Argthtjtr

Given an R x C grid of 1s and 0s (or True and False values), I need a function that can find the size of the largest connected component of 1s. For example, for the following grid,

grid = [[0, 1, 0, 1],

        [1, 1, 1, 0],

        [0, 1, 0, 0],

        [0, 0, 0, 1]]

The answer is 5.

Here is my implementation:

def largest_connected_component(nrows, ncols, grid):

    """Find largest connected component of 1s on a grid."""



    def traverse_component(i, j):

        """Returns no. of unseen elements connected to (i,j)."""

        seen[i][j] = True

        result = 1



        # Check all four neighbours

        if i > 0 and grid[i-1][j] and not seen[i-1][j]:

            result += traverse_component(i-1, j)

        if j > 0 and grid[i][j-1] and not seen[i][j-1]:

            result += traverse_component(i, j-1)

        if i < len(grid)-1 and grid[i+1][j] and not seen[i+1][j]:

            result += traverse_component(i+1, j)

        if j < len(grid[0])-1 and grid[i][j+1] and not seen[i][j+1]:

            result += traverse_component(i, j+1)

        return result



    seen = [[False] * ncols for _ in range(nrows)]



    # Tracks size of largest connected component found

    component_size = 0



    for i in range(nrows):

        for j in range(ncols):

            if grid[i][j] and not seen[i][j]:

                temp = traverse_component(i, j)

                if temp > component_size:

                    component_size = temp



    return component_size

Feel free to use the following code to generate random grids to test the function,

from random import randint



N = 20

grid = [[randint(0,1) for _ in range(N)] for _ in range(N)]

Problem: My implementation runs too slow (by about a factor of 3). Since I wrote this as a naive approach by myself, I am guessing there are clever optimizations that can be made.

Context: This is for solving the Gridception problem from Round 2 of Google Codejam 2018. My goal is to solve the problem in Python 3. As a result, there is a hard constraint of using only the Python 3 standard library.

I have figured out that this particular portion of the full solution is my performance bottleneck and thus, my solution fails to clear the Large Input due to being too slow.

Thank you so much!

Edit: Adding some timeit benchmarks

For a randomly generated 20 x 20 grid, my implementation takes 219 +/- 41 μs (a grid of 0s takes 30 μs, and a grid of 1s takes 380 μs).

In programming challenges, proper performances improvement usually come from a smarter algorithm. Unfortunately, I have no algorithm better than the one you have implemented.

I have found a single trick to shave off some time.

Remove all the logic around seen

In all places where you access elements from grid and seen, we have basically: if grid[pos] and not seen[pos].

An idea could be to update grid in place to remove seen elements from it. From an engineering point of view, it is not very nice: I would not expect an function computing the size of the biggest connected components to update the provided input. For a programming challenge, we can probably accept such a thing...

We'd get:

def largest_connected_component(nrows, ncols, grid):

    """Find largest connected component of 1s on a grid."""



    def traverse_component(i, j):

        """Returns no. of unseen elements connected to (i,j)."""

        grid[i][j] = False

        result = 1



        # Check all four neighbours

        if i > 0 and grid[i-1][j]:

            result += traverse_component(i-1, j)

        if j > 0 and grid[i][j-1]:

            result += traverse_component(i, j-1)

        if i < len(grid)-1 and grid[i+1][j]:

            result += traverse_component(i+1, j)

        if j < len(grid[0])-1 and grid[i][j+1]:

            result += traverse_component(i, j+1)

        return result



    # Tracks size of largest connected component found

    component_size = 0



    for i in range(nrows):

        for j in range(ncols):

            if grid[i][j]:

                temp = traverse_component(i, j)

                if temp > component_size:

                    component_size = temp



    return component_size

Another idea in order to do the same type of things without changing grid could be to store "positive" elements in a set. This also remove the need to check for edge cases of the grid. The great thing is that we can populate that set with less array accesses. This is still pretty hackish:

def largest_connected_component(nrows, ncols, grid):

    """Find largest connected component of 1s on a grid."""



    def traverse_component(pos):

        """Returns no. of unseen elements connected to (i,j)."""

        elements.remove(pos)

        i, j = pos

        result = 1



        # Check all four neighbours

        for new_pos in [(i-1, j), (i+1, j), (i, j-1), (i, j+1)]:

            if new_pos in elements:

                result += traverse_component(new_pos)

        return result



    # Tracks size of largest connected component found

    elements = set()



    for i, line in enumerate(grid):

        for j, cell in enumerate(line):

            if cell:

                elements.add((i, j))



    return max(traverse_component(pos) for pos in set(elements) if pos in elements)

Edit: rewriting the solution to avoid the copy of elements, we have a slightly faster solution:

def largest_connected_component(nrows, ncols, grid):

    """Find largest connected component of 1s on a grid."""



    def traverse_component(pos):

        """Returns no. of unseen elements connected to (i,j)."""

        i, j = pos

        result = 1



        # Check all four neighbours

        for new_pos in [(i-1, j), (i+1, j), (i, j-1), (i, j+1)]:

            if new_pos in elements:

                elements.remove(new_pos)

                result += traverse_component(new_pos)

        return result



    # Tracks size of largest connected component found

    elements = set((i, j) for i, line in enumerate(grid) for j, cell in enumerate(line) if cell)

    largest = 0

    while elements:

        pos = elements.pop()

        largest = max(largest, traverse_component(pos))

    return largest

Are you absolutely sure this is the bottleneck? Looking at the linked problem and the solution analysis, I'm not sure if this component is even needed to solve the given problem. It's possible your overall algorithm is inefficient in some way, but obviously I couldn't really tell unless I saw the whole program that this is part of.

@Josay's already given a good improvement, but in the grand scheme of things, this doesn't really shave off that much measurable time for larger grids. The original solution was a pretty good algorithm for solving the problem of largest connected subsections.

General comments

Having three lines here is unnecessary:

temp = traverse_component(i, j)

if temp > component_size:

    component_size = temp

because of one nice Python built-in, max:

component_size = max(component_size, traverse_component(i,j))

component_size could be named more descriptively as largest_size.