What is the time complexity of Minimum Limit of Balls in a Bag?

The optimal solution has a time complexity of O(nlog(m)).

What data structures are used to solve Minimum Limit of Balls in a Bag?

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

What are the key insights for solving Minimum Limit of Balls in a Bag?

Insight 1: The problem can be solved using binary search. The search space is the possible values of the maximum number of balls in a bag (the 'penalty'). Insight 2: For a given 'penalty' value (the middle value in our binary search), we can determine how many operations are needed to achieve that penalty by iterating through the `nums` array and calculating the number of splits needed for each bag.

Which companies ask LeetCode 1760. Minimum Limit of Balls in a Bag in interviews?

This problem is frequently asked at Flipkart, Intuit.

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Minimum Limit of Balls in a Bag - LeetCode 1760 Solution

Minimum Limit of Balls in a Bag - Complete Solution Guide

Minimum Limit of Balls in a Bag is LeetCode problem 1760, 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 array nums where the i th bag contains nums[i] balls. You are also given an integer maxOperations . You can perform the following operation at most maxOperations times: Take any bag of balls and divide it into two new bags with a positive number of balls. For example, a bag of 5 balls can become two new bags of 1 and 4 balls, or two new bags of 2 and 3 balls. Your penalty is the maximum number of balls in a bag. You want to minimize your penalty after the operations. Ret

Detailed Explanation

The problem asks us to minimize the maximum number of balls in any bag after performing at most `maxOperations` operations. Each operation allows us to split a bag into two new bags with positive integer numbers of balls. The 'penalty' is defined as the maximum number of balls in any bag after performing the operations, and we want to find the minimum possible penalty.

Solution Approach

The solution employs a binary search algorithm. We search for the minimum possible penalty within the range [1, max(nums)]. For each potential penalty value (the `mid` value in binary search), we calculate the number of operations required to reduce the size of each bag to be at most `mid`. If the total operations required is less than or equal to `maxOperations`, it means we can achieve this penalty. Therefore, we try to reduce the penalty further (search in the lower half). Otherwise, we need to increase the penalty (search in the upper half).

Step-by-Step Algorithm

Step 1: Initialize `low` to 1 and `high` to the maximum value in the `nums` array. This defines our search space for the minimum penalty.
Step 2: Perform binary search while `low <= high`.
Step 3: Calculate the middle value `mid = low + (high - low) // 2`. This is our potential penalty value.
Step 4: Calculate the number of operations needed to reduce each bag's size to at most `mid`. This is done by iterating through the `nums` array and summing `(num - 1) // mid` for each `num` in `nums`.
Step 5: If the total operations needed is less than or equal to `maxOperations`, it means we can achieve a penalty of `mid` or less. Update the `ans` to `mid` and narrow the search space to the lower half by setting `high = mid - 1`.
Step 6: If the total operations needed is greater than `maxOperations`, it means the penalty `mid` is too low, and we need to increase it. Narrow the search space to the upper half by setting `low = mid + 1`.
Step 7: After the binary search completes, `ans` will hold the minimum possible penalty.

Key Insights

Insight 1: The problem can be solved using binary search. The search space is the possible values of the maximum number of balls in a bag (the 'penalty').
Insight 2: For a given 'penalty' value (the middle value in our binary search), we can determine how many operations are needed to achieve that penalty by iterating through the `nums` array and calculating the number of splits needed for each bag.
Insight 3: The number of splits needed for a bag with `num` balls to achieve a maximum size of `mid` is `(num - 1) // mid`. This is because if `num` is divisible by `mid`, then `num/mid - 1` splits are needed to create `num/mid` bags each of size `mid`. If `num` is not divisible by `mid`, the same formula applies.
Insight 4: Understanding the ceiling division concept `(num-1)//mid+1` which is equal to `(num+mid-1)//mid`

Complexity Analysis

Time Complexity: O(nlog(m))

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search.

Companies

Asked at: Flipkart, Intuit.

Minimum Limit of Balls in a Bag | LeetCode 1760

Medium

Array Binary Search

Time: O(nlog(m))Space: O(1)

Solution Code

Available in:

class Solution {
    public int minimumSize(int[] nums, int maxOperations) {
        int low = 1;
        int high = 0;
        for (int num : nums) {
            high = Math.max(high, num);
        }
        int ans = high;

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

            if (mid == 0) {
                low = 1;
                continue;
            }

            int ops = 0;
            for (int num : nums) {
                ops += (num - 1) / mid;
            }

            if (ops <= maxOperations) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }

        return ans;
    }
}

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Problem Info

Description

Problem 1760: Minimum Limit of Balls in a Bag - Detailed explanation and solution approaches.

Topics

Array Binary Search

Companies

FlipkartIntuit

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1760

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 and `high` to the maximum value in the `nums` array. This defines our search space for the minimum penalty.
2
Step 2: Perform binary search while `low <= high`.
3
Step 3: Calculate the middle value `mid = low + (high - low) // 2`. This is our potential penalty value.
4
Step 4: Calculate the number of operations needed to reduce each bag's size to at most `mid`. This is done by iterating through the `nums` array and summing `(num - 1) // mid` for each `num` in `nums`.
5
Step 5: If the total operations needed is less than or equal to `maxOperations`, it means we can achieve a penalty of `mid` or less. Update the `ans` to `mid` and narrow the search space to the lower half by setting `high = mid - 1`.
6
Step 6: If the total operations needed is greater than `maxOperations`, it means the penalty `mid` is too low, and we need to increase it. Narrow the search space to the upper half by setting `low = mid + 1`.
7
Step 7: After the binary search completes, `ans` will hold the minimum possible penalty.

💡

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(nlog(m))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The binary search takes O(log(m)) time, where m is the maximum value in `nums`. Inside the binary search loop, we iterate through the `nums` array, which takes O(n) time. Therefore, the overall time complexity is O(nlog(m)). The space complexity is O(1) because we only use a few integer variables.

📖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

Linear Search

We could iterate through the possible penalty values from 1 to max(nums) and check if each value is achievable within `maxOperations`. However, this approach would be significantly slower.

Time: O(n*m), where n is the length of nums and m is the maximum element in nums.Space: O(1)

Trade-offs: This is much simpler to implement, but it is far less efficient than binary search, especially when max(nums) is large. Avoid this for larger input sizes.

✨

Key Insights

1Insight 1: The problem can be solved using binary search. The search space is the possible values of the maximum number of balls in a bag (the 'penalty').
2Insight 2: For a given 'penalty' value (the middle value in our binary search), we can determine how many operations are needed to achieve that penalty by iterating through the `nums` array and calculating the number of splits needed for each bag.
3Insight 3: The number of splits needed for a bag with `num` balls to achieve a maximum size of `mid` is `(num - 1) // mid`. This is because if `num` is divisible by `mid`, then `num/mid - 1` splits are needed to create `num/mid` bags each of size `mid`. If `num` is not divisible by `mid`, the same formula applies.
4Insight 4: Understanding the ceiling division concept `(num-1)//mid+1` which is equal to `(num+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!

Edge Cases

Edge case 1: `nums` is empty. The problem statement states that `1 <= nums.length`, so this case doesn't need to be explicitly handled. However, in a real-world scenario, handling an empty `nums` array with a check like `if not nums: return 0` would be prudent.
Edge case 2: `maxOperations` is 0. In this case, no operations can be performed, and the minimum penalty is simply the maximum value in `nums`. The code handles this correctly because if we start with a `mid` and operations exceed the limit, we increase `low`, and binary search effectively determines that max(nums) is the correct value.
Edge case 3: All elements in `nums` are already small (e.g., nums = [1,1,1]). The binary search still works correctly because the number of required operations will be 0 for any `mid` greater than or equal to 1, and thus low will converge on 1.
Edge case 4: When calculating operations needed to split the bags, using `(num-1)//mid` avoids division by zero if mid is larger than num.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using binary search. A linear search approach might seem simpler at first but will lead to a Time Limit Exceeded error for larger inputs.

✗Common mistake 2: Integer overflow when calculating `mid`. Using `mid = low + (high - low) // 2` instead of `mid = (low + high) // 2` prevents integer overflow issues, especially in languages like Java and C++.

✗Common mistake 3: Incorrectly calculating the number of operations needed for each bag. The formula `(num - 1) // mid` is crucial for calculating the minimum number of operations to achieve a maximum bag size of `mid`.

✗Common mistake 4: Incorrect initialization of `low` or `high`. Setting low to 0 can lead to division by zero errors within the loop. `low` should always be initialized to 1.

✅

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: Imagine dividing tasks (represented by `nums`) among workers (`maxOperations`). We want to minimize the maximum workload any worker has to handle.

🏦Real-world scenario 2: Resource allocation: Consider distributing resources (e.g., memory or storage) among different processes. We want to minimize the maximum resource usage by any single process after splitting large chunks into smaller ones.

🎮Real-world scenario 3: Bandwidth allocation: In network management, dividing large data streams (represented by nums) across multiple channels (maxOperations) to minimize the peak bandwidth usage on any single channel.

📱Real-world scenario 4: Cutting materials: Cutting raw materials of different lengths into pieces, subject to a constraint of the number of cuts that can be made. The goal is to minimize the length of the longest piece.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: October 19, 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 "Minimum Limit of Balls in a Bag"?

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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