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Problem solution · C++

Minimum Moves to Move a Box to Their Target Location

Minimum Moves to Move a Box to Their Target Location: a C++ solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Breadth-first search
Source
walkccc LeetCode Solutions
Length
91 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

Breadth-first search

For Minimum Moves to Move a Box to Their Target Location, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 91 lines of C++ from the credited upstream file 1263.cpp.
  • The implementation visibly relies on sequence storage, work queue.
  • 7 loop blocks detected.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

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 codeMinimum Moves to Move a Box to Their Target Location · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int minPushBox(vector<vector<char>>& grid) {    const int m = grid.size();    const int n = grid[0].size();    vector<int> box;    vector<int> player;    vector<int> target;     for (int i = 0; i < m; ++i)      for (int j = 0; j < n; ++j)        if (grid[i][j] == 'B')          box = {i, j};        else if (grid[i][j] == 'S')          player = {i, j};        else if (grid[i][j] == 'T')          target = {i, j};     // (boxX, boxY, playerX, playerY)    queue<tuple<int, int, int, int>> q{        {{box[0], box[1], player[0], player[1]}}};    vector<vector<vector<vector<bool>>>> seen(        m, vector<vector<vector<bool>>>(               n, vector<vector<bool>>(m, vector<bool>(n))));    seen[box[0]][box[1]][player[0]][player[1]] = true;     for (int step = 0; !q.empty(); ++step)      for (int sz = q.size(); sz > 0; --sz) {        const auto [boxX, boxY, playerX, playerY] = q.front();        q.pop();        if (boxX == target[0] && boxY == target[1])          return step;        for (int k = 0; k < 4; ++k) {          const int nextBoxX = boxX + kDirs[k % 4][0];          const int nextBoxY = boxY + kDirs[k % 4][1];          if (isInvalid(grid, nextBoxX, nextBoxY))            continue;          if (seen[nextBoxX][nextBoxY][boxX][boxY])            continue;          const int fromX = boxX + kDirs[(k + 2) % 4][0];          const int fromY = boxY + kDirs[(k + 2) % 4][1];          if (isInvalid(grid, fromX, fromY))            continue;          if (canGoTo(grid, playerX, playerY, fromX, fromY, boxX, boxY)) {            seen[nextBoxX][nextBoxY][boxX][boxY] = true;            q.emplace(nextBoxX, nextBoxY, boxX, boxY);          }        }      }     return -1;  }  private:  static constexpr int kDirs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};   // Returns true if (playerX, playerY) can go to (fromX, fromY).  bool canGoTo(const vector<vector<char>>& grid, int playerX, int playerY,               int fromX, int fromY, int boxX, int boxY) {    queue<pair<int, int>> q{{{playerX, playerY}}};    vector<vector<bool>> seen(grid.size(), vector<bool>(grid[0].size()));    seen[playerX][playerY] = true;     while (!q.empty()) {      const auto [i, j] = q.front();      q.pop();      if (i == fromX && j == fromY)        return true;      for (const auto& [dx, dy] : kDirs) {        const int x = i + dx;        const int y = j + dy;        if (isInvalid(grid, x, y))          continue;        if (seen[x][y])          continue;        if (x == boxX && y == boxY)          continue;        q.emplace(x, y);        seen[x][y] = true;      }    }     return false;  }   bool isInvalid(const vector<vector<char>>& grid, int playerX, int playerY) {    return playerX < 0 || playerX == grid.size() || playerY < 0 ||           playerY == grid[0].size() || grid[playerX][playerY] == '#';  }};

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