[CodeChef] Starters 79 — K-Beautiful Subarrays: two bucket sizes, two stars-and-bars
Published 30 September 2026
Problem: CodeChef — K-Beautiful Subarrays (Starters 79, difficulty 2753). Official editorial: KBEAUTIFUL editorial.
I haven’t finished this one. I got the structure quickly — the array has to be periodic, there is a forced minimum cost, and the leftover operations are what you count — and then I stalled on the counting itself. I knew it was stars and bars, and I could not see how to apply stars and bars to an inequality whose coefficients aren’t all 1. This post is that one step, worked slowly on small numbers, plus a TODO for the rest.
The problem¶
You’re given an array \(A\) of size \(N\) and may perform at most \(M\) operations, each one increasing a single element by \(1\). An array is \(K\)-beautiful if every subarray of size \(K\) has the same sum. Count the distinct \(K\)-beautiful final arrays you can reach, modulo \(998244353\).
Constraints: \(1 \le K \le N \le 10^6\), \(1 \le M \le 10^6\), \(A_i \le 10^{18}\), and the sums of \(N\) and of \(M\) over all test cases are each at most \(10^6\).
Equal window sums means periodic¶
Two neighbouring windows share everything except their end elements:
So \(K\)-beautiful just means every residue class of positions mod \(K\) holds a single value. I’ll call a residue class a bucket: there are \(K\) of them, and the final array is completely described by one number per bucket.
The forced cost¶
Operations only increase, so a bucket can’t end below its current maximum. Raising every element of a bucket to that maximum costs
and nothing cheaper makes the bucket constant. Add this up over all \(K\) buckets to get the baseline. If the baseline exceeds \(M\) the answer is \(0\). Otherwise
operations are left over, and they’re optional.
What the leftover buys¶
After the baseline every bucket is already constant. The only thing left to do is raise a whole bucket further, and raising a bucket of size \(c\) by \(1\) costs \(c\) operations, because each of its elements has to move together.
If bucket \(j\) has size \(c_j\) and I raise it by \(x_j \ge 0\), the spend is \(\sum_j c_j x_j\). Different tuples \((x_1, \ldots, x_K)\) give different final arrays, and every reachable final array comes from exactly one tuple. So the answer is the number of non-negative integer solutions of
Only two bucket sizes exist¶
This is the fact that makes it tractable. Write
Then exactly \(t\) buckets have size \(q + 1\) (the first \(t\) residues, which get one extra element), and the other \(K - t\) buckets have size \(q\). There are no other sizes. Since \(K \le N\), \(q \ge 1\).
So the \(K\) coefficients collapse into two groups, and the inequality is really
with the \(x\)’s being the raises of the size-\(q\) buckets and the \(y\)’s the raises of the size-\((q+1)\) buckets.
Where I got stuck¶
Stars and bars counts solutions of \(z_1 + \cdots + z_g = S\): there are
of them. Every coefficient is \(1\). My inequality has coefficients \(q\) and \(q + 1\), and I kept trying to find a single stars-and-bars expression for the whole thing. There isn’t one.
What works is to stop looking at the individual variables and name the two group totals:
The inequality only ever sees \(X\) and \(Y\):
And inside a group every coefficient is the same, so once \(X\) is fixed, “how do I split \(X\) among the \(K - t\) small buckets” is plain stars and bars, with no coefficients in sight. Same for \(Y\) among the \(t\) big buckets. The two splits are independent, so they multiply:
So it’s two stars-and-bars, one per bucket size, glued together by enumerating the pair of totals. The coefficients never enter a binomial; they only decide which pairs \((X, Y)\) are allowed.
One edge: when \(t = 0\) there are no big buckets, so \(Y\) must be \(0\) and its factor is \(1\). The small group is never empty, since \(t < K\).
Example 1: three buckets¶
Take \(N = 5\), \(K = 3\), and suppose \(R = 4\) operations are left after the baseline. Then \(q = 1\), \(t = 2\):
- positions \(\{1, 4\}\) — size 2
- positions \(\{2, 5\}\) — size 2
- position \(\{3\}\) — size 1
Call the raises \(u, v\) for the two size-2 buckets and \(w\) for the size-1 bucket. The constraint is \(2u + 2v + w \le 4\). Listing by hand, for each \((u, v)\) count the \(w\)’s that still fit:
| \(u\) | \(v\) | spent \(2u + 2v\) | possible \(w\) | count |
|---|---|---|---|---|
| 0 | 0 | 0 | 0, 1, 2, 3, 4 | 5 |
| 0 | 1 | 2 | 0, 1, 2 | 3 |
| 1 | 0 | 2 | 0, 1, 2 | 3 |
| 0 | 2 | 4 | 0 | 1 |
| 1 | 1 | 4 | 0 | 1 |
| 2 | 0 | 4 | 0 | 1 |
Total \(14\).
Now the formula. Here \(X = w\) (one small bucket) and \(Y = u + v\) (two big buckets), with \(X + 2Y \le 4\):
| \(Y\) | ways to split \(Y\): \(\binom{Y+1}{1}\) | allowed \(X\) | ways per \(X\): \(\binom{X}{0}\) | contribution |
|---|---|---|---|---|
| 0 | 1 | 0..4 | 1 | \(1 \cdot 5 = 5\) |
| 1 | 2 | 0..2 | 1 | \(2 \cdot 3 = 6\) |
| 2 | 3 | 0 | 1 | \(3 \cdot 1 = 3\) |
\(5 + 6 + 3 = 14\). The rows of the first table with the same \(u + v\) are exactly what \(\binom{Y+1}{1}\) is counting: \(Y = 1\) is the two rows \((0,1), (1,0)\), and \(Y = 2\) is the three rows \((0,2), (1,1), (2,0)\).
Example 2: ten buckets¶
With more buckets the hand enumeration stops being possible and the formula doesn’t change. Take \(N = 23\), \(K = 10\), \(R = 4\). Then \(q = 2\), \(t = 3\): seven buckets of size 2 and three of size 3.
\(Y\) can only be \(0\) or \(1\). If \(Y = 0\) then \(X \in \{0, 1, 2\}\); if \(Y = 1\) then \(X = 0\). Four pairs:
| \((X, Y)\) | \(\binom{X+6}{6}\) | \(\binom{Y+2}{2}\) | product |
|---|---|---|---|
| \((0, 0)\) | 1 | 1 | 1 |
| \((1, 0)\) | 7 | 1 | 7 |
| \((2, 0)\) | 28 | 1 | 28 |
| \((0, 1)\) | 1 | 3 | 3 |
Total \(39\).
The \(28\) is worth unpacking once, because it’s the binomial doing real work: \(X = 2\) spread over seven buckets is either a \(2\) in one bucket (\(7\) ways) or a \(1\) in each of two buckets (\(\binom{7}{2} = 21\) ways), and \(7 + 21 = 28 = \binom{8}{6}\).
Checking against sample 1¶
The first sample is \(N = 5\), \(M = 4\), \(K = 3\), \(A = [1, 2, 1, 2, 1]\), expected answer \(5\).
Buckets: \(\{A_1, A_4\} = \{1, 2\}\) costs \(1\) to level, \(\{A_2, A_5\} = \{2, 1\}\) costs \(1\), \(\{A_3\}\) costs \(0\). Baseline \(2\), so \(R = 2\), and with the same shape as Example 1 the constraint is \(X + 2Y \le 2\):
- \(Y = 0\): \(X \in \{0, 1, 2\}\), contribution \(1 \cdot 3 = 3\)
- \(Y = 1\): \(X = 0\), contribution \(2 \cdot 1 = 2\)
\(3 + 2 = 5\), matching the five arrays listed in the statement.
The double sum as code¶
A direct transcription of the formula. It reproduces all five sample answers, and it agrees with a brute force over every tuple of raises on a few thousand small random cases.
import sys
from math import comb
MOD = 998244353
def ways(total, groups):
# non-negative solutions of z_1 + ... + z_groups = total
if groups == 0:
return 1 if total == 0 else 0
return comb(total + groups - 1, groups - 1)
def solve(n, m, k, a):
baseline = 0
for r in range(k):
bucket = a[r::k]
baseline += len(bucket) * max(bucket) - sum(bucket)
if baseline > m:
return 0
rem = m - baseline
q, t = divmod(n, k)
small, big = k - t, t # buckets of size q, of size q + 1
ans = 0
for y in range(rem // (q + 1) + 1):
for x in range((rem - (q + 1) * y) // q + 1):
ans += ways(x, small) * ways(y, big)
return ans % MOD
def main():
data = sys.stdin.buffer.read().split()
p = 1
out = []
for _ in range(int(data[0])):
n, m, k = int(data[p]), int(data[p + 1]), int(data[p + 2]); p += 3
a = [int(v) for v in data[p:p + n]]; p += n
out.append(solve(n, m, k, a))
print("\n".join(map(str, out)))
main()
This is correct and far too slow. The number of \((X, Y)\) pairs is about
\(\dfrac{R^2}{2q(q+1)}\), and when \(K > N/2\) we have \(q = 1\), so \(R = 10^6\) gives roughly
\(2.5 \cdot 10^{11}\) pairs — before even counting the cost of exact comb on big arguments.
TODO¶
- Collapse the inner loop. For a fixed \(Y\), \(X\) runs over a full prefix \(0..A\) with \(A = \left\lfloor \frac{R - (q+1)Y}{q} \right\rfloor\), and the hockey-stick identity \(\sum_{X=0}^{A} \binom{X + g - 1}{g - 1} = \binom{A + g}{g}\) should turn it into one binomial. That would leave a single sum over \(Y\) with about \(R/(q+1)\) terms, which fits in the \(\sum M \le 10^6\) budget. Work through why the identity holds (it’s “at most \(A\) split among \(g\) buckets” with a slack bucket added) instead of just using it.
- Re-do Example 1 with the single sum and confirm it still gives \(14\), to be sure the collapse doesn’t lose the split between buckets of the same size.
- Precompute factorials and inverse factorials mod \(998244353\) so each binomial is \(O(1)\). The largest top argument is about \(R + K \le M + N\).
- Write the fast version, test it against the double sum above on random small cases, and submit.
- Read the editorial’s derivation and compare.
The lesson so far. Stars and bars needs every coefficient to be \(1\). When a constraint has a few distinct coefficients, group the variables by coefficient, enumerate the group totals, and use stars and bars inside each group. The coefficients pick which totals are allowed; the binomials count the splits.