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Strings Differ by One Character #12 Most Asked

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ArraysStrings

Given a list of strings dict where all the strings are the same length.

Return true if there are two strings that only differ by one character in the same index, otherwise return false.

Example 1:

Input: dict = ["abcd","acbd","aacd"]
Output: true
Explanation:
Strings "abcd" and "aacd" differ only by one character in the index 1.

Example 2:

Input: dict = ["ab","cd","yz"]
Output: false

Example 3:

Input: dict = ["abcd","cccc","abyd","abab"]
Output: true

Constraints:

  • 1 <= dict.length <= 104
  • 1 <= dict[i].length <= 100
  • dict[i] consists of lowercase English letters.
  • All the strings in the dict have the same length.

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What are the constraints on the length of the input array `strs` and the length of the strings within `strs`?
  2. Are the strings in `strs` guaranteed to contain only lowercase English letters, or can they contain other characters?
  3. If no two strings differ by exactly one character, what should I return (e.g., `false`, `null`, or throw an exception)?
  4. Are duplicate strings allowed in the input array `strs`? If so, should I consider a string differing from *itself* by one character as a valid pair?
  5. Is the input array guaranteed to be non-empty?

Brute Force Solution

Approach

The brute force approach to finding if strings differ by one character is to compare each string against every other string in the list. We will look at all possible pairs of strings and check how many characters are different between them.

Here's how the algorithm would work step-by-step:

  1. Take the first string from the list.
  2. Compare it to the second string, counting how many characters are different.
  3. If exactly one character is different, we have found our pair, and we can stop.
  4. If not, compare the first string to the third string, the fourth string, and so on, until we have compared the first string with every other string in the list.
  5. If we didn't find a match with the first string, move to the second string in the list.
  6. Compare the second string with all the strings after it in the list (the third string, the fourth string, and so on).
  7. Again, check if exactly one character is different between the strings.
  8. Keep repeating this process, moving through the list of strings one by one, and comparing each string to all the strings that come after it, until we find a pair that differs by only one character or we have compared all possible pairs.

Code Implementation

def strings_differ_by_one_character(input_strings):
    for first_word_index in range(len(input_strings)):
        first_word = input_strings[first_word_index]

        # Compare the first word with the remaining words
        for second_word_index in range(first_word_index + 1, len(input_strings)):
            second_word = input_strings[second_word_index]

            # Ensure words are of the same length for comparison
            if len(first_word) != len(second_word):
                continue

            difference_count = 0
            for char_index in range(len(first_word)):
                if first_word[char_index] != second_word[char_index]:
                    difference_count += 1

            # Check if the words differ by exactly one character
            if difference_count == 1:
                return True

    # No pair of words differs by exactly one character
    return False

Big(O) Analysis

Time Complexity
O(n²)The provided algorithm iterates through each string in the input list of n strings. For each string, it compares it with all the strings that come after it in the list. In the worst-case scenario, where no two strings differ by only one character, the algorithm compares approximately n * (n-1) / 2 pairs of strings. This simplifies to approximately n²/2, therefore, the time complexity is O(n²).
Space Complexity
O(1)The provided algorithm iterates through the input list of strings, comparing pairs of strings. It doesn't use any auxiliary data structures such as lists, hash maps, or sets to store intermediate results or track visited elements. Only a few constant space variables are used for loop indices and character comparison. Therefore, the space complexity is constant, regardless of the input size N, where N is the number of strings in the input list.

Optimal Solution

Approach

The optimal approach involves systematically comparing each word with every other word to quickly identify if they differ by only one character. We use a trick to make these comparisons very efficient by focusing on potential differences rather than checking every character in every word.

Here's how the algorithm would work step-by-step:

  1. First, check if all words have the same length. If they don't, no words can differ by only one character, so you're done.
  2. For each word, create a slightly modified version of it. This modified version removes one character at a time, replacing it with a special symbol.
  3. Store these modified versions in a way that makes it easy to look them up. For example, put them in groups based on what the modified version looks like.
  4. Now, go through each group of modified versions. If you find a group where the same modified version appears twice, it means the original words only differed in the character that was replaced by the symbol. These are your differing words.
  5. If you find no such groups, it means no words differ by only one character.

Code Implementation

def strings_differ_by_one_character(input_strings):
    altered_strings_map = {}

    for original_string in input_strings:
        for index in range(len(original_string)):
            # Create the altered string by removing one char
            altered_string = original_string[:index] + original_string[index+1:]

            # Store the altered string and its origin.
            if altered_string in altered_strings_map:
                if original_string not in altered_strings_map[altered_string]:
                    #Strings differ by one character if present
                    return True
                
            else:
                altered_strings_map[altered_string] = [original_string]

    # No differing strings found.
    return False

Big(O) Analysis

Time Complexity
O(n*m)The algorithm iterates through each of the n words in the input list. For each word, it generates m variations, where m is the length of the word (number of characters). Each variation is created by replacing one character with a special symbol. The creation and hashing (grouping) of these variations take constant time per variation. Since each of the n words generates m variations, the total time complexity is O(n*m), assuming hash operations are O(1) on average and the word lengths are relatively consistent. Therefore, the nested operation results in a time complexity of O(n*m).
Space Complexity
O(N*L)The primary auxiliary space is used to store the modified versions of the words. In the worst case, we create L modified versions for each of the N words, where L is the length of the word. These modified versions are stored in a data structure (e.g., a hash map) to group identical modified words together. Therefore, the space used by this data structure is proportional to N * L. Consequently, the space complexity is O(N*L), where N is the number of words and L is the length of the word.

Edge Cases

strs is null or empty
How to Handle:
Return false if strs is null or empty, as there are no strings to compare.
strs contains fewer than 2 strings
How to Handle:
Return false if strs contains fewer than 2 strings, as no comparison is possible.
Strings in strs have different lengths
How to Handle:
Return false or throw an exception, as strings must have the same length for a valid comparison.
strs contains only identical strings
How to Handle:
Return false if all strings in strs are identical, as they don't differ by any character.
strs contains very long strings (performance)
How to Handle:
Ensure that the chosen algorithm (e.g., hashing with wildcards) has reasonable time complexity for long strings to avoid timeouts.
Large input array strs (memory considerations)
How to Handle:
The solution should be memory-efficient, avoiding unnecessary copying or storing of large substrings; consider using iterators instead.
Strings contain non-ASCII or Unicode characters
How to Handle:
Ensure the character comparison logic is Unicode-aware if the strings might contain non-ASCII characters.
Strings with only one character
How to Handle:
Handle appropriately, comparing single character strings for difference is a valid scenario
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