# Check if a string is the typed name of the given name in Python

PythonServer Side ProgrammingProgramming

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