Problem Description

In this problem, we are given a square binary matrix grid where each cell is either 0 (representing water) or 1 (representing land). The task is to convert at most one 0 to 1, effectively "creating" additional land, and then determine the size of the largest "island" that results. An island is defined as a group of 1s that are connected horizontally or vertically (4-directionally). The size of an island is the number of 1s that make up the island. We are to return the size of the largest island after the conversion.

Flowchart Walkthrough

Let's analyze LeetCode 827. Making A Large Island using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The grid in this problem can be conceptualized as a graph, where each cell is a node and edges exist between adjacent cells with value 1.

Is it a tree?

• No: The grid may contain multiple connected components of 1s. There isn't a unique parent-child relationship that characterizes a tree.

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

• No: The problem involves finding the largest connected island by potentially flipping one cell value, and there are cycles in these connections.

Is the problem related to shortest paths?

• No: The objective is to find the largest connected component, not the shortest path between nodes.

Does the problem involve connectivity?

• Yes: The challenge involves determining the size of connected groups of cells and the effect of changing one cell on these connections.

Is the graph weighted?

• No: There are no weights associated with the cells; each cell is simply marked as either land (1) or water (0).

Conclusion: According to the flowchart, for an unweighted connectivity problem, Depth-First Search (DFS) is the suitable approach. DFS is effective for exploring and marking all nodes in a connected component, and it can efficiently compute the sizes of these components, which is crucial for solving this problem by trying to create the largest possible island.

Intuition

The intuition behind the solution is to first understand what constitutes an island and how we can measure its size. Since an island is a group of connected 1s, we can use Depth First Search (DFS), Breadth First Search (BFS), or Union-Find algorithm to identify all the islands and their sizes before making any changes to the grid.

The Union-Find algorithm is an efficient algorithm to group elements into disjoint sets and is particularly useful in this scenario for keeping track of which cells (land tiles) belong to which island. Following this, we initialize each cell as its own parent (representative of its own set), and if a cell is part of an island, we perform a union operation to link it with its neighboring island cells.

After doing this for the entire grid, we can now consider each 0 and simulate changing it to a 1. When we convert a 0 to a 1, we must look at its 4-directional neighbors. If these neighbors are part of different islands, we can potentially join those islands together, thereby increasing the size of the resulting island. Therefore, for each 0, we determine the unique islands of its neighbors, sum up their sizes (since converting the 0 to 1 would connect them), and then track the largest size achieved by this operation.

The key steps within this approach are to use the Union-Find algorithm to group the islands and then iterate over the grid to find the best place to convert a 0 to 1 to maximize the island size.

We return the size of the largest island after considering the conversion of one 0 to 1. Note that it’s important to handle special cases separately, such as when there is no 0 in the grid or when the entire grid is water.

Learn more about Depth-First Search, Breadth-First Search and Union Find patterns.

Solution Approach

The provided solution implements the Union-Find algorithm which is well-suited for dynamic connectivity problems like this one. The Union-Find algorithm helps to efficiently join islands and track their size throughout the process.

Here are the key elements of the implementation:

1. Initialization:

   • The solution starts by initializing two lists: p and size.
   • p is the parent list where p[x] represents the parent (or root) of element x. Initially, each cell is its own parent.
   • size is a list that represents the size of the tree (or island in this case) for which the index is the root. Therefore, size[x] is the size of the set that has x as its root. Initially, each cell is its own island with size 1.

2. Union Operation:

   • The union function takes two elements a and b, finds their respective roots, and combines the sets they belong to (if they are in different sets).
   • When union is performed, the size of the root of the second element b is increased by the size of the root of the first element a.
   • This is done by setting the parent of pa to pb and updating the size of pb to include the size of pa.

3. Find Operation:

   • The find function is used to find the root of the given element x. It does this by traversing the parent pointers until it reaches the root.
   • Path compression is used here: during this traversal, the parent of all visited elements is updated to the root, which flattens the structure and speeds up future find operations.

4. Processing the Grid:

   • The solution iterates over the grid to update parent and size lists based on the initial 1's locations. It uses union operations to merge adjacent lands into single islands.
   • This is achieved by checking the North ([-1, 0]) and West ([0, -1]) neighbors of each cell, as Eastern and Southern neighbors will be checked when the iteration reaches those cells.

5. Simulation of Changing 0 to 1:

   • For each cell with a 0, the solution checks the possibility of converting it to 1 and calculates the potential size of the resulting island.
   • It examines the 4-directional neighbors and if any are 1, it finds the root of that neighboring island. If the root is not already considered (not in vis set), its size is added to a temporary variable t.
   • After considering all neighbors, the potential largest island size (t) is compared with the current answer (ans), and ans is updated if t is larger.

6. Result:

   • After all grid positions have been assessed, the maximum value found is returned as the answer.

The algorithm ensures that even if multiple 0 cells are connected to the same island, the island's size is only counted once thanks to the usage of the set vis. This approach yields the size of the largest possible island that could be created by turning one 0 into a 1.

Example Walkthrough

Let's use a small 3x3 example grid to illustrate the solution approach:

Grid:
1 0 1
0 1 0
1 0 1

We will follow the key elements of the implementation:

1. Initialization: The parent list p and size list size are initialized:

   p = [0, 1, 2, 3, 4, 5, 6, 7, 8]    // index represents the cell, value represents the parent
   size = [1, 1, 1, 1, 1, 1, 1, 1, 1]  // Each cell is initially considered its own island of size 1

2. Union Operation: The grid is processed, and union operations are performed for adjacent land cells (i.e., cells with 1):

   • For cell (0,0), 1, it has no Northern or Western land neighbors so it remains a single-island cell.
   • For cell (0,2), 1, it has no Northern or Western land neighbors so it remains a single-island cell.
   • For cell (1,1), 1, it has land neighbors at (0,1) and (1,0) but those are 0, so no union operation is needed.
   • For cell (2,0), 1, it has no Northern or Western land neighbors so it remains a single-island cell.
   • For cell (2,2), 1, it has no Northern or Western land neighbors so it remains a single-island cell.

   After considering all land cells, our p and size lists remain unchanged as each land cell has no adjacent land cells to union with.

3. Find Operation: Not needed yet since we have not performed any unions and each cell is still its own parent.

4. Processing the Grid: While iterating through the grid, we found no adjacent islands so the p and size don't change.

5. Simulation of Changing 0 to 1: Now we try to change each 0 to 1 and calculate the potential size of larger islands:

   • For cell (0,1), if we change 0 to 1, it connects two islands at (0,0) and (0,2). The potential island size will be size[find(0)] + size[find(2)] + 1 = 1 + 1 + 1 = 3.

   • For cell (1,0), if we change 0 to 1, it connects the island at (0,0) and (1,1). The potential island size will be size[find(0)] + size[find(4)] + 1 = 1 + 1 + 1 = 3.

   • For cell (1,2), if we change 0 to 1, it connects the islands at (0,2) and (1,1). The potential island size will be size[find(2)] + size[find(4)] + 1 = 1 + 1 + 1 = 3.

   • For cell (2,1), if we change 0 to 1, it connects the islands at (2,0) and (2,2). The potential island size will be size[find(6)] + size[find(8)] + 1 = 1 + 1 + 1 = 3.

   No other 0 can improve the potential island size past 3. ans is updated accordingly during the process.

6. Result: The algorithm finishes evaluating all the cells where 0 could be converted to 1. The maximum potential island size remains 3, which is the final result returned by the algorithm.

This small example illustrates that by changing any single 0 to a 1 on the grid, we can achieve an island of maximum size 3, considering horizontal and vertical connections only.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code implements a union-find algorithm to solve the problem of finding the largest island after changing a 0 to a 1 on a grid.

• First, it initializes union-find structures such as the parent array p and size array size, which takes O(n^2) time since there are n * n elements in the grid.
• The union-find operations (find and union functions) are called for each cell connected to its neighboring cells. There are at most 2 calls for union per cell (for the top and left neighbours if they are part of an island), each taking effectively O(1) time on average due to path compression. So, this step costs O(n^2) in total.
• The last nested for loops check all cells again to calculate the size of the largest possible island by changing one 0 to a 1. For each 0, up to 4 neighboring cells are checked, and find is called to determine the unique root of each neighboring 1-cell. These find operations are also effectively O(1) on average, and the number of checks is 4 per cell at most, leading to another O(n^2) operation.

Considering these steps, the dominant term in the time complexity is O(n^2), with the constant factors from the union-find operations being relatively low due to path compression and union by size/rank heuristics.

The time complexity is \(\mathcal{O}(n^2)\).

Space Complexity

• The parent array p and size array size each require O(n^2) space.
• The vis set in the last nested for loops can contain at most 4 elements in any one iteration, however, it's cleared after considering a single cell, so it does not contribute significantly to the space complexity.
• The space taken by the stack due to recursive calls in the find function could be O(n^2) in the worst case without path compression, however with path compression, the depth of the recursion tree is greatly reduced. This also means the effective space complexity due to recursion is much less on average.

In conclusion, the primary space complexity is due to the storage of elements in the union-find structure: \(\mathcal{O}(n^2)\).

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