Isomorphic Strings

To solve this coding challenge of determining if two strings are isomorphic, the approach involves ensuring that there is a one-to-one correspondence between characters of the two strings while maintaining the order. We'll use two mappings - one to track the mapping from the first string to the second string, and another to track the reverse mapping from the second string to the first string.

Explanation

The problem provides two strings

s

and

t

, and we need to determine if they are isomorphic. Two strings are isomorphic if each character in

s

can map to a unique character in

t

and vice versa, maintaining the order and ensuring no two characters map to the same character in any direction.

Example Explanation

• Example 1 :
  s = "egg"
  ,
  t = "add"
• 'e' maps to 'a'
• 'g' maps to 'd' (both occurrences in 's' map to both occurrences in 't')
• Result: True
• Example 2 :
  s = "foo"
  ,
  t = "bar"
• 'f' maps to 'b'
• 'o' maps to 'a'
• 'o' cannot map to 'r' because it was already mapped to 'a'
• Result: False

Detailed Steps in Pseudocode:

1. Length Check : If the lengths of
   s
   and
   t
   are not the same, return
   False
   immediately since they can't be isomorphic.
2. Initialize Dictionaries : Use two dictionaries (or hash maps):
• mapping_s_to_t
  to store the mapping of characters from
  s
  to
  t
  .
• mapping_t_to_s
  to store the mapping of characters from
  t
  to
  s
  .
3. Character Mapping :
• Iterate through each character pair
  (char_s, char_t)
  in
  s
  and
  t
  .
• For each pair:
• Check if
  char_s
  is already mapped in
  mapping_s_to_t
  :
• If it is and the mapped value isn't
  char_t
  , return
  False
  .
• If it isn't, create the mapping.
• Similarly, check if
  char_t
  is already mapped in
  mapping_t_to_s
  :
• If it is and the mapped value isn't
  char_s
  , return
  False
  .
• If it isn't, create the mapping.
4. If all characters are processed successfully and the mappings are consistent, return
   True
   .

Pseudocode:

                                            
# Function to check if two strings are isomorphic
function isIsomorphic(string s, string t):

# Length check
if length of s is not equal to length of t:
return False

# Initialize mapping dictionaries
mapping_s_to_t = empty dictionary
mapping_t_to_s = empty dictionary

# Iterate through characters of both strings
for i from 0 to length of s - 1:
char_s = s[i]
char_t = t[i]

# Check mapping from s to t
if char_s is in mapping_s_to_t:
if mapping_s_to_t[char_s] is not equal to char_t:
return False
else:
mapping_s_to_t[char_s] = char_t

# Check mapping from t to s
if char_t is in mapping_t_to_s:
if mapping_t_to_s[char_t] is not equal to char_s:
return False
else:
mapping_t_to_s[char_t] = char_s

# If all checks passed, strings are isomorphic
return True

                                        

Step-by-Step Explanation:

5. Length Check :
• If
  s
  and
  t
  don’t have the same length, there is no way they can be isomorphic. This is the first, simple, and fast check.
6. Initialize Dictionaries :
• Prepare two dictionaries to keep track of the mappings in both directions. This ensures that no two characters in
  s
  map to the same character in
  t
  and vice versa.
7. Iteration and Mapping :
• Loop through each character in
  s
  and
  t
  at the same positions. This helps in comparing characters from both strings one-by-one.
• For each character, check if it already has a mapping in the respective dictionary:
• If it does, verify if it maps to the correct counterpart as per the requirement.
• If not, create the new mapping.
• Handle both mappings simultaneously to ensure that both strings respect the one-to-one character mapping requirement.
8. Final Check and Result :
• If all character mappings are verified consistently without conflicts, the strings are isomorphic, and we return
  True
  .
• Otherwise, if any conflict arises during the mapping process, return
  False
  immediately.

This approach ensures that the given strings are verified to be isomorphic or not efficiently and effectively.