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Problem solution · C++

Bricks Falling When Hit

Bricks Falling When Hit: a C++ solution using disjoint set union. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Disjoint set union
Source
walkccc LeetCode Solutions
Length
103 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Disjoint set union

For Bricks Falling When Hit, the implementation maintains connected components and merges them as relationships are processed.

  1. Give each element a component representative.
  2. Merge representatives when a connection is accepted.
  3. Answer connectivity or component queries from the compressed representatives.

Code notes

  • 103 lines of C++ from the credited upstream file 803.cpp.
  • The implementation visibly relies on sequence storage.
  • 5 loop blocks detected.

Complexity

Account for every find and union operation; with path compression and ranked merging, the amortized cost is nearly constant per operation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeBricks Falling When Hit · C++C++
Use this to learn the idea, then write your own version.
class UnionFind { public:  UnionFind(int n) : id(n), sz(n, 1) {    iota(id.begin(), id.end(), 0);  }   void unionBySize(int u, int v) {    const int i = find(u);    const int j = find(v);    if (i == j)      return;    if (sz[i] < sz[j]) {      sz[j] += sz[i];      id[i] = j;    } else {      sz[i] += sz[j];      id[j] = i;    }  }   int getStableSize() {    // Bricks connected with 0 (top) are stable.    return sz[find(0)];  }  private:  vector<int> id;  vector<int> sz;   int find(int u) {    return id[u] == u ? u : id[u] = find(id[u]);  }}; class Solution { public:  vector<int> hitBricks(vector<vector<int>>& grid, vector<vector<int>>& hits) {    m = grid.size();    n = grid[0].size();     UnionFind uf(m * n + 1);  // 0 := top (stable)     // Mark cells to hit as 2.    for (const vector<int>& hit : hits) {      const int i = hit[0];      const int j = hit[1];      if (grid[i][j] == 1)        grid[i][j] = 2;    }     // Union all the 1s.    for (int i = 0; i < m; ++i)      for (int j = 0; j < n; ++j)        if (grid[i][j] == 1)          unionNeighbors(grid, uf, i, j);     vector<int> ans(hits.size());    int stableSize = uf.getStableSize();     for (int i = hits.size() - 1; i >= 0; --i) {      const int x = hits[i][0];      const int y = hits[i][1];      if (grid[x][y] == 2) {  // cells marked from 1 to 2        grid[x][y] = 1;       // Unhit and restore it back to 1.        unionNeighbors(grid, uf, x, y);        const int newStableSize = uf.getStableSize();        if (newStableSize > stableSize)          ans[i] = newStableSize - stableSize - 1;  // 1 := the hit cell        stableSize = newStableSize;      }    }     return ans;  }  private:  static constexpr int kDirs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};  int m;  int n;   void unionNeighbors(const vector<vector<int>>& grid, UnionFind& uf, int i,                      int j) {    const int hash = getHash(i, j);     for (const auto& [dx, dy] : kDirs) {      const int x = i + dx;      const int y = j + dy;      if (x < 0 || x == m || y < 0 || y == n)        continue;      if (grid[x][y] != 1)        continue;      uf.unionBySize(hash, getHash(x, y));    }     if (i == 0)      uf.unionBySize(hash, 0);  }   int getHash(int i, int j) {    return i * n + j + 1;  }};

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