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Program to Find Out the Special Nodes in a Tree in Python
Suppose we have a 2D list of values called 'tree' which represents an n-ary tree and another list of values called 'color'. The tree is represented as an adjacency list and its root is tree[0].
The characteristics of an i-th node −
tree[i] is its children and parent.
color[i] is its color.
We call a node N "special" if every node in the subtree whose root is at N has a unique color. So we have this tree, we have to find out the number of special nodes.
So, if the input is like tree = [ [1,2], [0], [0,3], [2] ]
colors = [1, 2, 1, 1], then the output will be 2.
To solve this, we will follow these steps −
result := 0
dfs(0, -1)
return result
Define a function check_intersection() . This will take colors, child_colors
if length of (colors) < length of (child_colors) , then
for each c in colors, do
if c in child_colors is non-zero, then
return True
otherwise,
for each c in child_colors, do
if c is present in child_colors, then
return True
Define a function dfs() . This will take node, prev
colors := {color[node]}
for each child in tree[node], do
if child is not same as prev, then
child_colors := dfs(child, node)
if colors and child_colors are not empty, then
if check_intersection(colors, child_colors) is non-zero, then
colors := null
otherwise,
if length of (colors) < length of (child_colors),then,
child_colors := child_colors OR colors
colors := child_colors
otherwise,
colors := colors OR child_colors
otherwise,
colors := null
if colors is not empty, then
result := result + 1
return colors
Example
Let us see the following implementation to get better understanding −
import collections
class Solution:
def solve(self, tree, color):
self.result = 0
def dfs(node, prev):
colors = {color[node]}
for child in tree[node]:
if child != prev:
child_colors = dfs(child, node)
if colors and child_colors:
if self.check_intersection(colors, child_colors):
colors = None
else:
if len(colors) < len(child_colors):
child_colors |= colors
colors = child_colors
else:
colors |= child_colors
else:
colors = None
if colors:
self.result += 1
return colors
dfs(0, -1)
return self.result
def check_intersection(self, colors, child_colors):
if len(colors) < len(child_colors):
for c in colors:
if c in child_colors:
return True
else:
for c in child_colors:
if c in colors:
return True
ob = Solution()
print(ob.solve( [
[1,2],
[0],
[0,3],
[2]
], [1, 2, 1, 1]))Input
[ [1,2], [0], [0,3], [2] ], [1, 2, 1, 1]
Output
2
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