One Edit Distance

class Solution {
public:
    bool checkOneEditDistance(string firstStr, string secondStr) {
        int firstLen = firstStr.length();
        int secondLen = secondStr.length();
    
        if (firstLen > secondLen) return checkOneEditDistance(secondStr, firstStr);
    
        if (secondLen - firstLen > 1) return false;
    
        for (int index = 0; index < firstLen; index++) {
            if (firstStr[index] != secondStr[index]) {
                if (firstLen == secondLen) return firstStr.substr(index + 1) == secondStr.substr(index + 1);
                else return firstStr.substr(index) == secondStr.substr(index + 1);
            }
        }
    
        return (firstLen + 1 == secondLen);
    }
};

This solution, written in C++, aims to determine if two strings differ by exactly one modification, where modifications can include insertions, deletions, or substitutions. The key function, checkOneEditDistance, compares the lengths of two strings to ensure the length difference is no more than 1. If one string is longer than the other, the function recursively checks the converse scenario.

The function then iterates over the strings. If a mismatching character is found:

• For equal-length strings, it checks if the substrings following the mismatch are identical, which would indicate a single substitution.
• For strings of differing lengths, it checks if ignoring the current character in the longer string makes the remainder of the two strings identical, suggesting a single insertion or deletion.

If no mismatches are found during the iteration, the function checks if the length of the first string is exactly one less than the second, indicating a single insertion.

This approach ensures that we check for all possible one-edit modifications effectively. Below outlines the choice of data structures and methods used in the implementation:

• Strings and their substrings are utilized to compare segments efficiently.
• The length() function helps in promptly determining the size differences between the two strings.
• The substr() function enables checking of remaining string segments post a detected difference.

class Solution {
    public boolean singleEditCheck(String first, String second) {
        int lenFirst = first.length();
        int lenSecond = second.length();
    
        // Shorten ensuring first is less than or equal to second in length
        if (lenFirst > lenSecond) return singleEditCheck(second, first);
    
        // Return false immediately if the length difference is more than 1
        if (lenSecond - lenFirst > 1) return false;
    
        for (int i = 0; i < lenFirst; i++) {
            if (first.charAt(i) != second.charAt(i)) {
                if (lenFirst == lenSecond) {
                    // When lengths are identical
                    return first.substring(i + 1).equals(second.substring(i + 1));
                } else {
                    // When second string is longer by exactly one character
                    return first.substring(i).equals(second.substring(i + 1));
                }
            }
        }
    
        // One additional length in second signifies a possible single edit by addition
        return (lenFirst + 1 == lenSecond);
    }
}

The Java solution provided is designed to determine if two given strings are exactly one edit away from each other. This "one edit" can be either:

• Inserting a character,
• Deleting a character, or
• Replacing a character.

Here's an outline of how the solution works:

• Compare Lengths: First, it ensures that the first string is always the shorter or equal in length to the second string by swapping if necessary.

• Quick Rejection: If the length difference between the two strings is more than one, the function immediately returns false.

• Detailed Comparison: The strings are then compared character by character:

  • If a mismatch is found and the strings are of equal length, it checks if the rest of the strings (after the mismatch) are equal — this would mean the edit was a replacement.
  • If the first string is shorter by one character, it checks if skipping the current character in the longer string makes the two strings identical — this would mean the edit was an insertion.

• End of String Consideration: If no mismatches are found until the end of the shorter string, it simply checks if the second string is exactly one character longer, suggesting that the last character was added.

This algorithm efficiently determines the one edit distance by examining each character at most once, leading to a linear complexity relative to the length of the strings. This method is effective for promptly validating or invalidating potential minor differences between two strings.

bool isSingleEditDistance(char* str1, char* str2) {
    int len1 = strlen(str1);
    int len2 = strlen(str2);
    
    if (len1 > len2) return isSingleEditDistance(str2, str1);
    if (len2 - len1 > 1) return false;
    
    for (int index = 0; index < len1; index++)
        if (str1[index] != str2[index])
            if (len1 == len2) return strcmp(str1 + index + 1, str2 + index + 1) == 0;
            else return strcmp(str1 + index, str2 + index + 1) == 0;
    
    return (len1 + 1 == len2);
}

This C program addresses the problem of determining if two strings (str1 and str2) are exactly one edit distance apart. An edit can be inserting, deleting, or replacing a character. Here's a step-by-step explanation of the provided solution:

1. Obtain the lengths of the two strings, len1 and len2.

2. If str1 is longer than str2, swap the strings to ensure str1 is always the shorter or equal length string. This simplifies further comparisons.

3. If the difference in lengths is more than 1, immediately return false. Two strings with a length difference greater than 1 cannot be one edit distance apart.

4. Iterate through each character of the shorter string (str1):

   • Compare characters of both strings at each index.
   • If a mismatch is detected and the lengths of the strings are the same, check if the rest of the strings after the differing character are identical. If so, return true.
   • If the lengths are different, check if the remainder of str1 starting from the current index matches the remainder of str2 starting from the next index. If this condition holds, return true.

5. If no mismatch was found within the loop and str1 is one character shorter than str2, return true. This covers the case where an edit is an insert at the end of str1.

6. If none of the above conditions satisfied, return false.

This algorithm efficiently addresses the "One Edit Distance" problem by explicitly handling each type of edit operation possible between two strings of similar or marginally different lengths.

var isSingleEditDistance = function (src, tgt) {
    let lenSrc = src.length;
    let lenTgt = tgt.length;
    
    if (lenSrc > lenTgt) return isSingleEditDistance(tgt, src);
    
    if (lenTgt - lenSrc > 1) return false;
    
    for (let index = 0; index < lenSrc; index++)
        if (src[index] != tgt[index])
            if (lenSrc == lenTgt)
                return src.slice(index + 1) === tgt.slice(index + 1);
            else return src.slice(index) === tgt.slice(index + 1);
    
    return lenSrc + 1 === lenTgt;
};

The solution provided as JavaScript code outlines an approach to determine if two strings (src and tgt) are exactly one edit distance apart. An edit can be an insertion, deletion, or substitution of a single character.

• Begin by checking if the sizes of both strings differ by more than one character. If so, return false since more than one edit would be necessary.
• Iterate through the characters of both strings:
  • If a mismatch is identified:
    • If the strings are of equal length, check for substitution by comparing the substrings after the index of mismatch. If equal, only one substitution is needed, hence return true.
    • If lengths differ, compare the source string from the mismatch index against the target string from the next index. A match implies one insertion in the source or deletion in the target is enough, so return true.
• After iterating, if no mismatch is found, check if the length of the target is one greater than the source to account for a single character insertion at the end.
• For efficiency, if the source string is longer, the function calls itself with parameters swapped to benefit from the structure designed primarily for handling shorter source strings.

This efficient method ensures the function only allows for the minimum operations, thereby determining the one-edit distance in a consistent and predictable manner.

class Solution:
    def singleEditDifference(self, string1: "str", string2: "str") -> "bool":
        len1, len2 = len(string1), len(string2)
    
        # Make sure string1 is shorter or the same length as string2
        if len1 > len2:
            return self.singleEditDifference(string2, string1)
    
        # Return False if the difference in length is greater than 1
        if len2 - len1 > 1:
            return False
    
        for i in range(len1):
            if string1[i] != string2[i]:
                # If both strings have the same length
                if len1 == len2:
                    return string1[i + 1 :] == string2[i + 1 :]
                # If both strings have different lengths
                else:
                    return string1[i:] == string2[i + 1 :]
    
        # Return True if all chars are the same in len1 and there's an extra char in string2
        return len1 + 1 == len2

The given Python program defines a method within a Solution class to determine if two strings are exactly one edit distance apart. An edit distance of one means that either you can insert, delete, or replace a character in the first string to make it equal to the second string.

Here is how the solution is structured:

• Firstly, the program calculates the lengths of both input strings.

• If the first string is longer than the second, the method calls itself with the strings reversed to ensure the first string is always the shorter one.

• If the difference in length between the two strings exceeds one, the function immediately returns False as no single edit could bridge a gap wider than that.

• The solution then iterates over the characters of the shorter string. When it finds a mismatching character between the two strings:

  • If the lengths of the strings are equal, it checks if the substring after the current character in both strings is the same.

  • If the lengths are unequal, indicating a potential insert or delete operation, it checks if the substring starting from the current character in the shorter string matches the substring starting from the next character in the longer string.

• If none of the characters mismatch for the length of the shorter string and there is exactly one extra character at the end of the longer string, the function returns True.

• The iteration through the shorter string with checks after the first mismatch ensures that the function runs in linear time relative to the length of the strings.

This succinct approach efficiently determines if two strings are one edit distance apart while ensuring minimal computational overhead.