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 check string is palindrome or not with equivalent pairs in Python
Suppose we have a lowercase alphabet string called s and also have a list of pairs called 'pairs'. Each element in pairs has two strings [a, b] where the character 'a' and 'b' are considered same. If there are two pairs like [a, b] and [b, c], then we can say a and b are equivalent also b and c are equivalent, so a and c are also equivalent. And any value a or b is equivalent to itself. We have to check whether s is a palindrome or not with the given equivalence relations.
So, if the input is like s = "raceckt" pairs = [["r", "t"], ["a", "k"], ["z", "x"]], then the output will be True, because "a" = "k", and "r" = "t" so the string can be "racecar" which is palindrome.
To solve this, we will follow these steps −
- g := adjacency list of a graph where list may contain duplicate elements
- G := adjacency list of a graph where will not contain duplicate elements
- for each x, y in pairs, do
- insert x at the end of g[x]
- insert y at the end of g[y]
- insert y at the end of g[x]
- insert x at the end of g[y]
- Define a function dfs() . This will take a, so_far
- insert a into so_far
- for each elem in g[a], do
- if elem is not in so_far, then
- dfs(elem, so_far)
- if elem is not in so_far, then
- From the main method, do the following −
- for each key in g, do
- dfs(key, G[key])
- for i in range 0 to floor of(size of s / 2), do
- if s[i] is same as s[size of s -1-i] or (s[i] is in G[s[size of s - 1-i]] or s[-1 - i] in G[s[i]]) , then
- go for next iteration
- otherwise,
- return False
- if s[i] is same as s[size of s -1-i] or (s[i] is in G[s[size of s - 1-i]] or s[-1 - i] in G[s[i]]) , then
- return True
Example
Let us see the following implementation to get better understanding −
from collections import defaultdict def solve(s, pairs): g = defaultdict(list) G = defaultdict(set) for x, y in pairs: g[x].append(x) g[y].append(y) g[x].append(y) g[y].append(x) def dfs(a, so_far): so_far.add(a) for elem in g[a]: if elem not in so_far: dfs(elem, so_far) for key in g: dfs(key, G[key]) for i in range(0, len(s) // 2): if s[i] == s[-1 - i] or (s[i] in G[s[-1 - i]] or s[-1 - i] in G[s[i]]): continue else: return False return True s = "raceckt" pairs = [["r", "t"], ["a", "k"], ["z", "x"]] print(solve(s, pairs))
Input
"raceckt", [["r", "t"], ["a", "k"], ["z", "x"]]
Output
True
102 Views
