Problem Description

You have n candies that need to be distributed among 3 children. Each child can receive between 0 and limit candies (inclusive). Your task is to find the total number of different ways to distribute all n candies such that no child receives more than limit candies.

For example, if you have 5 candies and the limit is 2, you need to count all valid distributions where:

• Child 1 gets a candies (0 ≤ a ≤ 2)
• Child 2 gets b candies (0 ≤ b ≤ 2)
• Child 3 gets c candies (0 ≤ c ≤ 2)
• And a + b + c = 5

The solution uses combinatorial mathematics with the principle of inclusion-exclusion. The approach treats this as a "stars and bars" problem where we distribute n identical items into 3 distinct groups.

The formula C(n+2, 2) represents all possible ways to distribute n candies among 3 children without any restrictions. This uses the stars and bars method where we need 2 dividers to create 3 groups from n items.

However, we must subtract cases where any child gets more than limit candies. If one child gets more than limit candies, we subtract 3 × C(n-limit+1, 2) (accounting for each of the 3 children potentially exceeding the limit).

Since we might over-subtract cases where two children simultaneously exceed the limit, we add back 3 × C(n-2×limit, 2) using the inclusion-exclusion principle.

The final answer gives the exact count of valid distribution methods.

Intuition

The key insight is recognizing this as a classic "stars and bars" combinatorial problem with constraints. When we need to distribute n identical items (candies) among k distinct groups (children), we can visualize this as arranging n stars and k-1 bars in a line.

First, let's think about the unconstrained version. If we have n candies and 3 children with no limits, we're essentially asking: "How many ways can we write n as a sum of 3 non-negative integers?" This is equivalent to arranging n stars and 2 bars, where the bars separate the stars into 3 groups. The formula for this is C(n+2, 2) because we're choosing 2 positions for bars among n+2 total positions.

But our problem has a constraint: no child can get more than limit candies. This is where the inclusion-exclusion principle becomes powerful.

Think of it this way:

• Start with all possible distributions: C(n+2, 2)
• Some of these are "bad" because at least one child gets more than limit candies
• We need to subtract these bad cases

To count the bad cases where child 1 gets more than limit candies, we can "pre-allocate" limit+1 candies to child 1, then distribute the remaining n-(limit+1) candies freely among all 3 children. This gives us C(n-limit+1, 2) bad cases. Since any of the 3 children could be the one exceeding the limit, we multiply by 3.

However, when we subtract 3 × C(n-limit+1, 2), we've double-counted cases where TWO children both exceed the limit. These cases were subtracted twice (once for each child exceeding), so we need to add them back once. If two children each get limit+1 candies minimum, we have n-2(limit+1) candies left to distribute, giving us C(n-2×limit, 2) such cases. Again, there are 3 ways to choose which two children exceed the limit.

This inclusion-exclusion pattern ensures we count each valid distribution exactly once.

Learn more about Math and Combinatorics patterns.

Solution Approach

The implementation follows the inclusion-exclusion principle directly. Here's how the solution works step by step:

Step 1: Handle Edge Case

if n > 3 * limit:
    return 0

If the total candies n exceeds 3 * limit (the maximum possible distribution), it's impossible to distribute them within the constraints, so we return 0.

Step 2: Calculate Total Unrestricted Distributions

ans = comb(n + 2, 2)

Using the stars and bars formula, we calculate C(n+2, 2) which represents all possible ways to distribute n candies among 3 children without any restrictions. This is equivalent to choosing 2 positions for dividers among n+2 total positions (n candies + 2 dividers).

Step 3: Subtract Invalid Cases (One Child Exceeds Limit)

if n > limit:
    ans -= 3 * comb(n - limit + 1, 2)

When n > limit, it's possible for a child to receive more than limit candies. For each child that gets at least limit + 1 candies:

• We "reserve" limit + 1 candies for that child
• Distribute the remaining n - (limit + 1) = n - limit - 1 candies among all 3 children
• This gives C(n - limit - 1 + 2, 2) = C(n - limit + 1, 2) invalid distributions
• Since any of the 3 children could be the one exceeding, multiply by 3

Step 4: Add Back Over-Subtracted Cases (Two Children Exceed Limit)

if n - 2 >= 2 * limit:
    ans += 3 * comb(n - 2 * limit, 2)

When n >= 2 * limit + 2, it's possible for two children to simultaneously exceed the limit. These cases were subtracted twice in Step 3, so we add them back once:

• Two children each get at least limit + 1 candies
• We have n - 2(limit + 1) = n - 2 * limit - 2 candies left to distribute
• This gives C(n - 2 * limit - 2 + 2, 2) = C(n - 2 * limit, 2) such distributions
• There are 3 ways to choose which 2 children exceed the limit (3 choose 2 = 3)

The final answer ans gives the exact count of valid distributions where each child receives between 0 and limit candies inclusive.

Note: The case where all three children exceed the limit is automatically handled because if n > 3 * limit, we already returned 0 at the beginning.

Example Walkthrough

Let's walk through a concrete example with n = 5 candies and limit = 2.

We need to find all valid ways to distribute 5 candies among 3 children where each child gets at most 2 candies.

Step 1: Check if distribution is possible

• Is n > 3 * limit? Is 5 > 6? No, so we continue.

Step 2: Calculate total unrestricted distributions

• Using stars and bars: C(n+2, 2) = C(7, 2) = 21
• This counts ALL ways to distribute 5 candies among 3 children without any limit.
• This includes invalid cases like (5,0,0), (4,1,0), (3,2,0), etc.

Step 3: Subtract cases where one child exceeds the limit

• Since n = 5 > limit = 2, we need to subtract invalid cases.
• Cases where child 1 gets > 2 candies (i.e., at least 3):
  • Pre-allocate 3 candies to child 1
  • Distribute remaining 5-3=2 candies among all 3 children
  • Number of ways: C(2+2, 2) = C(4, 2) = 6
  • These cases are: (3,2,0), (3,1,1), (3,0,2), (4,1,0), (4,0,1), (5,0,0)
• Similarly for child 2 and child 3 exceeding the limit
• Total to subtract: 3 * 6 = 18
• Running total: 21 - 18 = 3

Step 4: Add back cases where two children exceed the limit

• Check if n - 2 >= 2 * limit: Is 3 >= 4? No.
• So no cases where two children both get > 2 candies are possible.
• Nothing to add back.

Final Answer: 3

Let's verify by listing all valid distributions:

1. (2, 2, 1) - child 1 gets 2, child 2 gets 2, child 3 gets 1
2. (2, 1, 2) - child 1 gets 2, child 2 gets 1, child 3 gets 2
3. (1, 2, 2) - child 1 gets 1, child 2 gets 2, child 3 gets 2

Indeed, we have exactly 3 valid ways to distribute 5 candies with a limit of 2 per child!

The inclusion-exclusion principle correctly handled the overlapping constraints by:

• Starting with all 21 possible distributions
• Removing the 18 invalid cases where at least one child gets too many candies
• Resulting in the 3 valid distributions

Solution Implementation

Time and Space Complexity

The time complexity is O(1) because the algorithm performs a fixed number of operations regardless of the input size. The function executes at most:

• One comparison (n > 3 * limit)
• One call to comb(n + 2, 2)
• Two additional comparisons and conditional comb calculations

The comb function (combination calculation) with arguments like comb(n + 2, 2) computes (n + 2) * (n + 1) / 2, which involves a constant number of arithmetic operations. Each comb call takes O(1) time since we're always choosing 2 items (the second parameter is fixed at 2).

The space complexity is O(1) because the algorithm uses only a constant amount of extra space. It stores:

• The variable ans to accumulate the result
• Temporary values during arithmetic calculations
• No additional data structures that scale with input size

All operations are performed using a fixed amount of memory regardless of the value of n or limit.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Boundary Condition for Adding Back Over-Subtracted Cases

One of the most common mistakes is using the wrong condition for when to add back the over-subtracted cases in the inclusion-exclusion principle.

Incorrect Implementation:

if n - 2 >= 2 * limit:  # Wrong condition!
    ans += 3 * comb(n - 2 * limit, 2)

The Problem: The condition n - 2 >= 2 * limit is checking if n >= 2 * limit + 2, but this doesn't correctly identify when two children can simultaneously exceed the limit. For two children to each get more than limit candies, they need at least 2 * (limit + 1) = 2 * limit + 2 candies total.

Correct Implementation:

if n > 2 * limit + 1:  # Correct condition
    ans += 3 * comb(n - 2 * limit - 2, 2)

Or equivalently:

if n >= 2 * limit + 2:  # Also correct
    ans += 3 * comb(n - 2 * limit - 2, 2)

2. Forgetting to Handle the Combination Function's Edge Cases

The comb(n, k) function returns 0 when n < k or when n < 0, but relying on this behavior without explicit checks can lead to confusion and potential bugs.

Risky Implementation:

ans -= 3 * comb(n - limit + 1, 2)  # What if n - limit + 1 < 2?

Better Implementation with Explicit Checks:

if n > limit:
    # Only subtract when it's actually possible for a child to exceed limit
    ans -= 3 * comb(n - limit + 1, 2)

if n > 2 * limit + 1:
    # Only add back when it's possible for two children to exceed limit
    ans += 3 * comb(n - 2 * limit - 2, 2)

3. Off-by-One Errors in the Combinatorial Formula

A frequent mistake is incorrectly calculating the parameters for the combination function.

Common Mistakes:

# Wrong: forgetting to account for the "at least limit+1" constraint
ans -= 3 * comb(n - limit, 2)  # Should be comb(n - limit - 1 + 2, 2)

# Wrong: incorrect stars and bars formula
ans = comb(n + 3, 3)  # Should be comb(n + 2, 2) for 3 children

Solution: Remember that when using stars and bars:

• For distributing n items into k groups: C(n + k - 1, k - 1)
• When a child must get at least m candies, "reserve" those m candies first, then distribute the remaining n - m candies

4. Not Considering All Edge Cases

Incomplete Edge Case Handling:

def distributeCandies(self, n: int, limit: int) -> int:
    from math import comb
  
    # Missing edge case check!
    ans = comb(n + 2, 2)
    ans -= 3 * comb(n - limit + 1, 2)
    ans += 3 * comb(n - 2 * limit - 2, 2)
    return ans

Complete Solution:

def distributeCandies(self, n: int, limit: int) -> int:
    from math import comb
  
    # Essential edge case: impossible distribution
    if n > 3 * limit:
        return 0
  
    ans = comb(n + 2, 2)
  
    # Only subtract if it's possible to exceed limit
    if n > limit:
        ans -= 3 * comb(n - limit - 1 + 2, 2)
  
    # Only add back if two children can exceed limit
    if n > 2 * limit + 1:
        ans += 3 * comb(n - 2 * limit - 2 + 2, 2)
  
    return ans

These pitfalls highlight the importance of careful analysis when implementing inclusion-exclusion principle solutions and properly handling boundary conditions in combinatorial problems.