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 among the 3 children 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, child 2, and child 3 each get at most 2 candies, and the sum of candies equals 5.

The problem uses combinatorial mathematics and the inclusion-exclusion principle to solve this. The solution treats this as a "stars and bars" problem where we need to place n identical items into 3 distinct boxes with capacity constraints.

The formula calculates:

1. Start with all possible ways to distribute n candies among 3 children without restrictions: C(n+2, 2)
2. Subtract cases where at least one child gets more than limit candies: 3 * C(n-limit+1, 2)
3. Add back cases where at least two children get more than limit candies (since they were subtracted twice): 3 * C(n-2*limit, 2)

The function returns 0 if n > 3 * limit since it's impossible to distribute the candies within the constraints.

Intuition

The key insight is recognizing this as a classic "stars and bars" combinatorial problem with upper bound constraints.

First, let's think about the simpler version without any limits. If we just want to distribute n identical candies among 3 children, we can visualize this as arranging n stars and 2 bars, where bars separate the candies for each child. For example, **|*|*** means child 1 gets 2, child 2 gets 1, and child 3 gets 3 candies. The number of ways to arrange these is C(n+2, 2).

Why C(n+2, 2)? We have n candies and need 2 dividers to create 3 groups. We're choosing 2 positions for dividers among n+2 total positions.

Now comes the tricky part - we have an upper limit. Some of our distributions give a child more than limit candies, which violates our constraint. We need to remove these invalid cases.

Using the inclusion-exclusion principle:

• First, count all distributions (including invalid ones)
• Then subtract distributions where at least one child exceeds the limit
• But wait! When we subtract cases where "child 1 exceeds" and separately subtract "child 2 exceeds", we've double-subtracted cases where both children exceed the limit
• So we need to add those back

If one child must get more than limit candies (at least limit+1), we can "pre-allocate" limit+1 candies to that child, leaving n-(limit+1) candies to distribute freely. This gives us C(n-limit+1, 2) ways. Since any of the 3 children could be the one exceeding, we multiply by 3.

Similarly, for cases where two children both exceed the limit, we pre-allocate 2*(limit+1) candies and distribute the remaining n-2*limit-2 candies, giving us C(n-2*limit, 2) ways for each pair of children.

Learn more about Math and Combinatorics patterns.

Solution Approach

The implementation uses the inclusion-exclusion principle with combinatorial mathematics to count valid distributions.

Step 1: Check Impossible Case

if n > 3 * limit:
    return 0

If we have more candies than the maximum possible distribution (3 children × limit candies each), it's impossible to distribute them within constraints.

Step 2: Count All Possible Distributions Without Constraints

ans = comb(n + 2, 2)

Using the stars and bars method, we count all ways to distribute n candies among 3 children. This is equivalent to choosing 2 positions for dividers among n + 2 total positions, giving us C(n+2, 2) combinations.

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

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

When at least one child gets more than limit candies:

• We "fix" limit + 1 candies to one child (the minimum to exceed)
• 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) ways
• Since any of the 3 children could be the one exceeding, multiply by 3

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

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

In Step 3, we double-counted cases where two children both exceed the limit:

• Fix 2 * (limit + 1) candies to two children
• Distribute the remaining n - 2 * (limit + 1) = n - 2 * limit - 2 candies
• This gives C(n - 2 * limit - 2 + 2, 2) = C(n - 2 * limit, 2) ways
• There are 3 ways to choose which 2 children exceed the limit

Note: Cases where all 3 children exceed the limit are impossible since that would require n > 3 * limit, which we already handled in Step 1.

The final answer is the result after applying all inclusion-exclusion adjustments.

Example Walkthrough

Let's walk through a concrete example: n = 5 candies, limit = 2

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

Step 1: Check if distribution is possible

• Maximum possible candies = 3 × 2 = 6
• Since 5 ≤ 6, distribution is possible ✓

Step 2: Count all distributions without constraints Using stars and bars: C(5+2, 2) = C(7, 2) = 21

This counts arrangements like:

• (5, 0, 0) - child 1 gets 5 candies ❌ exceeds limit
• (3, 1, 1) - child 1 gets 3 candies ❌ exceeds limit
• (2, 2, 1) - valid ✓
• (2, 1, 2) - valid ✓
• (1, 2, 2) - valid ✓

Step 3: Subtract cases where at least one child exceeds limit (gets > 2) A child exceeding means getting at least 3 candies.

• Fix 3 candies to one child, distribute remaining 5-3=2 candies
• Ways to distribute 2 candies among 3 children: C(2+2, 2) = C(4, 2) = 6
• Any of 3 children could exceed: 3 × 6 = 18

These 18 cases include distributions like:

• Child 1 exceeds: (3,2,0), (3,1,1), (3,0,2), (4,1,0), (4,0,1), (5,0,0)
• Child 2 exceeds: (2,3,0), (1,3,1), (0,3,2), (1,4,0), (0,4,1), (0,5,0)
• Child 3 exceeds: (2,0,3), (1,1,3), (0,2,3), (1,0,4), (0,1,4), (0,0,5)

Step 4: Add back cases where at least two children exceed limit Two children exceeding means each gets at least 3 candies = 6 total minimum. Since 5 < 6, it's impossible for two children to both exceed the limit.

• Check: 5 - 2 = 3, and 3 < 2×2 = 4, so this step contributes 0

Final Answer: 21 - 18 + 0 = 3

The 3 valid distributions are:

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

Each distribution sums to 5 and no child gets more than 2 candies ✓

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 only executes:

• A constant number of comparisons (if statements)
• A fixed number of calls to the comb function, where each call computes comb(a, 2) which equals a * (a - 1) / 2 and can be calculated in constant time
• Basic arithmetic operations (additions, subtractions, multiplications)

Since all operations take constant time and there are no loops or recursive calls that depend on the input size, the overall time complexity is O(1).

The space complexity is O(1) because the algorithm only uses a fixed amount of extra space:

• A single variable ans to store the result
• Temporary variables for intermediate calculations in the comb function calls
• No additional data structures that grow with input size

The space usage remains constant regardless of the value of n or limit, resulting in O(1) space complexity.

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

Common Pitfalls

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

One of the most common mistakes is using the wrong condition to determine 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)

Why it's wrong: The condition n - 2 >= 2 * limit doesn't correctly check whether two children can each receive more than limit candies. The correct condition should verify if we have enough candies to give limit + 1 to two children.

Correct Implementation:

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

Explanation: For two children to each exceed the limit, we need at least 2 * (limit + 1) = 2 * limit + 2 candies. So the condition should be n >= 2 * limit + 2 or equivalently n > 2 * limit + 1.

2. Misunderstanding the Combinatorial Formula

Another pitfall is incorrectly calculating the combinations when applying inclusion-exclusion.

Incorrect Approach:

# Wrong: Using n - limit - 1 instead of n - limit + 1
if n > limit:
    ans -= 3 * comb(n - limit - 1, 2)  # Wrong formula

Correct Approach:

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

Why: When one child gets limit + 1 candies (minimum to exceed), we have n - (limit + 1) = n - limit - 1 candies left to distribute among 3 children. Using stars and bars, this gives us C(n - limit - 1 + 2, 2) = C(n - limit + 1, 2).

3. Not Handling Edge Cases Properly

Problem: Forgetting to check if the combination arguments are valid (non-negative).

Example of problematic code:

# This could fail if n - 2 * limit - 2 < 0
ans += 3 * comb(n - 2 * limit - 2, 2)

Solution: Always ensure the first argument to comb() is non-negative:

if n >= 2 * limit + 2:  # Ensure n - 2 * limit - 2 >= 0
    ans += 3 * comb(n - 2 * limit - 2, 2)

4. Off-by-One Errors in Limit Calculations

Common Mistake: Confusing "at most limit" with "exactly limit + 1" when calculating invalid cases.

Wrong thinking: "If a child gets more than limit candies, they get at least limit candies" Correct thinking: "If a child gets more than limit candies, they get at least limit + 1 candies"

This subtle difference affects all the formulas in the inclusion-exclusion calculation and can lead to completely incorrect results.