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For LeetCode 2929, count ordered triples of nonnegative integers (a, b, c) with a + b + c = n and each value at most limit. Inclusion-exclusion applied to the stars-and-bars count gives C(n+2,2) − 3C(n−limit+1,2) + 3C(n−2limit,2) − C(n−3limit−1,2). Here, define C(x,2) = x(x−1)/2 when x ≥ 2, and C(x,2) = 0 otherwise.
What the formula counts
LeetCode 2929 asks for the number of ways to distribute n candies among three children, with no child receiving more than limit. The children are distinguishable, so changing which child receives an amount creates a different allocation. In symbols, count the ordered triples (a, b, c) satisfying a + b + c = n and 0 ≤ a, b, c ≤ limit. The problem statement gives 1 ≤ n ≤ 106 and 1 ≤ limit ≤ 106 (LeetCode problem statement).
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How inclusion-exclusion produces the formula
Start by counting every nonnegative triple, then remove allocations in which one or more children exceed the cap. A child exceeds the cap by receiving at least limit + 1 candies. Inclusion-exclusion subtracts single-child violations, restores pairwise overlaps, and subtracts the triple overlap.
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By stars and bars, the number of nonnegative integer triples summing to n is C(n + 2, 2). Imagine arranging n candies and inserting two dividers to split them among three children; a section may be empty. This is the unrestricted starting count.
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2. Subtract allocations where one child exceeds the cap
Choose one specified child, say A, and reserve limit + 1 candies for A. After reserving them, n − limit − 1 candies remain to distribute freely among all three children. Stars and bars counts these allocations as:
C((n − limit − 1) + 2, 2) = C(n − limit + 1, 2).
There are three possible children to choose, so subtract 3C(n − limit + 1, 2). The term is zero when too few candies remain to form a distribution.
3. Restore allocations counted as two violations
If two specified children each receive at least limit + 1, the previous step subtracts that allocation twice. Reserve limit + 1 candies for each of the pair. The leftover is n − 2(limit + 1), and distributing it among three children gives:
C(n − 2(limit + 1) + 2, 2) = C(n − 2limit, 2).
There are three pairs of children, so add 3C(n − 2limit, 2).
4. Subtract the triple overlap
If all three children exceed the cap, the preceding steps restore that allocation once too many. Reserving limit + 1 candies for each leaves n − 3(limit + 1). Its unrestricted distribution count is:
C(n − 3(limit + 1) + 2, 2) = C(n − 3limit − 1, 2).
Subtract this overlap once. Combining the four stages yields:
answer = C(n + 2, 2) − 3C(n − limit + 1, 2) + 3C(n − 2limit, 2) − C(n − 3limit − 1, 2).
Use C(x, 2) = x(x − 1)/2 for x ≥ 2 and zero for x < 2. This convention makes shifted terms vanish when the reserved candies leave too little to distribute, without requiring separate boundary-case formulas. The stars-and-bars and inclusion-exclusion approach is also described in the LeetCode.ca solution explanation.
Check the formula with the official examples
n = 5, limit = 2
C(7, 2) − 3C(4, 2) + 3C(1, 2) − C(−2, 2) = 21 − 18 + 0 − 0 = 3. The valid allocations are the three permutations of (1, 2, 2).
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The cap excludes no allocation, so the count is C(5, 2) = 10. Both results match the examples in the LeetCode statement.
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An alternate derivation: fix one child’s share
You can also count by choosing the first child’s amount, then counting the possible amounts for the second. This makes the valid ranges explicit, though summing over the first child is not a constant-time formula.
Find the feasible range for the first child
Call the first child’s share i. It cannot exceed limit, and the other two children can hold at most 2limit together. Therefore:
max(0, n − 2limit) ≤ i ≤ min(n, limit).
Count the second child’s choices for each share
Once i is fixed, the second child’s share j must leave a valid amount for the third child while staying within its own cap:
max(0, n − i − limit) ≤ j ≤ min(limit, n − i).
Each integer in this inclusive interval gives one valid choice for j and a corresponding third share n − i − j. Thus, for each feasible i, the number of choices is the upper endpoint minus the lower endpoint plus one. Sum that count over the feasible values of i. The interval-based approach is presented in the CodeJeet solution explanation.
Quick Recap
Which derivation should you use?
| Approach | What it emphasizes | Counting work | Useful when |
|---|---|---|---|
| Stars and bars plus inclusion-exclusion | How the closed form follows from unrestricted counts and overlapping violations | Constant time: a fixed number of arithmetic operations | You want the formula or a direct constant-time implementation |
| Fix the first child’s share | The exact feasible interval for the remaining choices | Sum over feasible values of i | You want an intuitive enumeration of valid allocations |
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