Check If a String is the Typed Name in Python

Suppose we have two lowercase strings s and t. Sometimes, when we type a vowel, the key might get long pressed, and the vowel will be repeated 1 or more times. We have to check whether it is possible that t is typed that indicates s or not.

So, if the input is like s = "mine" t = "miiine", then the output will be True as vowel 'i' is repeated three times, other letters are fine.

To solve this, we will follow these steps −

• s_len := size of s
• t_len := size of t
• j := 0
• for i in range 0 to s_len - 1, do
  • if s[i] is not same as t[j], then
    • return False
  • if s[i] is not vowel, then
    • j := j + 1
    • go for next iteration
  • cnt_1 := 1
  • while i < s_len - 1 and s[i] is same as s[i + 1], do
    • cnt_1 := cnt_1 + 1
    • i := i + 1
  • cnt_2 := 1
  • while j < t_len - 1 and t[j] is same as s[i], do
    • cnt_2 := cnt_2 + 1
    • j := j + 1
  • if cnt_1 > cnt_2, then
    • return False
• return True

Let us see the following implementation to get better understanding −

Example

 Live Demo

def isVowel(c):
   vowel = "aeiou"
   return c in vowel
def solve(s, t):
   s_len = len(s)
   t_len = len(t)
   j = 0
   for i in range(s_len):
      if s[i] != t[j]:
         return False
      if isVowel(s[i]) == False:
         j = j + 1
         continue
      cnt_1 = 1
      while i < s_len - 1 and (s[i] == s[i + 1]):
         cnt_1 = cnt_1 + 1
         i = i + 1
      cnt_2 = 1
      while j < t_len - 1 and t[j] == s[i]:
         cnt_2 = cnt_2 + 1
         j = j + 1
      if cnt_1 > cnt_2:
         return False
   return True
s = "mine"
t = "miiine"
print(solve(s, t))

Input

"mine", "miiine"

Output

True