Graph walking in python - detailed explanation with video

  • 0

    This is deeply explained in this youtube video, but the summary is:

    Given a / b = 2.0, b / c = 3.0 we can write following equations:

    • a = 2b
    • b = 3c
    • b = .5a
    • c = 1/3*b

    This can be represented as a hash-of-hashes:

                'a' : {
                    'b' : 2
                'b' : {
                    'a' : .5,
                    'c' : 3,
                'c' : {
                    'b' : .333333,

    And the problem becomes easier, since now we can walk thru a "virtual graph", where nodes are a, b, c, and edges have weights 2, .5, 3, .33333.
    The path shouldn't be exactly the shortest, just any.
    So we can write a recursive function which calculates the "cost" of a path a~~~>b by checking following 3 cases:

    • a=b => cost is 1 if a is b
    • a->b => cost is hash[a][b] if a is connected to b directly
    • a->key~~~>b => cost is hash[a][key] * recursive_call(key, b) if a is connected to key, which hash a path to b

    During the implementation of this recursive function, we'd have to use seen set to avoid loops.

    Special cases, like a/a, x/x, a/e (where a is known, but x and e are not) we can do extra checks to cut off unnecessary recursive calls.

    Final solution:

    from collections import defaultdict
    class Solution(object):
        def calcEquation(self, equations, values, queries):
            self.hash = self.prepHash(equations, values)
            return [self.solve(x[0], x[1]) for x in queries]
        def prepHash(self, equations, values):
            hash = defaultdict(lambda: {})
            for i in xrange(len(values)):
                eq = equations[i]
                val = values[i]
                hash[eq[0]][eq[1]] = 0.0+val
                hash[eq[1]][eq[0]] = 1/(0.0+val)
            return hash
        def solve(self, a, b):
            res = self.rec(a, b, set([a]))
            if res is None:
                return -1.0
            return res
        # solves a/b
        # a ~~~~> b
        # a=b
        # a->b
        # a->key~~~>b
        def rec(self, a, b, seen):
            if a == b:
                return 1.0 if a in self.hash else None
            if a not in self.hash or b not in self.hash:
                return None
            if b in self.hash[a]: 
                return self.hash[a][b]
                for key in self.hash[a]:
                    if key not in seen:
                        res = self.rec(key, b, seen)
                        if res is not None:
                            return self.hash[a][key] * res
            return None

    I hope that the video was not too boring and long, but I made it to explain the whole process of solving this task as if you were coding on a whiteboard. This can be helpful for those who goes to onsite interviews soon. Anyways, please, leave comments here or below the video if you have them - I'll answer and improve! Good luck!

Log in to reply

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.