Hard
Minimum Number of Visited Cells in a Grid — C++
Full explanation · Time O(m * n) · Space O(m * n)
// Time: O(m * n * alpha(m + n)) = O(m + n)
// Space: O(m * n)
// bfs, union find
class Solution {
public:
int minimumVisitedCells(vector<vector<int>>& grid) {
const int m = size(grid), n = size(grid[0]);
vector<UnionFind> uf1(m, UnionFind(n + 1));
vector<UnionFind> uf2(n, UnionFind(m + 1));
int d = 1, i = 0, j = 0;
vector<pair<int, int>> q = {{i, j}};
uf1[i].union_set(j, j + 1);
uf2[j].union_set(i, i + 1);
while (!empty(q)) {
vector<pair<int, int>> new_q;
for (const auto& [i, j] : q) {
if (i == m - 1 && j == n - 1) {
return d;
}
while (uf1[i].right_set(j) <= min(j + grid[i][j], n - 1)) {
const int k = uf1[i].right_set(j);
new_q.emplace_back(i, k);
uf2[k].union_set(i, i + 1);
uf1[i].union_set(k, k + 1);
}
while (uf2[j].right_set(i) <= min(i+grid[i][j], m - 1)) {
const int k = uf2[j].right_set(i);
new_q.emplace_back(k, j);
uf1[k].union_set(j, j + 1);
uf2[j].union_set(k, k + 1);
}
}
q = move(new_q);
++d;
}
return -1;
}
private:
class UnionFind {
public:
UnionFind(int n)
: set_(n)
, rank_(n)
, right_(n) { // added
iota(begin(set_), end(set_), 0);
iota(begin(right_), end(right_), 0); // added
}
int find_set(int x) {
if (set_[x] != x) {
set_[x] = find_set(set_[x]); // Path compression.
}
return set_[x];
}
bool union_set(int x, int y) {
x = find_set(x), y = find_set(y);
if (x == y) {
return false;
}
if (rank_[x] > rank_[y]) {
swap(x, y);
}
set_[x] = y; // Union by rank.
if (rank_[x] == rank_[y]) {
++rank_[y];
}
right_[y] = max(right_[x], right_[y]); // added
return true;
}
int right_set(int x) { // added
return right_[find_set(x)];
}
private:
vector<int> set_;
vector<int> rank_;
vector<int> right_; // added
};
};
// Time: O(m * n * alpha(m + n)) = O(m + n)
// Space: O(m * n)
// bfs, bst
class Solution2 {
public:
int minimumVisitedCells(vector<vector<int>>& grid) {
const int m = size(grid), n = size(grid[0]);
vector<set<int>> bst1(m);
vector<set<int>> bst2(n);
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
bst1[i].emplace(j);
bst2[j].emplace(i);
}
}
int d = 1, i = 0, j = 0;
vector<pair<int, int>> q = {{i, j}};
while (!empty(q)) {
vector<pair<int, int>> new_q;
for (const auto& [i, j] : q) {
if (i == m - 1 && j == n - 1) {
return d;
}
for (auto it = bst1[i].lower_bound(j + 1);
it != end(bst1[i]) && *it <= j + grid[i][j];
it = bst1[i].erase(it)) {
new_q.emplace_back(i, *it);
bst2[*it].erase(i);
}
for (auto it = bst2[j].lower_bound(i + 1);
it != end(bst2[j]) && *it <= i + grid[i][j];
it = bst2[j].erase(it)) {
new_q.emplace_back(*it, j);
bst1[*it].erase(j);
}
}
q = move(new_q);
++d;
}
return -1;
}
};