Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Program to find length of longest possible stick in Python?
Suppose we have a list of integers sticks. Here each element in the list represents a stick with two ends, these values are between 1 and 6. These are representing each end. We can connect two sticks together if any of their ends are same. The resulting stick's ends will be the leftover ends and its length is increased. We have to find the length of the longest stick possible.
So, if the input is like sticks = [ [2, 3], [2, 4], [3, 5], [6, 6] ], then the output will be 3, as we can connect [2, 3] and [2, 4] to get [3, 4], which we can connect it with [3, 5] to get [4, 5].
To solve this, we will follow these steps:
Define a function dfs() . This will take node, edge_idx, and a set visited
if edge_idx is not null, then
if edge_idx is visited, then
return 0
mark edge_idx as visited
res := 0
for each e_idx in g[node], do
n_node := sticks[e_idx, 0] when sticks[e_idx, 1] is same as node otherwise sticks[e_idx, 1]
res := maximum of res and 1 + dfs(n_node, e_idx, visited)
if edge_idx is non-zero, then
delete edge_idx from visited
return res
From the main method do the following:
sticks := a list for all s in sticks (s[0], s[1])
vertices := a new set
g := an empty map
for each index i and edge in sticks, do
insert i into g[edge[0]]
insert i into g[edge[1]]
insert edge[0] and edge[1] into vertices
res := 0
for each v in vertices, do
res := maximum of res and dfs(v, null, and empty set)
return res - 1
Let us see the following implementation to get better understanding:
Example
from collections import defaultdict class Solution: def solve(self, sticks): def dfs(node, edge_idx, visited): if edge_idx is not None: if edge_idx in visited: return 0 visited.add(edge_idx) res = 0 for e_idx in g[node]: n_node = sticks[e_idx][0] if sticks[e_idx][1] == node else sticks[e_idx][1] res = max(res, 1 + dfs(n_node, e_idx, visited)) if edge_idx: visited.remove(edge_idx) return res sticks = [(s[0], s[1]) for s in sticks] vertices = set() g = defaultdict(set) for i, edge in enumerate(sticks): g[edge[0]].add(i) g[edge[1]].add(i) vertices.add(edge[0]) vertices.add(edge[1]) res = 0 for v in vertices: res = max(res, dfs(v, None, set())) return res - 1 ob = Solution() sticks = [ [2, 3], [2, 4], [3, 5], [6, 6] ] print(ob.solve(sticks))
Input
sticks = [ [2, 3], [2, 4], [3, 5], [6, 6] ]
Output
3
167 Views
