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198 changes: 112 additions & 86 deletions solutions/platinum/cses-2429.mdx
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Expand Up @@ -2,12 +2,12 @@
id: cses-2429
source: CSES
title: Grid Completion
author:
author:Alexis Tao
---

# Solution
## Implementation

Time complexity: $\mathcal O(N^3)$
**Time Complexity:** $\mathcal{O}(N^3)$

<LanguageSection>

Expand All @@ -20,94 +20,120 @@ Time complexity: $\mathcal O(N^3)$
using namespace std;
typedef long long ll;
const ll MOD = 1e9 + 7;
const int maxN = 505; // Maximum grid size

int N, p[maxN], q[maxN], C[6];
bool inp[maxN], inq[maxN];
char S[maxN];
ll fact[maxN], inv[maxN];

// Compute modular inverse using Fermat's Little Theorem
ll inverse(ll x) {
ll res = 1;
ll b = MOD - 2; // Theorem: x^(MOD-2) ≡ x^(-1) (mod MOD)
while (b) {
if (b & 1) res = (res * x) % MOD;
x = (x * x) % MOD;
b >>= 1;
}
return res;
}

// Function to compute binomial coefficient C(x, y) modulo MOD
ll choose(int x, int y) { return (fact[x] * inv[y] % MOD) * inv[x - y] % MOD; }

// Precompute factorials and their modular inverses
void init() {
fact[0] = inv[0] = 1;
for (int i = 1; i <= N; i++) {
fact[i] = (fact[i - 1] * i) % MOD;
inv[i] = (inv[i - 1] * inverse(i)) % MOD;
}
}

// Function to calculate the number of ways to complete the grid based on
// parameters i, j, k
ll f(int i, int j, int k) {
ll res = (choose(C[0], i) * choose(C[1], j) % MOD * choose(C[2], k) % MOD *
choose(C[3], i) % MOD);

// Multiply by the factorials of the number of rows and columns left
res = (res * fact[i] % MOD * fact[C[4] - i - j] % MOD * fact[C[5] - i - k] % MOD);

// Apply inclusion-exclusion principle to adjust the result
if ((i + j + k) % 2 == 1) res = (MOD - res);

return res;
ll modInverse(ll x, ll MOD) {
ll res = 1;
ll b = MOD - 2; // Theorem: x^(MOD-2) ≡ x^(-1) (mod MOD)
while (b) {
if (b & 1) {
res = (res * x) % MOD;
}
x = (x * x) % MOD;
b >>= 1;
}
return res;
}

int main() {
scanf("%d", &N);

init();

// Read grid and fill arrays p, q, inp, and inq
for (int i = 0; i < N; i++) {
scanf(" %s", S);
p[i] = q[i] = -1;
for (int j = 0; j < N; j++) {
if (S[j] == 'A') {
p[i] = j;
inp[j] = true;
}
if (S[j] == 'B') {
q[i] = j;
inq[j] = true;
}
}
}

// Count various configurations based on positions of 'A' and 'B'
for (int i = 0; i < N; i++) {
if (p[i] == -1 && q[i] == -1) C[0]++;
if (p[i] == -1 && q[i] != -1 && !inp[q[i]]) C[1]++;
if (p[i] != -1 && q[i] == -1 && !inq[p[i]]) C[2]++;
}

// Count rows and columns without 'A' or 'B'
for (int i = 0; i < N; i++) {
if (!inp[i] && !inq[i]) C[3]++;
if (!inp[i]) C[4]++;
if (!inq[i]) C[5]++;
}

// Calculate the number of valid grid configurations
ll ans = 0;
for (int i = 0; i <= min(C[0], C[3]); i++)
for (int j = 0; j <= C[1]; j++)
for (int k = 0; k <= C[2]; k++) ans = (ans + f(i, j, k)) % MOD;

printf("%lld\n", ans);
int N;
cin >> N;

vector<int> p(N, -1), q(N, -1), C(6, 0);
vector<bool> inp(N, false), inq(N, false);
vector<ll> fact(N + 1), inv(N + 1);
string S;

// Precompute factorials
fact[0] = 1;
for (int i = 1; i <= N; i++) {
fact[i] = (fact[i - 1] * i) % MOD;
}

// Compute the modular inverse of the largest factorial first
inv[N] = modInverse(fact[N], MOD);

// Compute modular inverses of all factorials in reverse order
for (int i = N - 1; i >= 0; i--) {
inv[i] = (inv[i + 1] * (i + 1)) % MOD;
}

// Read grid and fill arrays p, q, inp, and inq
for (int i = 0; i < N; i++) {
cin >> S;
for (int j = 0; j < N; j++) {
if (S[j] == 'A') {
p[i] = j;
inp[j] = true;
}
if (S[j] == 'B') {
q[i] = j;
inq[j] = true;
}
}
}

// Count various configurations based on positions of 'A' and 'B'
for (int i = 0; i < N; i++) {
if (p[i] == -1 && q[i] == -1) {
C[0]++;
}
if (p[i] == -1 && q[i] != -1 && !inp[q[i]]) {
C[1]++;
}
if (p[i] != -1 && q[i] == -1 && !inq[p[i]]) {
C[2]++;
}
}

// Count rows and columns without 'A' or 'B'
for (int i = 0; i < N; i++) {
if (!inp[i] && !inq[i]) {
C[3]++;
}
if (!inp[i]) {
C[4]++;
}
if (!inq[i]) {
C[5]++;
}
}

// Function to compute binomial coefficient C(x, y) modulo MOD
auto choose = [&](int x, int y) {
if (y > x || y < 0) return 0LL;
return (fact[x] * inv[y] % MOD) * inv[x - y] % MOD;
};

// Function to calculate the number of ways to complete the grid
auto f = [&](int i, int j, int k) {
ll res = (choose(C[0], i) * choose(C[1], j) % MOD *
choose(C[2], k) % MOD * choose(C[3], i) % MOD);

// Multiply by the factorials of the number of rows and columns left
res = (res * fact[i] % MOD * fact[C[4] - i - j] % MOD * fact[C[5] - i - k] % MOD);

// Apply inclusion-exclusion principle to adjust the result
if ((i + j + k) % 2 == 1) {
res = (MOD - res);
}

return res;
};

// Calculate the number of valid grid configurations
ll ans = 0;
for (int i = 0; i <= min(C[0], C[3]); i++) {
for (int j = 0; j <= C[1]; j++) {
for (int k = 0; k <= C[2]; k++) {
ans = (ans + f(i, j, k)) % MOD;
}
}
}

cout << ans << endl;

return 0;
}
```
</CPPSection>
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