Syntax error Find the minimum number of preprocess moves required to make two strings equal in Python

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 −

 Live Demo

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
Updated on: 2020-08-20T08:15:59+05:30

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