Medium
Number of Ways to Build Sturdy Brick Wall — C++
Full explanation · Time O(h * p^2) · Space O(p^2)
// Time: O(h * p^2), p is the number of patterns
// Space: O(p^2)
// bitmask, backtracking, dp
class Solution {
public:
int buildWall(int height, int width, vector<int>& bricks) {
static const int MOD = 1e9 + 7;
vector<int> patterns;
unordered_set<int> lookup;
backtracking(height, width, bricks, 0, 0, &lookup, &patterns);
vector<vector<int>> adj(size(patterns));
for (int i = 0; i < size(patterns); ++i) {
for (int j = 0; j < size(patterns); ++j) {
if (!(patterns[i] & patterns[j])) {
adj[i].emplace_back(j);
}
}
}
vector<vector<int>> dp = {vector<int>(size(patterns), 1), vector<int>(size(patterns), 0)};
for (int i = 0; i < height; ++i) {
for (int j = 0; j < size(patterns); ++j) {
dp[(i + 1) % 2][j] = 0;
for (const auto& k : adj[j]) {
dp[(i + 1) % 2][j] = (dp[(i + 1) % 2][j] + dp[i % 2][k]) % MOD;
}
}
}
return accumulate(cbegin(dp[(height - 1) % 2]), cend(dp[(height - 1) % 2]), 0,
[](const auto& total, const auto& x) {
return (total + x) % MOD;
}); // Time: O(p)
}
private:
void backtracking(
int height, int width, const vector<int>& bricks,
int total, int mask,
unordered_set<int> *lookup,
vector<int> *patterns) {
if (lookup->count(mask)) {
return;
}
lookup->emplace(mask);
if (total >= width) {
if (total == width) {
patterns->emplace_back(mask ^ (1 << width));
}
return;
}
for (const auto& x : bricks) {
backtracking(height, width, bricks, total + x, mask | (1 << (total + x)), lookup, patterns);
}
}
};
// Time: O(p^3 * logh), p is the number of patterns, p may be up to 512
// Space: O(p^3)
// bitmask, backtracking, matrix exponentiation
class Solution_TLE {
public:
int buildWall(int height, int width, vector<int>& bricks) {
vector<int> patterns;
unordered_set<int> lookup;
backtracking(height, width, bricks, 0, 0, &lookup, &patterns);
if (empty(patterns)) {
return 0;
}
vector<vector<int>> matrix;
vector<vector<int>> counts = {vector<int>(size(patterns), 1)};
for (const auto& mask1 : patterns) {
matrix.emplace_back();
for (const auto& mask2 : patterns) {
matrix.back().emplace_back(static_cast<int>((mask1 & mask2) == 0));
}
}
const auto& result = matrixMult(counts, matrixExpo(matrix, height - 1)); // Time: O(p^3 * logh), Space: O(p^2)
return accumulate(cbegin(result[0]), cend(result[0]), 0,
[](const auto& total, const auto& x) {
return (total + x) % MOD;
}); // Time: O(p)
}
private:
void backtracking(
int height, int width, const vector<int>& bricks,
int total, int mask,
unordered_set<int> *lookup,
vector<int> *patterns) {
if (lookup->count(mask)) {
return;
}
lookup->emplace(mask);
if (total >= width) {
if (total == width) {
patterns->emplace_back(mask ^ (1 << width));
}
return;
}
for (const auto& x : bricks) {
backtracking(height, width, bricks, total + x, mask | (1 << (total + x)), lookup, patterns);
}
}
vector<vector<int>> matrixExpo(const vector<vector<int>>& A, int pow) {
vector<vector<int>> result(A.size(), vector<int>(A.size()));
vector<vector<int>> A_exp(A);
for (int i = 0; i < A.size(); ++i) {
result[i][i] = 1;
}
while (pow) {
if (pow % 2 == 1) {
result = matrixMult(result, A_exp);
}
A_exp = matrixMult(A_exp, A_exp);
pow /= 2;
}
return result;
}
vector<vector<int>> matrixMult(const vector<vector<int>>& A, const vector<vector<int>>& B) {
vector<vector<int>> result(A.size(), vector<int>(B[0].size()));
for (int i = 0; i < A.size(); ++i) {
for (int j = 0; j < B[0].size(); ++j) {
int64_t entry = 0;
for (int k = 0; k < B.size(); ++k) {
entry = (static_cast<int64_t>(A[i][k]) * B[k][j] % MOD + entry) % MOD;
}
result[i][j] = static_cast<int>(entry);
}
}
return result;
}
static const int MOD = 1e9 + 7;
};