What is the time complexity of Distribute Candies Among Children II?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Distribute Candies Among Children II?

This problem uses Math, Combinatorics, Enumeration. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Distribute Candies Among Children II?

Insight 1: The core problem is a combinatorial problem of distributing identical items (candies) among distinct recipients (children). Insight 2: Directly iterating through all possible combinations will be inefficient due to the large input range (n <= 10^6).

Which companies ask LeetCode 2929. Distribute Candies Among Children II in interviews?

This problem is frequently asked at Rubrik, ZS Associates.

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Distribute Candies Among Children II - LeetCode 2929 Solution

Q: How to solve Distribute Candies Among Children II?

The provided solution employs the Inclusion-Exclusion Principle. First, it calculates the total number of ways to distribute 'n' candies among 3 children *without* any limits. Then, it subtracts the number of ways where at least one child receives more than 'limit' candies. After that, it adds back the number of ways where at least two children receive more than 'limit' candies (since these were subtracted twice), and finally subtracts the number of ways where all three children receive more than 'limit' candies (since these were added back incorrectly). This ensures a correct count of valid distributions.

Q: Which companies ask LeetCode 2929. Distribute Candies Among Children II in interviews?

This problem is frequently asked at Rubrik, ZS Associates.

Distribute Candies Among Children II - Complete Solution Guide

Distribute Candies Among Children II is LeetCode problem 2929, 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 two positive integers n and limit . Return the total number of ways to distribute n candies among 3 children such that no child gets more than limit candies. Example 1: Input: n = 5, limit = 2 Output: 3 Explanation: There are 3 ways to distribute 5 candies such that no child gets more than 2 candies: (1, 2, 2), (2, 1, 2) and (2, 2, 1). Example 2: Input: n = 3, limit = 3 Output: 10 Explanation: There are 10 ways to distribute 3 candies such that no child gets more than 3 candies: (0

Detailed Explanation

The problem asks us to find the number of ways to distribute 'n' identical candies among 3 children, with the constraint that each child can receive at most 'limit' candies. We need to count all possible non-negative integer solutions (x, y, z) to the equation x + y + z = n, where 0 <= x <= limit, 0 <= y <= limit, and 0 <= z <= limit. Essentially, we are counting valid combinations of candies given to the three children that sum up to 'n' without exceeding 'limit' for any child.

Solution Approach

The provided solution employs the Inclusion-Exclusion Principle. First, it calculates the total number of ways to distribute 'n' candies among 3 children *without* any limits. Then, it subtracts the number of ways where at least one child receives more than 'limit' candies. After that, it adds back the number of ways where at least two children receive more than 'limit' candies (since these were subtracted twice), and finally subtracts the number of ways where all three children receive more than 'limit' candies (since these were added back incorrectly). This ensures a correct count of valid distributions.

Step-by-Step Algorithm

Step 1: Calculate the total number of ways to distribute 'n' candies among 3 children without any limits. This is equivalent to finding the number of non-negative integer solutions to x + y + z = n, which is C(n + 3 - 1, 3 - 1) = C(n + 2, 2) = (n+2)*(n+1)/2.
Step 2: Calculate the number of ways where at least one child receives more than 'limit' candies. Let's say child 1 receives more than 'limit' candies. Then, x > limit, so let x' = x - (limit + 1). Now, x' + y + z = n - (limit + 1). The number of non-negative integer solutions to this is C(n - limit - 1 + 2, 2) = C(n - limit + 1, 2). Since any of the three children could receive more than 'limit' candies, we multiply this by 3: 3 * C(n - limit + 1, 2). We subtract this from the initial count.
Step 3: Calculate the number of ways where at least two children receive more than 'limit' candies. Let's say children 1 and 2 receive more than 'limit' candies. Then x > limit and y > limit. Let x' = x - (limit + 1) and y' = y - (limit + 1). Then x' + y' + z = n - 2*(limit + 1). The number of non-negative integer solutions is C(n - 2*(limit + 1) + 2, 2) = C(n - 2*limit, 2). Since any two of the three children could receive more than 'limit' candies, we multiply this by 3: 3 * C(n - 2*limit, 2). We add this to the running count.
Step 4: Calculate the number of ways where all three children receive more than 'limit' candies. Then x > limit, y > limit, and z > limit. Let x' = x - (limit + 1), y' = y - (limit + 1), and z' = z - (limit + 1). Then x' + y' + z = n - 3*(limit + 1). The number of non-negative integer solutions is C(n - 3*(limit + 1) + 2, 2) = C(n - 3*limit - 1, 2). We subtract this from the running count.
Step 5: The final result is the running count after applying the Inclusion-Exclusion Principle.

Key Insights

Insight 1: The core problem is a combinatorial problem of distributing identical items (candies) among distinct recipients (children).
Insight 2: Directly iterating through all possible combinations will be inefficient due to the large input range (n <= 10^6).
Insight 3: The problem is efficiently solved using the Inclusion-Exclusion Principle to handle the 'limit' constraint. This allows us to subtract invalid cases from the total possible combinations without limits.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Math, Combinatorics, Enumeration.

Companies

Asked at: Rubrik, ZS Associates.

Distribute Candies Among Children II | LeetCode 2929

Medium

Math Combinatorics Enumeration

Time: O(1)Space: O(1)

Solution Code

Available in:

class Solution {
    public long distributeCandies(int n, int limit) {
        
        // Helper function to calculate combinations (n choose 2)
        function combinations2(k: int) -> int:
            if k < 2:
                return 0
            return k * (k - 1) // 2

        private long combinations2(long k) {
            if (k < 2) {
                return 0;
            }
            return k * (k - 1) / 2;
        }

        // Using Inclusion-Exclusion Principle
        
        // Case 0: No constraints (total ways to distribute n candies to 3 children)
        // This is C(n + 3 - 1, 3 - 1) = C(n + 2, 2)
        long ans = combinations2(n + 2);
        
        // Case 1: At least one child gets more than 'limit' candies
        // Subtract C(3, 1) * C(n - (limit + 1) + 2, 2)
        ans -= 3 * combinations2(n - limit -1 + 2);
        
        // Case 2: At least two children get more than 'limit' candies
        // Add back C(3, 2) * C(n - 2*(limit + 1) + 2, 2)
        ans += 3 * combinations2(n - 2 * (limit + 1) + 2);
        
        // Case 3: All three children get more than 'limit' candies
        // Subtract C(3, 3) * C(n - 3*(limit + 1) + 2, 2)
        ans -= combinations2(n - 3 * (limit + 1) + 2);
        
        return ans;
    }
}

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Problem Info

Description

Problem 2929: Distribute Candies Among Children II - Detailed explanation and solution approaches.

Topics

Math Combinatorics Enumeration

Companies

RubrikZS Associates

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2929

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

5 steps

1
Step 1: Calculate the total number of ways to distribute 'n' candies among 3 children without any limits. This is equivalent to finding the number of non-negative integer solutions to x + y + z = n, which is C(n + 3 - 1, 3 - 1) = C(n + 2, 2) = (n+2)*(n+1)/2.
2
Step 2: Calculate the number of ways where at least one child receives more than 'limit' candies. Let's say child 1 receives more than 'limit' candies. Then, x > limit, so let x' = x - (limit + 1). Now, x' + y + z = n - (limit + 1). The number of non-negative integer solutions to this is C(n - limit - 1 + 2, 2) = C(n - limit + 1, 2). Since any of the three children could receive more than 'limit' candies, we multiply this by 3: 3 * C(n - limit + 1, 2). We subtract this from the initial count.
3
Step 3: Calculate the number of ways where at least two children receive more than 'limit' candies. Let's say children 1 and 2 receive more than 'limit' candies. Then x > limit and y > limit. Let x' = x - (limit + 1) and y' = y - (limit + 1). Then x' + y' + z = n - 2*(limit + 1). The number of non-negative integer solutions is C(n - 2*(limit + 1) + 2, 2) = C(n - 2*limit, 2). Since any two of the three children could receive more than 'limit' candies, we multiply this by 3: 3 * C(n - 2*limit, 2). We add this to the running count.
4
Step 4: Calculate the number of ways where all three children receive more than 'limit' candies. Then x > limit, y > limit, and z > limit. Let x' = x - (limit + 1), y' = y - (limit + 1), and z' = z - (limit + 1). Then x' + y' + z = n - 3*(limit + 1). The number of non-negative integer solutions is C(n - 3*(limit + 1) + 2, 2) = C(n - 3*limit - 1, 2). We subtract this from the running count.
5
Step 5: The final result is the running count after applying the Inclusion-Exclusion Principle.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The solution performs a fixed number of arithmetic operations regardless of the input size 'n' and 'limit'. The `combinations2` function and the main `distributeCandies` function involve only basic arithmetic, hence O(1) time complexity. It uses a fixed number of variables, leading to O(1) space complexity.

📖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

Dynamic Programming

A dynamic programming approach could be used to solve this problem. We create a dp table where dp[i][j] stores the number of ways to distribute 'i' candies among 'j' children, with the limit constraint. However, DP is not space-efficient.

Time: O(n*3*limit) = O(n*limit)Space: O(n*3) = O(n)

Trade-offs: DP is less efficient than the Inclusion-Exclusion Principle because of the nested for loops. The Inclusion-Exclusion approach is preferable because it offers O(1) time complexity.

✨

Key Insights

1Insight 1: The core problem is a combinatorial problem of distributing identical items (candies) among distinct recipients (children).
2Insight 2: Directly iterating through all possible combinations will be inefficient due to the large input range (n <= 10^6).
3Insight 3: The problem is efficiently solved using the Inclusion-Exclusion Principle to handle the 'limit' constraint. This allows us to subtract invalid cases from the total possible combinations without limits.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: n is small (e.g., n = 1, limit = 1). The Inclusion-Exclusion principle still holds, and the calculations will correctly return the number of valid combinations.
Edge case 2: limit is very small compared to n. The algorithm correctly handles situations where multiple children might exceed the limit simultaneously. If n < 0 at any step the combinations2 function will return 0.
Edge case 3: n is smaller than limit. The first calculation C(n+2, 2) accounts for all valid cases where each child could receive 0 to n candies.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the 'limit' constraint and simply calculating C(n + 2, 2). This will overcount the number of valid combinations.

✗Common mistake 2: Incorrectly applying the Inclusion-Exclusion Principle, such as forgetting to add back the cases where at least two children exceed the limit, or subtracting the cases where all three children exceed the limit.

✅

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: Resource allocation among different departments or teams, where there is a constraint on how much each entity can receive. For example, allocating a budget to different teams with a maximum spending limit per team.

🏦Real-world scenario 2: Dividing tasks among employees, where each employee has a maximum capacity or workload limit. The problem becomes determining the number of ways to assign tasks such that no employee is overloaded.

💡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

Always test your solution with edge cases: empty input, single element, and boundary values.

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: July 20, 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 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

Frequently Asked Questions

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Distribute Candies Among Children II"?

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

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