Codility 'WireBurnouts' 2012 Psi Solution

Martin Kysel · December 22, 2014

Short Problem Definition:

While removing edges from a mesh grid, find the moment when there ceases to be a connection between opposite corners.



expected worst-case time complexity is O(N2\*log(N))

expected worst-case space complexity is O(N2)


This problem screams for the use of a union-find. The hard part was to identify that union-find has no effective delete operation and you first need to add all remaining wires and start from the end. I created a wrapper class to effectively pack the input data into a set. Parsing the whole input for each union candidate (the not in burnouts part) would result in O(N²xN²). By using a set (hash O(1) in Python, RB-tree O(log(N)) in C++) you are down to N² x log(N). The union operation is also guaranteed to be log(N).

Beware though. (0,0) is the bottom left corner. First coordinate is X (right) and second coordinate is Y(up). I, for some reason, am used for the first coordinate to describe the row(down).

You can view my complete task timeline here. How can someone solve this in 30 minutes? It took me longer to type the solution…



Union-find data structure. Based on Josiah Carlson's code,
with significant additional changes by D. Eppstein.

class UnionFind:
    """Union-find data structure.

    Each unionFind instance X maintains a family of disjoint sets of
    hashable objects, supporting the following two methods:

    - X[item] returns a name for the set containing the given item.
      Each set is named by an arbitrarily-chosen one of its members; as
      long as the set remains unchanged it will keep the same name. If
      the item is not yet part of a set in X, a new singleton set is
      created for it.

    - X.union(item1, item2, ...) merges the sets containing each item
      into a single larger set.  If any item is not yet part of a set
      in X, it is added to X as one of the members of the merged set.

    def __init__(self):
        """Create a new empty union-find structure."""
        self.weights = {}
        self.parents = {}

    def __getitem__(self, object):
        """Find and return the name of the set containing the object."""

        # check for previously unknown object
        if object not in self.parents:
            self.parents[object] = object
            self.weights[object] = 1
            return object

        # find path of objects leading to the root
        path = [object]
        root = self.parents[object]
        while root != path[-1]:
            root = self.parents[root]

        # compress the path and return
        for ancestor in path:
            self.parents[ancestor] = root
        return root

    def __iter__(self):
        """Iterate through all items ever found or unioned by this structure."""
        return iter(self.parents)

    def union(self, *objects):
        """Find the sets containing the objects and merge them all."""
        roots = [self[x] for x in objects]
        heaviest = max([(self.weights[r],r) for r in roots])[1]
        for r in roots:
            if r != heaviest:
                self.weights[heaviest] += self.weights[r]
                self.parents[r] = heaviest

class Wrapper:
    """ Wrapper class used purely for set hashing """
    def __init__(self, A, B, C):
        self.A = A
        self.B = B
        self.C = C

    def __eq__(self, other):
        return self.A == other.A and self.B == other.B and self.C == other.C

    def __hash__(self):
        return hash((self.A, self.B, self.C))

    def __str__(self):
        return "Wrapper: a is %s, b is %s, c is %s" % (self.A, self.B, self.C)

def solution(N, A, B, C):
    uf = UnionFind()
    grid_size = N*N

    burnouts = set()
    for idx in xrange(len(A)):
        burnouts.add(Wrapper(B[idx], A[idx], C[idx]))

    # the problem could be solved without an actual grid, using only an range index
    grid = range(grid_size + 2)

    # grid_size (N*N+1) represents a pseudo position before the first element
    # grid_size+1 (N*N+2) represents a pseudo position after the last element
    # when those two are connected, the current flows
    uf.union(grid[0], grid[grid_size])
    uf.union(grid[grid_size-1], grid[grid_size+1])

    # add all grids that did not burn out
    for left_right in xrange(N):
        for top_down in xrange(N):
            if top_down != N-1 and Wrapper(top_down, left_right, 0) not in burnouts:
                uf.union(grid[top_down*N+left_right], grid[(top_down+1)*N+left_right])
            if left_right != N-1 and Wrapper(top_down, left_right, 1) not in burnouts:
                uf.union(grid[top_down*N+left_right], grid[top_down*N+(left_right+1)])

    # as per problem specification
    if uf[grid[grid_size]] == uf[grid[grid_size+1]]:
        return -1

    # start adding burned out wires in a reversed manner
    for idx in reversed(xrange(len(C))):
        if C[idx] == 0:
            uf.union(grid[B[idx]*N+A[idx]], grid[(B[idx]+1)*N+A[idx]])
            uf.union(grid[B[idx]*N+A[idx]], grid[(B[idx])*N+A[idx]+1])

        if uf[grid[grid_size]] == uf[grid[grid_size+1]]:
            return idx+1

    # this should never happen
    return 0

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