Does the O space complexity (number_rows + number_cols) for the width search first Grid. This is an attempt to show my reasoning:
For example, the filling question is described here:
The fill algorithm using the first width search (queue) has a space complexity: O (number_rows + number_cols).
Why? Suppose you start from the upper left corner (or the coordinate (0,0)). By going to the right, we will have at most O (number_cols) in the queue. When we reach the end of the column, we can then begin to descend from the (0, 0) coordinate by giving O (number_cows + number_cols) in the queue.
Then, is it true that most of the questions for which we use a deep search in width on a grid lead to a spatial complexity of O (number_rows + number_cols). For example:
- Fill in the question above,
- Maze where you have to find the shortest
way from beginning to end,
- Search the number of islands (reference below)
But for 3) to find the number of islets, it seems that some people say that the time complexity is O (number_rows * number_cols) from https://stackoverflow.com/questions/50901203/dfs-and-bfs-time -and-space-complexities-of-number-of-islands-on-leetetcode
On the other hand, I would assume that the spatial complexity of dfs on a grid is O (number_rows * number_cols)
The questions are as follows on the basis of the above:
- Is the spatial complexity of width first a search on a grid: O (number_rows + number_cols)?
- Is the spatial complexity of the depth first searched on a grid: O (number_rows * number_cols)?