Isomorphic Strings

We can determine if two strings are isomorphic by checking if there is a one-to-one correspondence between the characters in the two strings. We’ll use two dictionaries to keep track of this correspondence. Let’s break down the code step by step.

  1. Check the Length: If the lengths of the two strings are not the same, return False, as they cannot be isomorphic.

  2. Create Mappings: For each character in the two strings, create a mapping from characters in s to characters in t and a reverse mapping from characters in t to characters in s. This ensures that no two characters may map to the same character.

  3. Check Mappings: While creating the mappings, if we find a contradiction (e.g., a character in s is mapped to two different characters in t), then return False.

Here’s the code implementing this logic:

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class Solution(object):
    def isIsomorphic(self, s: str, t: str) -> bool:
        if len(s) != len(t):
            return False
        
        mapping_s_t = {}
        mapping_t_s = {}
        
        for char_s, char_t in zip(s, t):
            # Check if mapping for char_s exists, if not, add it
            if char_s not in mapping_s_t:
                mapping_s_t[char_s] = char_t
            else:
                # If a mapping exists, it should map to the same char_t
                if mapping_s_t[char_s] != char_t:
                    return False
            
            # Check reverse mapping for char_t
            if char_t not in mapping_t_s:
                mapping_t_s[char_t] = char_s
            else:
                if mapping_t_s[char_t] != char_s:
                    return False
        
        return True

Explanation:

  • The zip(s, t) function allows us to iterate over corresponding characters in both strings.
  • The mapping_s_t dictionary keeps track of how characters in s map to characters in t.
  • The mapping_t_s dictionary keeps track of how characters in t map to characters in s.
  • If a contradiction is found in either of the mappings, we return False.
  • If no contradictions are found, we return True.

This solution has a time complexity of O(n), where n is the length of the input strings, and a space complexity of O(c), where c is the number of unique characters in the strings.

“Isomorphic Strings” aligns well with “Word Pattern”. In both problems, we need to map each character of one string to the corresponding character of another string, ensuring that the mapping is a bijection.

In “Isomorphic Strings”, you’re given two strings s and t and asked to determine if they are isomorphic. Two strings are isomorphic if the characters in s can be replaced to get t. All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character, but a character may map to itself.

In “Word Pattern”, you’re given a pattern and a string str, you have to determine if str follows the same pattern. Here, bijection correspondence occurs between a letter in pattern and a non-empty word in str.

The core idea of character mapping is shared between these two problems. Thus, if you can solve one problem, it is highly likely that you can solve the other with minor adjustments. Regarding simplicity, both problems share similar complexity and problem structure, though “Word Pattern” might be slightly simpler due to dealing with whole words instead of individual characters.

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class Solution(object):
    def isIsomorphic(self, s: str, t: str) -> bool:
        map1 = []
        map2 = []
        for idx in s:
            map1.append(s.index(idx))
        for idx in t:
            map2.append(t.index(idx))
        if map1 == map2:
            return True
        return False

Problem Classification

The problem falls under the classification of:

  • String Processing: The problem involves iterating over characters of two strings, s and t.

  • Mapping: The task is to determine if there’s a one-to-one correspondence (isomorphism) between the characters in the two strings.

  • Pattern Recognition: The aim is to identify a pattern in how the characters in two strings relate to each other.

This could be classified as a string processing problem involving mapping and pattern recognition.

Language Agnostic Coding Drills

  1. Understanding the problem: The problem requires us to determine whether two strings are isomorphic, which means each character in one string can be replaced to get the second string.

  2. Manipulating Strings: The characters of the string are accessed one by one. Drill: Write a code to iterate over a string and print each character.

  3. Understanding and Using List Data Structure: A list is used to map the first occurrence of a character in the string. Drill: Create a list and perform basic operations like addition and access of elements.

  4. Understanding the concept of Indexing in Strings: The index of the first occurrence of a character in a string is stored in the list. Drill: Write a function to find the index of the first occurrence of a character in a string.

  5. Building the mapping: For each string, create a list where each character’s index is stored at the corresponding position. Drill: Write a function that takes a string and returns a list with the index of the first occurrence of each character in the string.

  6. Comparing lists: Compare the two lists and if they are equal, return true, else return false. Drill: Write a function that takes two lists as input and returns true if they are equal, else returns false.

The overall problem-solving approach would be to first understand the concept of isomorphic strings and how the first occurrence index of characters can be used to identify if two strings are isomorphic. Next, for each string, we create a mapping of the index of the first occurrence of each character in that string. Finally, if both mappings are equal, we can say that the two strings are isomorphic.

Targeted Drills in Python

Drill 1: Write a code to iterate over a string and print each character.

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def print_chars(s):
    for char in s:
        print(char)

print_chars("hello")

Drill 2: Create a list and perform basic operations like addition and access of elements.

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def list_operations():
    my_list = [1, 2, 3]
    print(my_list)
    my_list.append(4)
    print(my_list)
    print(my_list[2])

list_operations()

Drill 3: Write a function to find the index of the first occurrence of a character in a string.

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def first_occurrence(s, char):
    return s.index(char)

print(first_occurrence("hello", "l"))

Drill 4: Write a function that takes a string and returns a list with the index of the first occurrence of each character in the string.

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def build_map(s):
    char_map = [s.index(char) for char in s]
    return char_map

print(build_map("hello"))

Drill 5: Write a function that takes two lists as input and returns true if they are equal, else returns false.

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def compare_lists(list1, list2):
    return list1 == list2

print(compare_lists([1, 2, 3], [1, 2, 3]))
print(compare_lists([1, 2, 3], [1, 2, 4]))

After understanding these drills, you can integrate them to solve the problem in a more structured way.