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
Find the minimum number of preprocess moves required to make two strings equal in Python
Suppose we have two strings P and Q of same lengths with only lower case letters, we have to count the minimum number of pre-processing moves on string P are needed to make it equal to string Q after applying below operations −
Select any index i and swap characters pi and qi.
Select any index i and swap characters pi and pn – i – 1.
Select any index i and swap characters qi and qn – i – 1.
Note − The value of i in range (0 ≤ i < n)
In one move we can change a character in P with any other character of English alphabet.
So, if the input is like P = “pqprpqp”, Q = “qprpqpp”, then the output will be 4 as if we set P0 = ‘q’, P2 = ‘r’, P3 = ‘p’ and P4 = ‘q’ and P will be “qqrpqqp”. After that we can get equal strings by the following sequence of operations: swap(P1, Q1) and swap(P1, P5).
To solve this, we will follow these steps −
n := size of P
res := 0
for i in range 0 to n / 2, do
my_map := a new map
my_map[P[i]] := 1
if P[i] is same as P[n - i - 1], then
my_map[P[n - i - 1]] := my_map[P[n - i - 1]] + 1
if Q[i] is in my_map, then
my_map[Q[i]] := my_map[Q[i]] + 1
otherwise,
my_map[Q[i]] := 1
if Q[n - i - 1] is in my_map, then
my_map[Q[n - 1 - i]] := my_map[Q[n - 1 - i]] + 1
otherwise,
my_map[Q[n - 1 - i]] := 1
size := size of my_map
if size is same as 4, then
res := res + 2
otherwise when size is same as 3, then
res := res + 1 + (1 when (P[i] is same as P[n - i - 1]), otherwise 0)
otherwise when size is same as 2, then
res := res + my_map[P[i]] is not same as 2
if n mod 2 is same as 1 and P[n / 2] is not same as Q[n / 2], then
res := res + 1
return res
Example
Let us see the following implementation to get better understanding −
def count_preprocess(P, Q): n = len(P) res = 0 for i in range(n // 2): my_map = dict() my_map[P[i]] = 1 if P[i] == P[n - i - 1]: my_map[P[n - i - 1]] += 1 if Q[i] in my_map: my_map[Q[i]] += 1 else: my_map[Q[i]] = 1 if Q[n - i - 1] in my_map: my_map[Q[n - 1 - i]] += 1 else: my_map[Q[n - 1 - i]] = 1 size = len(my_map) if (size == 4): res += 2 elif (size == 3): res += 1 + (P[i] == P[n - i - 1]) elif (size == 2): res += my_map[P[i]] != 2 if (n % 2 == 1 and P[n // 2] != Q[n // 2]): res += 1 return res A = "pqprpqp" B = "qprpqpp" print(count_preprocess(A, B))
Input
"pqprpqp", "qprpqpp"
Output
4
172 Views
