Frog position after t seconds

We will create a adjacency list to represent the graph. We will then use a DFS to traverse the graph. We will keep track of the number of children each node has. If the node has no children, we will add the probability of reaching that node to the total probability. If the node has children, we will divide the probability by the number of children and continue the DFS.

class Solution:
    def frogPosition(self, n: int, edges: List[List[int]], t: int, target: int) -> float:
        if n == 1:
            return 1.0

        graph = collections.defaultdict(list)
        for s, d in edges:
            graph[s].append(d)
            graph[d].append(s)

        visited = set()

        def dfs(i, t):
            if i != 1 and len(graph[i]) == 1 or t == 0:
                return i == target

            visited.add(i)
            res = sum(dfs(j, t-1) for j in graph[i] if j not in visited)
            return res * 1.0 / (len(graph[i]) - (i!=1))

        return dfs(1, t)

Time complexity: O(n)
Space complexity: O(n)