Problem Description

The problem presents a 3x3 matrix grid, where each cell of the grid contains a certain number of stones, with a total of 9 stones distributed across the matrix. Here, it's possible to have multiple stones in a single cell. The task is to move the stones such that there is exactly one stone in each cell. A move consists of transferring a stone from one cell to any adjacent cell (one that shares a side).

To solve this problem, we need to determine the minimum number of moves necessary to achieve the goal where each cell contains just one stone.

Flowchart Walkthrough

Let's dive into the LeetCode problem 2850. Minimum Moves to Spread Stones Over Grid using the Flowchart. Here’s a step-by-step analysis to determine the appropriate algorithm:

Is it a graph?

• Yes: The grid can be seen as a graph where each cell is a node and stones can move between adjacent cells, forming edges.

Is it a tree?

• No: The graph does not have hierarchical structure typical of trees, and multiple paths may exist between stones, allowing cycles.

Is the problem related to directed acyclic graphs (DAGs)?

• No: The problem is more focused on spreading stones uniformly rather than operations typically associated with DAGs, like topological sorting.

Is the problem related to shortest paths?

• Yes: Essentially, moving stones to ensure each cell has only one can be interpreted as ensuring each stone reaches an optimal position (node) using the minimum number of moves, resembling a shortest-path problem in this unique context.

Is the graph weighted?

• No: Each move of a stone from one cell to an adjacent one can be considered as having equal "cost" or weight.

Conclusion: Following the flowchart, the focus on an unweighted shortest-path problem without specific small constraints leads us to utilize BFS. This algorithm is suitable for exploring the grid to find minimum moves while maintaining equal distribution because BFS efficiently handles unweighted graphs to ensure level-by-level expansion, pertinent for equidistant spread of stones.

This approach using BFS will help in evenly spreading stones over the grid ensuring each move reaches the optimal cell in terms of distance and distribution.

Intuition

Determining the minimum number of moves required to distribute the stones one per cell is akin to solving a puzzle where we move pieces on a board to reach an end configuration.

We recognize that each move can only transfer one stone to an adjacent cell. This move is effectively a displacement by a certain "distance". We can define this distance as the total number of vertical and horizontal steps required to move a stone from its initial position to the target cell.

One way to approach this problem is through bit masking and dynamic programming, which can manage the permutations of moves to ensure each stone ends in a unique cell.

Here's the intuition behind the solution:

1. We need to pair each starting point with a unique endpoint; that is, each initial stone's location (where stones are more than one) with one of the empty cells.

2. To optimize the process, we can use a bit mask to represent each stone's possible destination cell choices. This provides a means to calculate the minimum number of moves cal(a, b) which uses the Manhattan distance (the sum of the absolute differences of their Cartesian coordinates).

3. By iterating through all possible distributions of stones to open cells, we keep track of the minimum number of moves required to reach each distribution from the starting point, using dynamic programming.

4. The state f[i] in the dynamic array represents the minimum number of moves required to achieve the distribution encoded by the bitmask i.

5. We iterate over all possible combinations of distributions of stones to empty cells ((1 << n) represents all combinations for n stones), calculating and updating the minimum number of moves for each combination.

In the given code, cal function calculates the Manhattan distance between a source and destination cell, left array holds the coordinates of empty cells, right array holds the coordinates of cells with more than one stone, and n is the number of empty cells. f is the dynamic programming array that holds the minimum moves required and bit_count is a method that counts the number of set bits (1 bits) in the binary representation of a number, which in this context, represents how many stones have been placed in their unique position so far.

The final answer is then the value of the last element in the dynamic programming array f, which represents the minimum number of moves required to reach the end configuration with unique stones.

Learn more about Breadth-First Search and Dynamic Programming patterns.

Solution Approach

To solve the minimum number of moves problem for distributing stones on a 3x3 grid, the Reference Solution Approach employs dynamic programming alongside bit masking to efficiently calculate the minimum moves required. Here's a walkthrough of how the implementation translates to an algorithm:

1. Data Structures Used:

   • left list: Stores the coordinates of empty cells (cells that need stones).
   • right list: Stores the coordinates corresponding to the position of each stone that needs to move.
   • A dynamic array f indexed by bitmasks representing subsets of stones moved.

2. Dynamic Programming (DP) State:

   • f[i]: Minimum number of moves needed to move k stones to the position dictated by the subset represented by the binary digits of i.

3. Bit Masking:

   • Each bit in a bitmask represents whether a stone from the right list has been moved to the corresponding cell in the left list (an empty position).
   • The 1 << n operation is used to iterate over all possible subsets of stones, where n represents the number of stones needing to be moved.
   • i.bit_count() gives the number of stones that have already been moved according to bitmask i.

4. Algorithm Steps:

   • Initialize the dynamic array f with infinity (inf) since initially, the number of moves required for each subset is unknown.
   • Set f[0] = 0 because no moves are required when no stones have been moved.
   • Iterate over all possible subsets of stone movements. For each subset i, iterate over all the positions j that could be the last move in the subset.
   • Update the dynamic programming state f[i] if moving a stone from position j in the right list to the position k-1 in the left list results in fewer moves. This is done using the Manhattan distance calculation cal(left[k - 1], right[j]).

5. Function cal(a, b):

   • This function calculates the Manhattan distance between two points, a and b, which is the sum of the horizontal and vertical distances and is computed as |a[0] - b[0]| + |a[1] - b[1]|.

6. DP Transition:

   • For each state i, we calculate the stone to be placed by counting the number of bits set in i (how many stones are already placed). We call this k.
   • Transition from every state i ^ (1 << j) (where i ^ (1 << j) means the state before placing the j-th stone) to state i, and the cost of this transition is the Manhattan distance between the stone's initial and target positions.

7. Final Step:

   • The answer to the problem is in f[-1], which represents the minimum number of moves required after all stones are placed in their correct positions, taking into account all the permutations and combinations.

By implementing bit masking and intelligent dynamic programming using Manhattan distances, this approach effectively solves a potentially computationally expensive combinatorial problem in a more manageable and optimized way.

Example Walkthrough

Let's say we have a 3x3 grid with the following distribution of stones:

2 0 1
1 0 1
0 3 2

In the grid, 2, 1, 0, 3, etc., represent the number of stones in each cell. Empty cells are our target cells where we need to move stones. Let's apply the steps described in the Solution Approach to this grid:

1. Data Structures Initialization:

   • left = [(0, 1), (1, 1), (2, 0)] - coordinates of empty cells.
   • right = [(0, 0), (2, 1), (2, 2), (0, 2), (1, 0)] - coordinates of cells with stones to move (each coordinate may repeat based on the number of stones).
   • f = [inf, inf, inf, ..., 0, inf, inf] - dynamic array initialized with infinity, except f[0].

2. DP Initialization:

   • We start with f[0] = 0 because if no stones are moved, no moves are needed.

3. Iterate over Subsets:

   • Now, we iterate over all combinations of moves. For example, when we consider a subset where only the first stone from the right list has been moved, the bitmask would be 00001 (assuming we have 5 stones to move).

4. Calculating Moves:

   • Suppose we are checking the subset where we have moved the first and third stones, according to our example lists. The bitmask is 00101.
   • Now we want to place the fourth stone. We check the cost of moving it to all empty positions while considering the already moved stones.

5. Distance Calculation with cal:

   • To update f[00101], we compute the Manhattan distances from the fourth stone's position (2, 2) to all possible empty positions. We estimate the cost for each possible move and keep the minimum.

6. DP Transition:

   • We calculate the new state by setting the bit corresponding to the fourth stone, resulting in a new bitmask, say 01101.
   • We update f[01101] if the cost of moving the fourth stone added to f[00101] is less than the current value of f[01101].

7. Repeat Steps:

   • We iterate the above steps, updating the DP array f until all stones are considered.

8. Final Step:

   • After considering all subsets and moves, f[-1] (or f[(1 << n) - 1] if n is the total number of stones) will give us the minimum number of moves required to achieve our goal. For our example, if n = 5, then f[11111] represents the state where all 5 stones have been moved to empty cells.

This example walkthrough provides an illustration of how the solution approach works step-by-step to calculate the minimum number of moves needed to distribute stones on the 3x3 grid using bit masking and dynamic programming.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code is O(n * 2^n). The reason for this time complexity is due to the combination of iterating over all subsets of "left" and calculating the minimum distances against all elements in "right". Specifically:

• There are n cities on the left side, resulting in 2^n subsets due to the binary representation used to enumerate these subsets.
• For each subset (represented by i), we use bit_count() which contributes O(n) as it counts the number of set bits in the binary representation of the integer.
• The inner loop runs n times for each subset to calculate distances and find the minimum after excluding elements using the XOR operation i ^ (1 << j).
• Each call to cal() function is constant time, however, it's called n times in the worst case.

Hence, combining these factors, we end up with O(n * 2^n).

Space Complexity

The space complexity of the code is O(2^n). The reasons are as follows:

• We maintain an array f of size 2^n that keeps track of the minimum distance for every subset of cities on the "left".
• The left and right lists have a linear space complexity based on the input size, in the worst case it would be O(n), which is eclipsed by the space needed for f.
• Auxiliary space used by the recursion stack for bit_count() or temporary variables in the loops are constant in comparison to the space used by f.

Overall, the dominant factor here is the f array, hence the space complexity is O(2^n).

Learn more about how to find time and space complexity quickly using problem constraints.