Problem Description

This problem asks you to determine if a string s matches a given pattern through a bijective mapping relationship.

The pattern consists of characters (like 'a', 'b', 'c'), and you need to map each unique character in the pattern to a non-empty substring of s. The mapping must satisfy two key conditions:

1. One-to-one mapping: Each character in the pattern must map to exactly one unique substring (no two different pattern characters can map to the same substring)
2. Consistent mapping: Once a pattern character is mapped to a substring, it must always map to that same substring throughout

When you replace each character in the pattern with its mapped substring, the concatenated result should exactly equal the string s.

For example:

• Pattern: "aba", String: "redredblue"

  • We could map: 'a' -> "red", 'b' -> "blue"
  • Replacing gives: "red" + "blue" + "red" = "redredblue" ✓
  • This is a valid match

• Pattern: "ab", String: "redred"

  • We cannot map both 'a' and 'b' to "red" (violates bijection)
  • No valid mapping exists

The solution uses backtracking with two pointers (i for pattern position, j for string position) to try different substring mappings. It maintains a dictionary d for character-to-substring mappings and a set vis to track used substrings, ensuring the bijective property is preserved.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: The problem does not involve nodes and edges or graph traversal. We're dealing with pattern matching between a string pattern and a target string.

Need to solve for kth smallest/largest?

• No: We're not looking for any kth smallest or largest element. We need to determine if a valid mapping exists.

Involves Linked Lists?

• No: The problem works with strings and pattern matching, not linked list data structures.

Does the problem have small constraints?

• Yes: Pattern matching problems typically have manageable constraints since we need to explore different possible mappings. Each character in the pattern needs to be mapped to substrings of s, and we need to try different combinations.

Brute force / Backtracking?

• Yes: We need to explore all possible ways to map pattern characters to substrings. When a mapping doesn't work, we need to backtrack and try different substring assignments. This is a classic backtracking scenario where we:
  • Try assigning a substring to a pattern character
  • Recursively check if the rest can be matched
  • If it fails, undo the assignment (backtrack) and try a different substring

Conclusion: The flowchart correctly leads us to use a Backtracking approach. This aligns perfectly with the solution, which uses DFS with backtracking to try different substring mappings, maintaining a dictionary for character-to-substring mappings and undoing failed attempts.

Intuition

The key insight is that we need to explore all possible ways to partition the string s and map these partitions to pattern characters. Since we don't know in advance which substrings should map to which pattern characters, we must try different possibilities systematically.

Think of it like solving a puzzle where you're trying to match pieces (substrings) to slots (pattern characters). You don't know which piece goes where, so you try different combinations until you find one that works or exhaust all possibilities.

The backtracking approach naturally emerges from this need to:

1. Make a choice: Pick a substring from the current position in s and try mapping it to the current pattern character
2. Check validity: Ensure the mapping respects our bijection rules (one-to-one correspondence)
3. Explore further: If valid, recursively try to match the rest
4. Backtrack if needed: If we hit a dead end, undo our choice and try a different substring

The brilliance lies in using two tracking mechanisms:

• A dictionary d to remember "pattern character → substring" mappings
• A set vis to ensure no two pattern characters map to the same substring

At each step, we face two scenarios:

• If the current pattern character already has a mapping, we check if the expected substring matches what we see in s
• If it's unmapped, we try assigning various substring lengths starting from the current position

The recursive nature with backtracking ensures we systematically explore all valid mapping combinations without missing any possibilities. We prune invalid branches early (like when remaining string length is less than remaining pattern length) to optimize the search.

Solution Approach

The solution implements a depth-first search with backtracking using two pointers to track positions in both the pattern and string.

Core Algorithm Structure:

The dfs(i, j) function explores whether we can match pattern[i:] with s[j:]:

1. Base Cases:

   • If both i == m and j == n: We've successfully matched everything, return True
   • If only one pointer reaches the end or n - j < m - i: Impossible to match remaining characters, return False

2. Try Different Substring Lengths: For each possible substring starting at position j, we iterate k from j to n-1, creating substring t = s[j:k+1].

3. Two Mapping Scenarios:

   Case 1: Pattern character already mapped

   • If d.get(pattern[i]) == t: The existing mapping matches our current substring
   • Continue with dfs(i + 1, k + 1) to match the rest

   Case 2: Pattern character not yet mapped

   • Check if pattern[i] not in d (unmapped) AND t not in vis (substring not used)
   • Create new mapping: d[pattern[i]] = t and mark substring as used: vis.add(t)
   • Recursively try dfs(i + 1, k + 1)
   • If unsuccessful, backtrack: d.pop(pattern[i]) and vis.remove(t)

Data Structures:

• d: Dictionary mapping pattern characters to their assigned substrings
• vis: Set tracking which substrings have been used (ensures bijection)
• m, n: Lengths of pattern and string respectively

Key Optimization: The condition n - j < m - i prunes branches where the remaining string is shorter than the remaining pattern, making it impossible to assign at least one character to each pattern symbol.

The algorithm systematically explores all valid partitioning possibilities through backtracking, ensuring we find a valid bijective mapping if one exists.

Example Walkthrough

Let's trace through the algorithm with pattern = "aba" and s = "catdogcat".

Initial State:

• d = {} (empty mappings)
• vis = set() (no substrings used)
• Start with dfs(0, 0) (i=0 points to 'a', j=0 points to 'c')

Step 1: dfs(0, 0) - Mapping first 'a'

• Pattern character 'a' is unmapped
• Try substring "c" (s[0:1]):
  • Map: d['a'] = "c", vis = {"c"}
  • Call dfs(1, 1)

Step 2: dfs(1, 1) - Mapping 'b'

• Pattern character 'b' is unmapped
• Try substring "a" (s[1:2]):
  • Map: d['b'] = "a", vis = {"c", "a"}
  • Call dfs(2, 2)

Step 3: dfs(2, 2) - Checking second 'a'

• Pattern character 'a' already mapped to "c"
• Check s[2:3] = "t" ≠ "c" (mismatch!)
• Return False, backtrack to Step 2

Step 2 (continued): Try longer substring for 'b'

• Try substring "at" (s[1:3]):
  • Map: d['b'] = "at", vis = {"c", "at"}
  • Call dfs(2, 3)

Step 4: dfs(2, 3) - Checking second 'a'

• Pattern character 'a' mapped to "c"
• Check s[3:4] = "d" ≠ "c" (mismatch!)
• Return False, continue trying longer substrings...

[After more backtracking...]

Eventually: dfs(0, 0) - Try longer substring for first 'a'

• Try substring "cat" (s[0:3]):
  • Map: d['a'] = "cat", vis = {"cat"}
  • Call dfs(1, 3)

Next: dfs(1, 3) - Mapping 'b'

• Try substring "dog" (s[3:6]):
  • Map: d['b'] = "dog", vis = {"cat", "dog"}
  • Call dfs(2, 6)

Final: dfs(2, 6) - Checking second 'a'

• Pattern character 'a' mapped to "cat"
• Check s[6:9] = "cat" ✓ (matches!)
• Call dfs(3, 9)

Base Case: dfs(3, 9)

• Both i=3 (end of pattern) and j=9 (end of string)
• Return True! ✓

Result: Found valid mapping: 'a' → "cat", 'b' → "dog"

The algorithm successfully backtracks through invalid mappings and finds the correct bijective mapping where "aba" becomes "cat" + "dog" + "cat" = "catdogcat".

Solution Implementation

Time and Space Complexity

Time Complexity: O(n^m) where n is the length of string s and m is the length of the pattern.

The analysis breaks down as follows:

• At each position i in the pattern, we try all possible substrings starting from position j in string s
• For each character in the pattern, we can potentially try up to n different substring lengths
• The recursion tree has a depth of at most m (the pattern length)
• At each level, we branch out to try different substring assignments
• In the worst case, we explore all possible ways to partition string s into m parts
• The number of ways to partition a string of length n into at most m parts is bounded by O(n^m)
• The pruning condition n - j < m - i helps eliminate some branches early, but doesn't change the worst-case complexity

Space Complexity: O(m + n)

The space usage comes from:

• The recursion call stack depth: O(m) in the worst case, as we recurse once for each pattern character
• The dictionary d stores at most m mappings (one per unique pattern character): O(m)
• The set vis stores at most m unique substrings from s: O(n) in the worst case, as each substring can be up to length n
• The substring creation s[j : k + 1] creates temporary strings: O(n) for the longest possible substring
• Overall space complexity is O(m + n)

Common Pitfalls

1. Forgetting to Check Bijection (One-to-One Mapping)

A critical mistake is only checking if a pattern character maps consistently to the same substring, while forgetting to ensure different pattern characters map to different substrings.

Incorrect Implementation:

def backtrack(pattern_idx, string_idx):
    # ... base cases ...
  
    for end_idx in range(string_idx, string_length):
        substring = s[string_idx:end_idx + 1]
      
        # WRONG: Only checking if pattern char is mapped correctly
        if current_pattern_char in pattern_to_string:
            if pattern_to_string[current_pattern_char] == substring:
                if backtrack(pattern_idx + 1, end_idx + 1):
                    return True
        else:
            # WRONG: Not checking if substring is already used by another pattern char
            pattern_to_string[current_pattern_char] = substring
            if backtrack(pattern_idx + 1, end_idx + 1):
                return True
            del pattern_to_string[current_pattern_char]

This would incorrectly accept pattern "ab" with string "redred" by mapping both 'a' and 'b' to "red".

Correct Solution: Always maintain a used_substrings set to track which substrings are already mapped, ensuring no two pattern characters can map to the same substring.

2. Incorrect Backtracking - Not Restoring State Properly

Incorrect Implementation:

# WRONG: Forgetting to remove from used_substrings during backtracking
if current_pattern_char not in pattern_to_string and substring not in used_substrings:
    pattern_to_string[current_pattern_char] = substring
    used_substrings.add(substring)
  
    if backtrack(pattern_idx + 1, end_idx + 1):
        return True
  
    del pattern_to_string[current_pattern_char]
    # MISSING: used_substrings.remove(substring)

This leaves the substring incorrectly marked as used, preventing valid mappings in subsequent branches of the search tree.

Correct Solution: Always restore both data structures (pattern_to_string and used_substrings) when backtracking to ensure the search space is properly explored.

3. Inefficient Pruning or Missing Early Termination

Incorrect Implementation:

def backtrack(pattern_idx, string_idx):
    if pattern_idx == pattern_length and string_idx == string_length:
        return True
    if pattern_idx == pattern_length or string_idx == string_length:
        return False
    # MISSING: Check if remaining string is too short for remaining pattern

Without checking string_length - string_idx < pattern_length - pattern_idx, the algorithm wastes time exploring branches where there aren't enough characters left in the string to assign at least one character to each remaining pattern symbol.

Correct Solution: Add the pruning condition early in the function to avoid unnecessary recursive calls and improve performance significantly for longer inputs.