2123. Minimum Operations to Remove Adjacent Ones in Matrix

Problem Description

In this problem, we are provided with a binary matrix grid, where grid is comprised of 0s and 1s. The matrix is 0-indexed, meaning that both row and column indices start from 0. The problem requires us to perform operations on this matrix to achieve a specific condition known as "well-isolated." A binary matrix is said to be well-isolated if no 1 in the matrix is 4-directionally connected to another 1. In simpler terms, there should be no horizontally or vertically adjacent 1s in the entire matrix.

The operation we can perform is flipping any 1 in grid to become a 0. The problem poses the question of determining the minimum number of operations needed to make the grid well-isolated.

Intuition

To solve this problem, we need to determine pairs of adjacent 1s and then decide which 1 to flip in order to minimize the total number of operations. This leans towards a graph theoretical approach wherein we can model our 1s as vertices in a graph and the connections (adjacencies) as edges. The task then resembles a problem of finding a matching in the graph that covers the maximum number of vertices with the minimum number of edges – this is known as a maximum matching problem in a bipartite graph.

The solution uses adjacency lists (where each list contains indices of adjacent 1s), a matching array to keep track of the matchings, and a recursive function find() that tries to find augmenting paths and improve the existing matching. The indices are computed using the formula i * n + j to uniquely represent the cell at position (i, j) in a single integer, giving us a straightforward way to refer to specific cells in a linear fashion rather than as a 2D grid.

Here is how we arrive at the solution approach:

1. Build a graph where each cell containing a 1 is considered a vertex.
2. Connect vertices with edges if the corresponding 1s are horizontally or vertically adjacent.
3. Use a bipartite matching algorithm to find the maximum number of matched edges – these represent pairs of 1s that can be flipped in a single operation.
4. Since each matched edge represents two 1s that are adjacent, we flip one 1 in each matched pair.
5. The minimum number of operations required is then equal to the number of matched edges.

So, we iteratively look for augmenting paths using the find() function, which follows the Ford-Fulkerson method in the bipartite matching context. Each time we find a new augmenting path, we increment our answer. At the end of the process, the ans variable holds the minimum number of operations required to make the grid well-isolated.

Learn more about Graph patterns.

Solution Approach

The solution involves implementing a maximum matching algorithm on a bipartite graph. Here's a walk-through of the solution approach, incorporating algorithms, data structures, and design patterns:

1. Initialization: First, a graph g is created using the defaultdict from Python's collections module, to hold the adjacency lists. An array match is initialized to -1 for all the cells in the grid, representing that initially, no 1 is matched to any other.

2. Building the Graph:

   • Iterate through each cell (i, j) of the grid.
   • For each cell that contains a 1, compute its unique index x = i * n + j.
   • Check all four directions around cell (i, j) to find any adjacent 1s. For each adjacent 1, add the corresponding unique index to the adjacency list of x.

3. Maximum Matching:

   • Iterate through each vertex in the graph g (which corresponds to each cell containing a 1).
   • Use a vis set to keep track of visited vertices during the search for an augmenting path to avoid revisiting them.
   • Call the recursive function find(i).

4. Finding Augmenting Paths (find function):

   • For each adjacent vertex j of vertex i (corresponding to adjacent 1s):
     • Check if j was not visited already (to avoid cycles).
     • If match[j] is -1, it means j is unmatched and can be matched with i.
     • If match[j] is not -1, try to find an alternative path for the vertex matched with j (recursively calling find(match[j])).
     • If an augmenting path is found or j is unmatched, set match[j] to i and return 1.

5. Counting Operations:

   • For each explored vertex i, the recursive call of find(i) will return 1 if a new match is found, incrementing the minimum number of operations (stored in ans).

6. Result:

   • The variable ans now holds the total number of matches found, which is also the minimum number of operations needed to make the grid well-isolated. This is returned as the solution.

The main data structures used include a list for match that associates cells of the matrix which contain 1s with their matchings (or -1 if unmatched), and a defaultdict to create the adjacency list g. The set vis is used to facilitate the find function's tracking of visited vertices during each invocation. The designed approach leverages the concepts of bipartite matching and augmenting paths, employing a depth-first search-like algorithm to find matches.

The solution ensures that the operations are minimal since it leverages maximum matching—a fundamental concept ensuring the maximum number of connections with the minimum number of selections, which directly correlates to the minimal operational requirement of the given problem.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Consider the 3x3 binary matrix grid:

1grid = [
2    [1, 0, 0],
3    [0, 1, 1],
4    [0, 0, 1]
5]

Initialization

• Create an empty graph g using a defaultdict for adjacency lists.
• Initialize the match array to -1 for all cells, indicating no matches yet.

Building the Graph

• We traverse the grid to build our graph. Here are the steps for each cell:
  • (0, 0): It's a 1. Its index is 0 * n + 0 = 0. No adjacent 1s.
  • (1, 1): It's a 1. Its index is 1 * n + 1 = 4. Adjacent to (1, 2) which has index 5, so we add 5 to g[4].
  • (1, 2): It's a 1. Its index is 1 * n + 2 = 5. Adjacent to (1, 1) and (2, 2), so we add 4 to g[5] and 8 to g[5].
  • (2, 2): It's a 1. Its index is 2 * n + 2 = 8. Adjacent to (1, 2), so add 5 to g[8].

Now, the adjacency list for g looks like this:

1g: {
2    4: [5],
3    5: [4, 8],
4    8: [5]
5}

Maximum Matching

• We iterate over each vertex (4, 5, 8) in the graph g and attempt to find a match for each.
• Initialize an empty set vis before each call to find().

Finding Augmenting Paths (find function)

• For vertex 4, find(4) is called. Since g[4] = [5], we try to match it with 5.
  • 5 is not visited and match[5] is -1, so we set match[5] = 4 and find(4) returns 1.
• Next, for vertex 5, find(5) finds adjacent vertices 4 and 8.
  • 4 is already matched with 5.
  • 8 is not visited and match[8] is -1, so we set match[8] = 5 and find(5) returns 1.
• For vertex 8, find(8) returns 0 since its only adjacent vertex 5 is already matched with 8.

Counting Operations

• For each match found, we increment our answer ans. In our case, find(4) and find(5) both returned 1, so ans is 2.

Result

• We found two matches, which means we only need two operations to make the grid well-isolated. We can flip the 1s at the indices that correspond to any end of the matched pairs. In our matrix, we can flip 1 at (1, 1) and (1, 2), or (1, 2) and (2, 2).

The ans variable now holds the value 2, which is the minimum number of operations to make the grid well-isolated. This final result would then be returned by the algorithm.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the solution is determined by several factors:

1. Building the graph: The graph g is constructed by iterating over each cell in the grid. There are m * n cells, where m is the number of rows and n is the number of columns in the grid. Therefore, constructing the graph has a time complexity of O(m * n).

2. Finding matches: The function find attempts to find a match for each vertex. The worst-case scenario for this function is when it has to visit each of the m * n vertices once for each of the m * n vertices in g.keys(), leading to a complexity of O((m * n) * (m * n)), which can be simplified to O(m^2 * n^2).

Combining these two, the total time complexity of the minimumOperations function is dominated by the matching process and is O(m^2 * n^2).

Space Complexity

The space complexity is determined by the data structures used:

1. Graph representation: The graph g and the match array collectively take up O(m * n) space.

2. Visited set: The vis set can potentially hold all vertices in the worst case, thus taking O(m * n) space.

3. The recursion stack: The find function is a recursive function, and in the worst-case scenario, the depth of the recursive stack can be the number of vertices in the graph, which is O(m * n).

Thus, the total space complexity is O(m * n) as it is the largest term and other terms are not dependent on different variables.

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