Closed JLeow00 closed 7 months ago
Yeah, apart from that, we can go one step further to allocate just enough memory in SpT so as to not waste space!
class SparseTable { // OOP style
private:
vi A, P2, L2;
vector<vi> SpT; // the Sparse Table
public:
SparseTable() {} // default constructor
SparseTable(vi &initialA) { // pre-processing routine
A = initialA;
int n = (int)A.size();
int L2_n = (int)log2(n) + 1;
P2.assign(L2_n + 1, 0);
L2.assign((1 << L2_n) + 1, 0);
for (int i = 0; i <= L2_n; ++i) {
P2[i] = (1 << i); // to speed up 2^i
L2[(1 << i)] = i; // to speed up log_2(i)
}
for (int i = 2; i < P2[L2_n]; ++i)
if (L2[i] == 0) L2[i] = L2[i - 1]; // to fill in the blanks
// the initialization phase
SpT = vector<vi>(L2[n] + 1, vi());
SpT[0] = vi(n, 0);
for (int j = 0; j < n; ++j) SpT[0][j] = j; // RMQ of sub array [j..j]
// the two nested loops below have overall time complexity = O(n log n)
for (int i = 1; P2[i] <= n; ++i) { // for all i s.t. 2^i <= n
SpT[i] = vi(n + 1 - P2[i]); // initialize SpT[i]
for (int j = 0; j + P2[i] - 1 < n; ++j) { // for all valid j
int x = SpT[i - 1][j]; // [j..j+2^(i-1)-1]
int y = SpT[i - 1][j + P2[i - 1]]; // [j+2^(i-1)..j+2^i-1]
SpT[i][j] = A[x] <= A[y] ? x : y;
}
}
}
int RMQ(int i, int j) {
int k = L2[j - i + 1]; // 2^k <= (j-i+1)
int x = SpT[k][i]; // covers [i..i+2^k-1]
int y = SpT[k][j - P2[k] + 1]; // covers [j-2^k+1..j]
return A[x] <= A[y] ? x : y;
}
};
Thanks for the bug fix. I was so busy recently and didn't check this repo that often.
P2
andL2
are not allocated enough space because the loop in the next line would assignP2[L2_n]
andL2[(1<<L2_n)]
which would be out of range