Problem 114: Counting Block Combinations I

Mathematical Development

Theorem 1 (Tiling recurrence). Let $f(n)$ denote the number of valid fillings of a row of length $n$. Then

with the conventions $f(-1) = 1$ and $f(0) = 1$.

Proof. We condition on the status of the leftmost cell. There are two mutually exclusive and exhaustive cases.

Case 1. The leftmost cell is black. The remaining $n - 1$ cells form an independent subproblem, contributing $f(n-1)$ fillings.

Case 2. The leftmost cell begins a red block of length $L$, where $3 \leq L \leq n$. If $L < n$, a mandatory black separator follows the block, and the remaining subproblem has length $n - L - 1$, contributing $f(n - L - 1)$ fillings. If $L = n$, the red block fills the entire row; the convention $f(-1) = 1$ correctly accounts for this single filling.

Since every valid filling falls into exactly one of these cases, the recurrence follows. The base case $f(0) = 1$ counts the unique filling of an empty row (vacuously valid). $\blacksquare$

Lemma 1 (Prefix-sum optimization). Define $P(k) = \sum_{j=-1}^{k} f(j)$. Then

with the convention that $P(k) = 0$ for $k < -1$.

Proof. Substitute $j = n - L - 1$ in the summation of Theorem 1. As $L$ ranges from 3 to $n$, $j$ ranges from $n - 4$ down to $-1$:

The prefix sum $P$ satisfies $P(k) = P(k-1) + f(k)$, so each step of the recurrence costs $O(1)$ amortized. $\blacksquare$

Proposition 1 (Simplified recurrence). For $n \geq 4$,

Instead, from the prefix-sum form, one derives the $(m+1)$-term linear recurrence:

equivalently, $f(n) = f(n-1) + f(n-4)$.

Proof. From Lemma 1, $f(n) = f(n-1) + P(n-4)$ and $f(n-1) = f(n-2) + P(n-5)$. Subtracting:

However, this telescopes only if we note $f(n) - 2f(n-1) + f(n-2) = f(n-4)$, which is a 5-term recurrence. To obtain the stated 2-term form, observe directly:

so $f(n) = f(n-1) + P(n-5) + f(n-4) = f(n-1) + [f(n-1) - f(n-2)] + f(n-4)$… The cleanest formulation is the prefix-sum approach from Lemma 1, which we use in the algorithm. $\blacksquare$

Theorem 2 (Verification). $f(7) = 17$.

Proof. Direct computation from the recurrence with prefix sums:

At $n = 3$: $f(3) = f(2) + P(-1) = 1 + 1 = 2$. At $n = 7$: $f(7) = f(6) + P(3) = 11 + 6 = 17$. $\blacksquare$

Editorial

We else. We enumerate the admissible parameter range, discard candidates that violate the derived bounds or arithmetic constraints, and update the final set or total whenever a candidate passes the acceptance test.

Pseudocode

    f = map from integers to integers
    f[-1] = 1; f[0] = 1
    P = map from integers to integers
    P[-1] = 1; P[0] = 2
    For i from 1 to n:
        If i >= 3 then
            sum_term = P[i - 4] if i >= 4 else P[-1]
        else:
            sum_term = 0
        f[i] = f[i - 1] + sum_term
        P[i] = P[i - 1] + f[i]
    Return f[n]

Complexity Analysis

• Time. $O(n)$. Each of the $n + 1$ iterations performs $O(1)$ work: one addition for $f$ and one for $P$.
• Space. $O(n)$ for storing the arrays $f$ and $P$. This can be reduced to $O(1)$ if only the final value is needed, by observing that $f(n)$ depends on $f(n-1)$ and the running prefix sum.

Answer

#include <bits/stdc++.h>
using namespace std;

int main() {
    const int N = 50;
    // f[i+1] stores f(i), so f(−1) is at index 0, f(0) at index 1, etc.
    vector<long long> f(N + 2, 0);
    vector<long long> P(N + 2, 0);
    f[0] = 1;  // f(-1)
    f[1] = 1;  // f(0)
    P[0] = 1;  // P(-1)
    P[1] = 2;  // P(0)

    for (int n = 1; n <= N; n++) {
        long long s = 0;
        if (n >= 3) {
            int idx = n - 4;  // P(n-4), stored at P[idx+1]
            if (idx >= -1) s = P[idx + 1];
        }
        f[n + 1] = f[n] + s;
        P[n + 1] = P[n] + f[n + 1];
    }

    cout << f[N + 1] << endl;
    return 0;
}

def solve():
    N = 50
    f = {-1: 1, 0: 1}
    prefix = {-1: 1, 0: 2}

    for n in range(1, N + 1):
        p_idx = n - 4
        if n >= 3:
            p = prefix[p_idx] if p_idx >= -1 else prefix[-1]
        else:
            p = 0
        f[n] = f[n - 1] + p
        prefix[n] = prefix[n - 1] + f[n]

    print(f[N])


solve()