Problem Description

You are given two strings s and t where:

• Every character appears exactly once in string s (no duplicates)
• String t is a permutation of string s (contains the same characters but possibly in a different order)

The permutation difference is calculated by:

1. For each character in the strings, find its position (index) in string s and its position in string t
2. Calculate the absolute difference between these two positions
3. Sum up all these absolute differences for all characters

Your task is to return this permutation difference value.

For example, if a character 'a' appears at index 2 in string s and at index 5 in string t, the contribution of 'a' to the permutation difference would be |2 - 5| = 3. You need to calculate this for all characters and return the total sum.

Intuition

Since we need to find the absolute difference between positions of the same character in two strings, the key insight is that we need a way to quickly look up where each character appears in string s.

Think about it this way: as we go through string t, for each character we encounter, we need to know "where was this character in string s?" This suggests we should preprocess string s to store the position of each character.

A hash table (dictionary) is perfect for this - we can map each character to its index in string s. Since each character appears exactly once in s, we don't have to worry about duplicate characters overwriting positions.

Once we have this mapping, the solution becomes straightforward:

1. Build a dictionary d where d[character] = index in s
2. Iterate through string t with its indices
3. For each character in t at position i, look up its position in s using our dictionary
4. Calculate |d[character] - i| and add it to our running sum

This approach is efficient because we only need to traverse each string once - once to build the dictionary from s, and once to calculate the sum while traversing t. The time complexity is O(n) where n is the length of the strings.

Solution Approach

The solution uses a hash table to efficiently map characters to their positions in string s.

Step 1: Build the position mapping

We create a dictionary d that stores the index of each character in string s:

d = {c: i for i, c in enumerate(s)}

This dictionary comprehension iterates through string s with enumerate(), which gives us both the index i and character c. For example, if s = "abc", the dictionary would be {'a': 0, 'b': 1, 'c': 2}.

Step 2: Calculate the permutation difference

We traverse string t and calculate the sum of absolute differences:

return sum(abs(d[c] - i) for i, c in enumerate(t))

This generator expression:

• Uses enumerate(t) to get each character c and its index i in string t
• Looks up the position of character c in string s using d[c]
• Calculates the absolute difference: abs(d[c] - i)
• The sum() function adds up all these absolute differences

Example walkthrough:

If s = "abc" and t = "bac":

1. Build dictionary: d = {'a': 0, 'b': 1, 'c': 2}
2. Process t:
   • Character 'b' at index 0 in t, at index 1 in s: |1 - 0| = 1
   • Character 'a' at index 1 in t, at index 0 in s: |0 - 1| = 1
   • Character 'c' at index 2 in t, at index 2 in s: |2 - 2| = 0
3. Total sum: 1 + 1 + 0 = 2

The solution has O(n) time complexity for building the dictionary and calculating the sum, and O(n) space complexity for storing the dictionary.

Example Walkthrough

Let's work through a concrete example with s = "acb" and t = "cab".

Step 1: Build the position mapping from string s

We create a dictionary that maps each character to its index in s:

• 'a' is at index 0
• 'c' is at index 1
• 'b' is at index 2

So our dictionary becomes: d = {'a': 0, 'c': 1, 'b': 2}

Step 2: Process string t and calculate differences

Now we traverse t = "cab" character by character:

• Index 0: Character 'c'

  • Position in t: 0
  • Position in s (from dictionary): d['c'] = 1
  • Contribution: |1 - 0| = 1

• Index 1: Character 'a'

  • Position in t: 1
  • Position in s (from dictionary): d['a'] = 0
  • Contribution: |0 - 1| = 1

• Index 2: Character 'b'

  • Position in t: 2
  • Position in s (from dictionary): d['b'] = 2
  • Contribution: |2 - 2| = 0

Step 3: Sum up all contributions

Total permutation difference = 1 + 1 + 0 = 2

This means the characters in t are collectively displaced by 2 positions compared to their original positions in s.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string s (or equivalently, the length of string t since they are permutations of each other). This is because:

• Building the dictionary d requires iterating through string s once: O(n)
• The sum comprehension iterates through string t once, and for each character, the dictionary lookup d[c] is O(1): O(n)
• Total: O(n) + O(n) = O(n)

The space complexity is O(|Σ|), where Σ is the character set. Since the problem involves permutations of the same characters, the dictionary d stores at most as many key-value pairs as there are unique characters in string s. For lowercase English letters, |Σ| ≤ 26, making the space complexity O(1) in practical terms.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Assuming Character Uniqueness Without Validation

The most common pitfall is trusting that the input strings truly have unique characters as stated in the problem. In real-world scenarios or modified problem variants, duplicate characters could cause incorrect results or runtime errors.

The Issue: If string s has duplicate characters, the dictionary comprehension {c: i for i, c in enumerate(s)} will only store the last occurrence of each character. This leads to incorrect position mappings.

Example of the problem:

s = "aba"  # 'a' appears twice
t = "baa"
# Dictionary becomes {'a': 2, 'b': 1} - first 'a' at index 0 is overwritten!

Solution: Add validation or handle duplicates explicitly:

class Solution:
    def findPermutationDifference(self, s: str, t: str) -> int:
        # Validate uniqueness (optional but safer)
        if len(s) != len(set(s)):
            raise ValueError("String s contains duplicate characters")
      
        char_to_index_in_s = {char: index for index, char in enumerate(s)}
        total_difference = sum(abs(char_to_index_in_s[char] - index) 
                              for index, char in enumerate(t))
        return total_difference

2. Missing Character in Translation

Another pitfall occurs when string t contains a character not present in string s, which would cause a KeyError when accessing char_to_index_in_s[char].

Example of the problem:

s = "abc"
t = "abd"  # 'd' is not in s
# This will raise KeyError: 'd'

Solution: Use .get() method with a default value or add validation:

class Solution:
    def findPermutationDifference(self, s: str, t: str) -> int:
        char_to_index_in_s = {char: index for index, char in enumerate(s)}
      
        # Option 1: Validate that t is truly a permutation
        if set(s) != set(t):
            raise ValueError("String t is not a permutation of string s")
      
        # Option 2: Use get() to handle missing keys gracefully
        # (though this changes the problem semantics)
        total_difference = sum(abs(char_to_index_in_s.get(char, index) - index) 
                              for index, char in enumerate(t))
      
        return total_difference

3. Performance Degradation with Large Strings

While the current solution is O(n), repeatedly accessing the dictionary in a tight loop can have overhead. For extremely large strings, consider whether the problem allows for optimization through vectorization or parallel processing.

Best Practice: The current solution is already optimal for the given constraints. However, always verify that the dictionary lookup approach is faster than alternative methods (like using s.index(char) directly) through benchmarking with your specific data sizes.