For LeetCode 2929, count ordered assignments of n candies to three children, allowing zero candies but capping each child at limit. The most useful edge checks are the total-capacity boundary, the unrestricted range, and the inclusion-exclusion terms that become zero when their arguments are negative.
What counts as a valid distribution?
The three children are distinct, so assignments are ordered: giving two candies to one child and one to another is different from swapping those recipients. A child may receive zero. The official statement gives 1 ≤ n ≤ 106 and 1 ≤ limit ≤ 106, with examples n = 5, limit = 2 yielding 3 and n = 3, limit = 3 yielding 10. See the official LeetCode 2929 statement.
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Which boundary cases should you check?
n > 3 × limit: Return 0. The three children together cannot hold enough candies.n = 3 × limit: Return 1. Only the allocation(limit, limit, limit)reaches the total.n ≤ limit: The cap cannot exclude any allocation. The answer is the unrestricted count,(n + 2)(n + 1) / 2.n = 1: There are 3 allocations, since exactly one of the three children receives the candy andlimit ≥ 1.
How does the direct counting method work?
Choose the first child’s amount, then count the feasible amounts for the second child; the third child’s amount is determined by the remainder. This directly expresses the constraints and can help when verifying a formula.
- Let the first child receive
i. Its feasible range ismax(0, n − 2 × limit)throughmin(n, limit), inclusive. - For each
i, let the second child’s amount range frommax(0, n − i − limit)throughmin(limit, n − i), inclusive. - Add the number of integers in that second range:
upper − lower + 1. The third child receivesn − i − second, which is within the cap by construction.
This takes linear time in the feasible range, at most about one million iterations for the stated constraints, and constant extra space.
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How does the constant-time inclusion-exclusion formula work?
Define W(x) as the number of ways to distribute x candies among three labeled children with no upper caps:
W(x) = 0whenx < 0.W(x) = (x + 2)(x + 1) / 2whenx ≥ 0.
Then set k = limit + 1 and calculate:
answer = W(n) − 3 × W(n − k) + 3 × W(n − 2k) − W(n − 3k)
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The shift is limit + 1 because a child first exceeds the cap by receiving that many candies. Subtract allocations where one selected child has this excess; add back allocations where two selected children do; subtract the cases where all three do. Any term with a negative argument contributes zero. The formula takes constant time and constant space.
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What implementation mistakes cause edge-case failures?
- Using
limitinstead oflimit + 1in the shifts: This removes valid allocations in which a child receives exactlylimit. - Keeping negative formula arguments: Define
W(x)to return zero for negativex; do not evaluate the triangular expression for them. - Forgetting the ordering: Count distributions by recipient, not merely the different candy amounts that appear in a triple.
- Assuming every child must receive candy: Zero is permitted by the statement.
- Using 32-bit arithmetic: At
n = 106, the unrestricted count is(106 + 2)(106 + 1) / 2, about5 × 1011. Use an integer type wide enough for intermediate products and the result.
Which method should you use?
Use the direct sum when you want each constraint to be visible and easy to trace. Use inclusion-exclusion when you want constant runtime; its main implementation risks are the limit + 1 shift and negative arguments. Both methods count labeled, ordered allocations and require constant extra space.
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