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Program to count number of unique paths that includes given edges in Python
Suppose we have a list of edges in the form (u, v) and these are representing a tree. For each edge we have to find the total number of unique paths that includes said edge, in the same order as given in the input.
So, if the input is like edges = [[0, 1],[0, 2],[1, 3],[1, 4]]
then the output will be [6, 4, 4, 4].
To solve this, we will follow these steps −
adj := adjacency list from given edges
count := an empty map
Define a function dfs() . This will take x, parent
count[x] := 1
for each nb in adj[x], do
if nb is same as parent, then
come out from the loop
count[x] := count[x] + dfs(nb, x)
return count[x]
From the main method do the following −
dfs(0, -1)
ans := a new list
for each edge (a, b) in edges, do
x := minimum of count[a] and count[b]
insert (x *(count[0] - x)) at the end of ans
return ans
Example
Let us see the following implementation to get better understanding −
from collections import defaultdict class Solution: def solve(self, edges): adj = defaultdict(list) for a, b in edges: adj[a].append(b) adj[b].append(a) count = defaultdict(int) def dfs(x, parent): count[x] = 1 for nb in adj[x]: if nb == parent: continue count[x] += dfs(nb, x) return count[x] dfs(0, -1) ans = [] for a, b in edges: x = min(count[a], count[b]) ans.append(x * (count[0] - x)) return ans ob = Solution() edges = [ [0, 1], [0, 2], [1, 3], [1, 4] ] print(ob.solve(edges))
Input
[ [0, 1], [0, 2], [1, 3], [1, 4] ]
Output
[6, 4, 4, 4]
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