Minimum Moves to Move a Box to Their Target Location - Solution & Explanation

This approach uses a BFS strategy, commonly used for exploring shortest paths, which also tracks states consisting of the player's and the box’s positions along with the number of pushes made. Each state can be represented as a tuple consisting of these values. The BFS runs until the box reaches its target position or all possible states are exhausted. To ensure efficiency, already visited states are stored and checked before exploration to prevent cyclic paths.

This Python solution uses a BFS approach to explore all possible ways the player can push the box to the target. It tracks the player's and the box's current positions and checks all possible moves. The BFS ensures the shortest path is found by exploring in layers. The main components include state validation, tracking visited states, and maintaining a queue for the BFS exploration.

This approach optimizes the BFS approach even further by using the A* search algorithm, which is a popular choice for pathfinding and graph traversal. In A*, each state is evaluated with a heuristic to estimate the minimum cost to reach the target. The heuristic can be the Manhattan distance from the box to the target which guides the search more directly towards optimal solutions more efficiently.

This C++ solution uses the A* algorithm which is advantageous over BFS by utilizing a heuristic approach. The heuristic guides the exploration to significantly reduce unnecessary steps by taking into consideration the Manhattan distance to the target. The state of the game is abstracted into a struct and processed through a priority queue, ensuring that the paths with the least cost are expanded first.

We consider the player's position and the box's position as a state, i.e., (s_i, s_j, b_i, b_j), where (s_i, s_j) is the player's position, and (b_i, b_j) is the box's position. In the code implementation, we define a function f(i, j), which maps the two-dimensional coordinates (i, j) to a one-dimensional state number, i.e., f(i, j) = i times n + j, where n is the number of columns in the grid. So the player and the box's state is (f(s_i, s_j), f(b_i, b_j)).

First, we traverse the grid to find the initial positions of the player and the box, denoted as (s_i, s_j) and (b_i, b_j).

Then, we define a double-ended queue q, where each element is a triplet (f(s_i, s_j), f(b_i, b_j), d), indicating that the player is at (s_i, s_j), the box is at (b_i, b_j), and d pushes have been made. Initially, we add (f(s_i, s_j), f(b_i, b_j), 0) to the queue q.

Additionally, we use a two-dimensional array vis to record whether each state has been visited. Initially, vis[f(s_i, s_j), f(b_i, b_j)] is marked as visited.

Next, we start the breadth-first search.

In each step of the search, we take out the queue head element (f(s_i, s_j), f(b_i, b_j), d), and check whether grid[b_i][b_j] = 'T' is satisfied. If it is, it means the box has been pushed to the target position, and now d can be returned as the answer.

Otherwise, we enumerate the player's next move direction. The player's new position is denoted as (s_x, s_y). If (s_x, s_y) is a valid position, we judge whether (s_x, s_y) is the same as the box's position (b_i, b_j):

• If they are the same, it means the player has reached the box's position and pushed the box forward by one step. The box's new position is (b_x, b_y). If (b_x, b_y) is a valid position, and the state (f(s_x, s_y), f(b_x, b_y)) has not been visited, then we add (f(s_x, s_y), f(b_x, b_y), d + 1) to the end of the queue q, and mark vis[f(s_x, s_y), f(b_x, b_y)] as visited.
• If they are different, it means the player has not pushed the box. Then we only need to judge whether the state (f(s_x, s_y), f(b_i, b_j)) has been visited. If it has not been visited, then we add (f(s_x, s_y), f(b_i, b_j), d) to the head of the queue q, and mark vis[f(s_x, s_y), f(b_i, b_j)] as visited.

We continue the breadth-first search until the queue is empty.

Note, if the box is pushed, the push count d needs to be incremented by 1, and the new state is added to the end of the queue q. If the box is not pushed, the push count d remains unchanged, and the new state is added to the head of the queue q.

Finally, if no valid push scheme is found, then return -1.

The time complexity is O(m^2 times n^2), and the space complexity is O(m^2 times n^2). Where m and n are the number of rows and columns in the grid, respectively.