step 2: While 'low' is less than 'high', calculate the middle value 'mid' as 'low + (high - low) // 2' to prevent potential overflow.
Step 3: Initialize 'stores_needed' to 0.
Step 4: Iterate through the 'quantities' array. For each quantity 'q', calculate the number of stores needed as '(q + mid - 1) // mid' and add it to 'stores_needed'. This is equivalent to ceiling division.
Step 5: If 'stores_needed' is less than or equal to 'n', it means 'mid' is a feasible maximized value. Therefore, update 'high' to 'mid' to search for a smaller maximized value.
Step 6: Otherwise, if 'stores_needed' is greater than 'n', it means 'mid' is not a feasible maximized value (too small). Therefore, update 'low' to 'mid + 1' to search for a larger maximized value.
Step 7: After the loop finishes (when 'low' equals 'high'), 'low' will contain the minimized maximum value. Return 'low'.

Key Insights

Insight 1: The problem asks to minimize the maximum value, which hints towards using binary search.
Insight 2: The key to the binary search condition is checking if a given 'mid' value (potential minimized maximum) allows us to distribute all products among the 'n' stores.
Insight 3: The number of stores needed for each product type can be calculated by dividing the quantity of the product type by 'mid' and rounding up (ceiling division): `(q + mid - 1) // mid` or `(q + mid - 1) / mid`

Complexity Analysis

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

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Greedy.

Companies

Asked at: Siemens.

Minimized Maximum of Products Distributed to Any Store | LeetCode 2064

Medium

Array Binary Search Greedy

Time: O(m*log(M))Space: O(1)

Solution Code

Available in:

class Solution {
    public int minimizedMaximum(int n, int[] quantities) {
        int low = 1;
        int high = 0;
        for (int q : quantities) {
            high = Math.max(high, q);
        }

        while (low < high) {
            int mid = low + (high - low) / 2;

            int stores_needed = 0;
            for (int q : quantities) {
                stores_needed += (q + mid - 1) / mid;
            }

            if (stores_needed <= n) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }

        return low;
    }
}

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Problem Info

Description

Problem 2064: Minimized Maximum of Products Distributed to Any Store - Detailed explanation and solution approaches.

Topics

Array Binary Search Greedy

Companies

Siemens

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2064

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 'low' to 1 (minimum possible maximized value) and 'high' to the maximum value in the 'quantities' array (maximum possible maximized value).
2
Step 2: While 'low' is less than 'high', calculate the middle value 'mid' as 'low + (high - low) // 2' to prevent potential overflow.
3
Step 3: Initialize 'stores_needed' to 0.
4
Step 4: Iterate through the 'quantities' array. For each quantity 'q', calculate the number of stores needed as '(q + mid - 1) // mid' and add it to 'stores_needed'. This is equivalent to ceiling division.
5
Step 5: If 'stores_needed' is less than or equal to 'n', it means 'mid' is a feasible maximized value. Therefore, update 'high' to 'mid' to search for a smaller maximized value.
6
Step 6: Otherwise, if 'stores_needed' is greater than 'n', it means 'mid' is not a feasible maximized value (too small). Therefore, update 'low' to 'mid + 1' to search for a larger maximized value.
7
Step 7: After the loop finishes (when 'low' equals 'high'), 'low' will contain the minimized maximum value. Return 'low'.

💡

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*log(M))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The binary search iterates log(M) times, where M is the max value in `quantities`. Inside the binary search loop, we iterate through the `quantities` array of size `m`. Therefore, the overall time complexity is O(m * log(M)). The space complexity is O(1) because we only use a few integer variables, regardless of the input size.

📖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

Iterate through all possible values of 'x' (from 1 to max(quantities)) and for each value, check if it's possible to distribute the products such that no store has more than 'x' products. Choose the minimum 'x' that satisfies this condition.

Time: O(M * m), where M is the max value in quantities and m is the length of the quantities array.Space: O(1)

Trade-offs: This is much slower than the binary search approach, especially when the maximum value in 'quantities' is large. The binary search approach is much more efficient.

✨

Key Insights

1Insight 1: The problem asks to minimize the maximum value, which hints towards using binary search.
2Insight 2: The key to the binary search condition is checking if a given 'mid' value (potential minimized maximum) allows us to distribute all products among the 'n' stores.
3Insight 3: The number of stores needed for each product type can be calculated by dividing the quantity of the product type by 'mid' and rounding up (ceiling division): `(q + mid - 1) // mid` or `(q + mid - 1) / mid`

📚

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!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: n = 1, quantities = [100000]. The algorithm correctly returns 100000, as all products must be assigned to the single store.
Edge case 2: quantities is empty. The algorithm still works because max() will handle this condition if properly initialized (high would initialize to 0). Although problem constraints specify m >= 1, it's a good practice.
Edge case 3: When the 'quantities' array has very large values that could potentially cause integer overflow during the calculation of `(q + mid - 1)`, the algorithm will still work as the operations in this problem will not likely cause the sum to exceed the maximum integer value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the ceiling division. It's essential to use '(q + mid - 1) // mid' to correctly calculate the number of stores needed.

✗Common mistake 2: Not initializing the 'low' and 'high' bounds correctly. 'low' should start from 1, and 'high' should be the maximum value in 'quantities'.

✗Common mistake 3: Integer overflow during the calculation of `mid = (low + high) / 2`. Use `mid = low + (high - low) // 2` to prevent it.

✅

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: Task scheduling where you need to distribute tasks (quantities) to workers (stores) to minimize the workload (maximum number of tasks) on any single worker.

🏦Real-world scenario 2: Resource allocation where you have different resources (quantities) to allocate among different servers (stores) to minimize the load (maximum resource usage) on any server.

💡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.

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 24, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

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

@DevExpert92Jul 7

❓

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 "Minimized Maximum of Products Distributed to Any Store"?

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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