## Thoughts and Talk

The problem presented here is a variation of the classic "Edit Distance" problem, which falls under the dynamic programming category. It asks whether two given strings can be transformed into each other with a single edit operation (insertion, deletion, or substitution). This problem has practical applications in areas such as spell checkers, DNA sequence analysis, and speech recognition systems. The problem requires understanding and implementing an efficient algorithm to determine whether two strings are one edit distance apart. While there are multiple approaches to solve this problem, it is essential to consider factors such as time complexity, space complexity, and code readability.

### Problem Statement

Given two strings

`s`

and `t`

, determine if they are one edit distance apart. Two strings are considered one edit distance apart if one of the following conditions is true:- The strings have the same length, and they differ by exactly one character at a specific position.

- One string can be obtained from the other by inserting/ deleting/ replacing a single character.

The function

`isOneEditDistance`

should return `True`

if the two strings are one edit distance apart, and `False`

otherwise.### Hints

- Consider the different cases where two strings can be one edit distance apart: same length with one character difference, or one string is one character longer than the other.

- Traverse the strings simultaneously and keep track of the first encountered difference.

- After finding the first difference, check if the remaining characters in both strings are identical.

- Handle edge cases, such as empty strings or strings with the same content.

### Approaches Explored

#### Approach 1: Brute Force

**Pros**:- Simple to understand and implement.
- Works for all cases.

**Cons**:- Time complexity of O(n^2), where n is the length of the longer string, as it involves generating and checking all possible edit combinations.
- Not efficient for large strings.

#### Approach 2: Optimized String Traversal

**Pros**:- Time complexity of O(n), where n is the length of the longer string.
- Space complexity of O(1), as it uses constant extra space.
- Efficient and optimized for most cases.

**Cons**:- Slightly more complex implementation compared to the brute force approach.

### Finalized Approach

The finalized approach follows the optimized string traversal technique, which has a time complexity of O(n) and a space complexity of O(1). This approach is more efficient and practical compared to the brute force method, especially for longer strings.

### Executable Code

TheTechCruise.com Pyodide Terminal

```
class Solution:
def isOneEditDistance(self, s: str, t: str) -> bool:
i, ls, lt = 0, len(s), len(t)
# Early return if the difference in length is more than 1, both strings are empty, or strings are equal
if (abs(ls - lt) > 1) or (ls == 0 and lt == 0) or s == t:
return False
# If one string is empty, the other must be exactly one character long to be true
elif ls == 0 or lt == 0:
return True
# Compare characters until the first difference is found
for i in range(min(ls, lt)):
if s[i] != t[i]:
break
else:
i += 1
# Handle the cases based on the length of strings and the position of difference
if ls == lt:
return s[i+1:] == t[i+1:]
elif lt > ls:
return s[i:] == t[i+1:]
elif lt < ls:
return s[i+1:] == t[i:]
return False
# Driver code to test the function with the given examples
def test_one_edit_distance():
solution = Solution()
test_cases = [
("ab", "acb"), # True: Insert 'c' in "ab"
("", ""), # False: Both strings are empty
("a", ""), # True: Remove 'a' from "a"
("a", "A"), # False: 'a' and 'A' are different (case-sensitive)
("a", "ac"), # True: Append 'c' to "a"
("c", "c"), # False: No changes needed (strings are the same)
("ba", "a"), # True: Remove 'b' from "ba"
("ab", "acd") # False: Requires two edits ('b' to 'c', append 'd')
]
for s, t in test_cases:
result = solution.isOneEditDistance(s, t)
print(f"isOneEditDistance({s}, {t}): {result}")
# Execute the test function
test_one_edit_distance()
```

### Code Analysis and Code Walkthrough

The

`isOneEditDistance`

function takes two strings `s`

and `t`

as input and returns a boolean value indicating whether they are one edit distance apart or not.- The function first initializes variables
`i`

,`ls`

, and`lt`

with values`0`

,`len(s)`

, and`len(t)`

, respectively.

- It then checks for edge cases:
- If the absolute difference between the lengths of
`s`

and`t`

is greater than 1, or both strings are empty, or both strings are equal, it returns`False`

. - If one of the strings is empty, it returns
`True`

.

- If none of the edge cases are met, the function iterates through the strings
`s`

and`t`

simultaneously using a`for`

loop. It compares the characters at each position and breaks out of the loop as soon as it encounters the first difference. The loop counter`i`

is updated to the index of the first difference.

- After the loop, the function checks the following cases:
- If the lengths of
`s`

and`t`

are equal, it checks if the remaining characters after the first difference are the same (i.e.,`s[i+1:] == t[i+1:]`

). - If the length of
`s`

is less than t, it checks if remaining characters of`s`

and`t`

are equal after removal of`t[i]`

i.e.,`s[i:] == t[i+1:]`

). - If the length of
`s`

is greater than`t`

, it checks if remaining characters of`s`

and`t`

are equal after removal of`s[i]`

i.e.,`s[i+1:] == t[i:]`

).

- If any of the above conditions are met, the function returns
`True`

, indicating that the strings are one edit distance apart. Otherwise, it returns`False`

.

### Example with Explanation

Let's consider an example with

`s = "random"`

and `t = "ransom"`

using the given code:- The function initializes
`i = 0`

,`ls = 6`

, and`lt = 6`

.

- It checks the edge cases:
- The absolute difference between the lengths of
`s`

and`t`

is 0, which is not greater than 1, so it proceeds. - The strings are not equal, so it proceeds.
- Neither string is empty, so it proceeds.

- The function iterates through the strings:
- At index 0,
`s[0] == 'r'`

and`t[0] == 'r'`

, so it continues. - At index 1,
`s[1] == 'a'`

and`t[1] == 'a'`

, so it continues. - At index 2,
`s[2] == 'n'`

and`t[2] == 'n'`

, so it continues. - At index 3,
`s[3] == 'd'`

and`t[3] == 's'`

, so it breaks out of the loop with`i = 3`

.

- Since the lengths of
`s`

and`t`

are the same (`ls = 6`

and`lt = 6`

), the function checks if the substrings of`s`

and`t`

after the first difference are equal: `s[i+1:] == t[i+1:] = "om" == "om"`

is`True`

.

- Since the condition is met, the function returns
`True`

, indicating that the strings`"random"`

and`"ransom"`

are one edit distance apart due to a single substitution of 'd' to 's'.

**Explanation for other Test cases**

**("ab", "acb")**: Inserting 'c' between 'a' and 'b' makes them one edit apart. Returns

.**True**

**("", "")**: Both strings are empty; no edits needed. Returns

.**False**

**("a", "")**: Removing 'a' from "a" makes it empty, which is one edit. Returns

.**True**

**("a", "A")**: Changing 'a' to 'A' involves a substitution, but since they're case-sensitive and identical otherwise, it's an edit. Returns

because the strings are not the same.**False**

**("a", "ac")**: Appending 'c' to "a" results in "ac", which is one edit. Returns

.**True**

**("c", "c")**: No changes needed as the strings are identical. Returns

.**False**

**("ba", "a")**: Removing 'b' from "ba" leaves "a", which is one edit. Returns

.**True**

**("ab", "acd")**: Two edits are needed ('b' to 'c' and appending 'd'), so it's more than one edit distance. Returns

.**False**

### Discussion on Complexity

**Time Complexity**: The time complexity of the above optimized string traversal approach is O(n), where n is the length of the longer string. This is because the algorithm iterates through the strings once, comparing characters at each position until it finds the first difference or reaches the end of the shorter string.

**Space Complexity**: The space complexity of the above solution is O(1), as it uses a constant amount of extra space to store the loop counter

`i`

and the lengths of the strings `ls`

and `lt`

.### Conclusion

The "One Edit Distance" problem is more than just a coding challenge; it's a gateway to mastering essential data structures and algorithms. This problem serves as a practical tool for enhancing critical thinking and problem-solving skills.

Our comprehensive blog post delves deep into the nuances of this problem, offering insights into its application and strategies for efficient coding in Python. By understanding and implementing the solutions discussed, learners and professionals can significantly advance their knowledge in algorithm design and data structure optimization.

Enhance your coding proficiency and prepare for technical interviews by exploring our detailed examples and complexity analysis.

Let's learn and grow together in the ever-evolving world of computer science! Do comment and let us know!