Problem Description

This problem asks you to determine if two strings are exactly one edit distance apart. Two strings are considered one edit distance apart if you can transform one string into the other by performing exactly one of these operations:

1. Insert one character: Add exactly one character at any position in string s to make it equal to string t. For example, "ab" → "adb" (insert 'd').

2. Delete one character: Remove exactly one character from string s to make it equal to string t. For example, "abc" → "ac" (delete 'b').

3. Replace one character: Change exactly one character in string s to a different character to make it equal to string t. For example, "abc" → "adc" (replace 'b' with 'd').

The key requirements are:

• Exactly one edit operation must be performed (not zero, not more than one)
• For replacement, the character must be changed to a different character (not the same)
• Return true if the strings are one edit distance apart, false otherwise

Examples of strings that are one edit distance apart:

• "ab" and "acb" (insert 'c')
• "cab" and "ab" (delete 'c')
• "1203" and "1213" (replace '0' with '1')

Examples of strings that are NOT one edit distance apart:

• "ab" and "ab" (zero edits needed - they're already equal)
• "ab" and "adb" (this would be one edit, so it IS one edit distance apart)
• "ab" and "adcb" (requires two insertions)

Intuition

The key insight is that when two strings differ by exactly one edit, there must be exactly one position where they first differ. Once we find this position, we can determine what type of edit operation would make them equal.

Let's think about what happens when we compare two strings character by character:

1. If the lengths differ by more than 1, it's impossible to make them equal with just one edit. We'd need multiple insertions or deletions.

2. If the lengths are equal, the only possible operation is a replacement. When we find the first mismatched character, we need to check if the remaining parts after this position are identical. If they are, then replacing this one character would make the strings equal.

3. If one string is longer by exactly 1, we could either delete from the longer string or insert into the shorter one (these are equivalent operations from different perspectives). When we find the first mismatch, we can "skip" the character in the longer string and check if the remaining parts match.

To simplify our logic, we can always ensure that s is the longer (or equal length) string by swapping if necessary. This way, we only need to handle the delete operation from s's perspective, rather than both insert and delete cases.

The elegance of this approach is that we don't need to actually perform any edit operations. We just need to:

• Find where the strings first differ
• Check if skipping or replacing that character would make the rest of the strings match
• If we never find a difference, the strings are identical except possibly for one extra character at the end of the longer string

This gives us a clean O(n) solution where we only need one pass through the strings.

Solution Approach

The implementation follows a systematic approach to handle all possible cases:

Step 1: Normalize the input

if len(s) < len(t):
    return self.isOneEditDistance(t, s)

We ensure that s is always the longer (or equal length) string by recursively calling the function with swapped parameters if needed. This simplifies our logic since we only need to handle deletion from s rather than both insertion and deletion cases.

Step 2: Check length difference

m, n = len(s), len(t)
if m - n > 1:
    return False

If the length difference is greater than 1, it's impossible to make them equal with exactly one edit, so we return False immediately.

Step 3: Find the first difference

for i, c in enumerate(t):
    if c != s[i]:

We iterate through the shorter string t and compare each character with the corresponding character in s. When we find the first mismatch at position i, we need to determine what type of edit would work:

Case 3a: Equal length strings (m == n)

return s[i + 1:] == t[i + 1:]

If the strings have equal length, the only option is replacement. We check if the remaining parts after position i are identical. If they are, replacing s[i] with t[i] would make the strings equal.

Case 3b: Different length strings (m == n + 1)

return s[i + 1:] == t[i:]

If s is longer by 1, we can delete the character at position i from s. We check if skipping s[i] and comparing the rest of s with the rest of t results in equal strings.

Step 4: Handle identical prefixes

return m == n + 1

If the loop completes without finding any difference, it means all characters in t match the corresponding characters in s. The strings are one edit apart only if s has exactly one extra character at the end (which can be deleted to make them equal).

The time complexity is O(min(m, n)) where m and n are the lengths of the strings, as we iterate through at most the shorter string. The space complexity is O(1) if we don't count the string slicing operations (which create new strings in Python), or O(m + n) if we do count them.

Example Walkthrough

Let's walk through the solution with the example where s = "cab" and t = "ab".

Step 1: Check if we need to swap

• len(s) = 3, len(t) = 2
• Since len(s) >= len(t), no swap needed
• s = "cab", t = "ab"

Step 2: Check length difference

• m = 3, n = 2
• m - n = 1, which is not greater than 1, so we continue

Step 3: Find the first difference

• Compare characters one by one:
  • i = 0: t[0] = 'a' vs s[0] = 'c' → Mismatch found!

Step 4: Determine the edit type

• Since m = 3 and n = 2, we have m == n + 1 (different lengths)
• We check if deleting s[0] would make the strings equal:
  • s[i + 1:] = s[1:] = "ab"
  • t[i:] = t[0:] = "ab"
  • "ab" == "ab" → True!

The function returns True because deleting the character 'c' at position 0 from "cab" gives us "ab", which equals t.

Let's trace through another example where s = "abc" and t = "adc":

Step 1: Check if we need to swap

• Both strings have length 3, no swap needed

Step 2: Check length difference

• m = 3, n = 3
• m - n = 0, so we continue

Step 3: Find the first difference

• Compare characters:
  • i = 0: t[0] = 'a' vs s[0] = 'a' → Match
  • i = 1: t[1] = 'd' vs s[1] = 'b' → Mismatch found!

Step 4: Determine the edit type

• Since m == n (equal lengths), we check if replacement would work:
  • s[i + 1:] = s[2:] = "c"
  • t[i + 1:] = t[2:] = "c"
  • "c" == "c" → True!

The function returns True because replacing 'b' with 'd' at position 1 makes the strings equal.

Solution Implementation

Time and Space Complexity

The time complexity is O(m), where m is the length of the longer string s. The algorithm ensures s is always the longer string through the recursive swap at the beginning. The main loop iterates through string t at most once, which has length n ≤ m. The string slicing operations s[i + 1:] and t[i + 1:] or t[i:] take O(m) time in the worst case for comparison. Since we return immediately upon finding the first difference, we perform at most one such slicing comparison, giving us O(m) overall time complexity.

The space complexity is O(m) due to the string slicing operations which create new string objects. When comparing s[i + 1:] == t[i + 1:] or s[i + 1:] == t[i:], Python creates temporary substring copies that can be up to O(m) in size. Additionally, there's a recursive call at the beginning when len(s) < len(t), but this happens at most once and doesn't add to the asymptotic space complexity.

Note: The reference answer states O(1) space complexity, which would be accurate if we consider string slicing as a comparison operation without creating copies, or if the implementation used index-based comparison instead of substring creation. However, in Python's standard implementation, string slicing creates new string objects.

Common Pitfalls

1. Forgetting to Handle the "Zero Edit" Case

The Pitfall: A common mistake is not properly checking whether the strings are already identical. The problem specifically requires exactly one edit, not "at most one" edit. Many developers might write code that returns True when the strings are already equal.

Incorrect Implementation:

def isOneEditDistance(self, s: str, t: str) -> bool:
    if len(s) < len(t):
        return self.isOneEditDistance(t, s)
  
    m, n = len(s), len(t)
    if m - n > 1:
        return False
  
    for i in range(n):
        if s[i] != t[i]:
            if m == n:
                return s[i + 1:] == t[i + 1:]
            else:
                return s[i + 1:] == t[i:]
  
    # Bug: Returns True even when strings are identical
    return True  # Wrong! This would return True for s="abc", t="abc"

The Fix: The correct implementation should return True only when s has exactly one extra character compared to t:

return len_s == len_t + 1  # Correct: only True if s has one extra character

2. Inefficient String Comparison Using Slicing

The Pitfall: While the current solution is correct, using string slicing (s[i+1:] and t[i+1:]) creates new string objects in Python, which can be inefficient for very long strings. This adds unnecessary space complexity.

More Efficient Solution: Instead of creating substrings, compare characters directly:

def isOneEditDistance(self, s: str, t: str) -> bool:
    if len(s) < len(t):
        return self.isOneEditDistance(t, s)
  
    len_s, len_t = len(s), len(t)
    if len_s - len_t > 1:
        return False
  
    for i in range(len_t):
        if s[i] != t[i]:
            if len_s == len_t:
                # Compare remaining characters one by one
                return all(s[j] == t[j] for j in range(i + 1, len_t))
            else:
                # Compare s[i+1:] with t[i:] without creating substrings
                return all(s[j + 1] == t[j] for j in range(i, len_t))
  
    return len_s == len_t + 1

3. Not Handling Edge Cases Properly

The Pitfall: Failing to consider edge cases like empty strings or single-character strings.

Examples that might be mishandled:

• s = "", t = "a" (should return True - insert one character)
• s = "a", t = "" (should return True - delete one character)
• s = "", t = "" (should return False - zero edits needed)

Why the current solution handles these correctly: The normalization step and length check naturally handle these cases:

• Empty strings work because the iteration and final check len_s == len_t + 1 correctly identify the one-character difference
• The recursive swap ensures consistent handling regardless of which string is empty