Odd String Difference

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You are given an array of equal-length strings words. Assume that the length of each string is n.

Each string words[i] can be converted into a difference integer array difference[i] of length n - 1 where difference[i][j] = words[i][j+1] - words[i][j] where 0 <= j <= n - 2. Note that the difference between two letters is the difference between their positions in the alphabet i.e. the position of 'a' is 0, 'b' is 1, and 'z' is 25.

For example, for the string "acb", the difference integer array is [2 - 0, 1 - 2] = [2, -1].

All the strings in words have the same difference integer array, except one. You should find that string.

Return the string in words that has different difference integer array.

Example 1:

Input: words = ["adc","wzy","abc"]
Output: "abc"
Explanation: 
- The difference integer array of "adc" is [3 - 0, 2 - 3] = [3, -1].
- The difference integer array of "wzy" is [25 - 22, 24 - 25]= [3, -1].
- The difference integer array of "abc" is [1 - 0, 2 - 1] = [1, 1]. 
The odd array out is [1, 1], so we return the corresponding string, "abc".

Example 2:

Input: words = ["aaa","bob","ccc","ddd"]
Output: "bob"
Explanation: All the integer arrays are [0, 0] except for "bob", which corresponds to [13, -13].

Constraints:

3 <= words.length <= 100
n == words[i].length
2 <= n <= 20
words[i] consists of lowercase English letters.

Odd String Difference

Easy

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Topics:

ArraysStrings

You are given an array of equal-length strings words. Assume that the length of each string is n.

Each string words[i] can be converted into a difference integer array difference[i] of length n - 1 where difference[i][j] = words[i][j+1] - words[i][j] where 0 <= j <= n - 2. Note that the difference between two letters is the difference between their positions in the alphabet i.e. the position of 'a' is 0, 'b' is 1, and 'z' is 25.

For example, for the string "acb", the difference integer array is [2 - 0, 1 - 2] = [2, -1].

All the strings in words have the same difference integer array, except one. You should find that string.

Return the string in words that has different difference integer array.

Example 1:

Input: words = ["adc","wzy","abc"]
Output: "abc"
Explanation: 
- The difference integer array of "adc" is [3 - 0, 2 - 3] = [3, -1].
- The difference integer array of "wzy" is [25 - 22, 24 - 25]= [3, -1].
- The difference integer array of "abc" is [1 - 0, 2 - 1] = [1, 1]. 
The odd array out is [1, 1], so we return the corresponding string, "abc".

Example 2:

Input: words = ["aaa","bob","ccc","ddd"]
Output: "bob"
Explanation: All the integer arrays are [0, 0] except for "bob", which corresponds to [13, -13].

Constraints:

3 <= words.length <= 100
n == words[i].length
2 <= n <= 20
words[i] consists of lowercase English letters.

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:

What are the constraints on the length of the `words` array, and the length of each individual word?
Can the input array `words` be empty, or contain null or empty strings?
If multiple strings have unique difference sequences, which one should I return? (e.g., the first one encountered?)
Are the characters in the strings guaranteed to be lowercase English letters, or can they be from a broader ASCII range?
If no string has a unique difference sequence, what should I return? (e.g., null, an empty string, or throw an exception?)

Brute Force Solution

Approach

The brute force approach in this scenario involves checking every possible word to find the 'odd' one out. We accomplish this by exhaustively calculating the difference between letter values within each word and comparing these differences across all words. This ensures no potential candidate is overlooked.

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

For each word, calculate the difference between the value of each letter and the value of the letter before it.
Store these differences for each word; this results in a set of differences for every word.
Compare the sets of differences between every pair of words. Do this for all possible pairs of words.
Count how many times each word has a unique set of differences compared to the others.
The word with the fewest matches (most unique differences) is the odd one out.

Code Implementation

def find_odd_string(words):
    string_differences = []
    for word in words:
        differences = []
        for i in range(1, len(word)):
            differences.append(ord(word[i]) - ord(word[i - 1]))
        string_differences.append(differences)

    odd_string_counts = {}
    for i in range(len(words)):
        odd_string_counts[words[i]] = 0

    # Compare all pairs of difference lists
    for i in range(len(string_differences)):

        for j in range(len(string_differences)):

            if i != j and string_differences[i] != string_differences[j]:
                odd_string_counts[words[i]] += 1

    # Find word with the most unique differences
    odd_string = None
    max_unique_differences = -1
    for word, count in odd_string_counts.items():
        if count > max_unique_differences:

            # Find the word with the maximum differences
            max_unique_differences = count

            odd_string = word

    #Return odd string if we found it
    if odd_string is not None:
        return odd_string

    # Return the first word if no odd string is found.
    return words[0]

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Big(O) Analysis

Time Complexity

O(n*m) – Let n be the number of words in the input array and m be the maximum length of a word. Step 1 involves calculating differences between letters in each word which takes O(m) time for each word. Since we perform this operation for all n words, the total time for Step 1 is O(n*m). Step 3 involves comparing the difference sets of every pair of words. Comparing two sets of differences for two words takes O(m) time. Because we compare all pairs of words, i.e. n * (n-1) / 2 pairs, the total time for Step 3 is approximately O(n^2 * m). Since O(n^2 * m) dominates O(n*m), the overall time complexity is O(n^2 * m), but since m is typically small and bounded it can be considered a constant factor. Therefore, the effective runtime is O(n*m).

Space Complexity

O(N*M) – The solution stores differences for each word, requiring a list of differences per word where each word can have up to M-1 differences if M is the length of longest word. Given N words as input, this results in a list of size N*M (in the worst case scenario) to store all the differences. Furthermore, temporary variables used during comparisons contribute negligibly to the overall space complexity. Therefore, the dominant factor is the space used to store the differences between letters, leading to O(N*M) space complexity, where N is the number of words, and M is the max length of any word.

Optimal Solution

Approach

The key insight is that we need to compute a 'difference sequence' for each word. This difference sequence is based on the numerical distance between consecutive letters. We then compare these sequences to find the 'odd' word.

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

For each word, transform it into a sequence of numbers.
To transform each word, calculate the difference between the position of each pair of consecutive letters in the alphabet (e.g., 'ac' becomes 2 because 'c' is two positions after 'a').
Compare the numerical sequences. Count how many times we see each unique sequence.
The sequence that appears only once corresponds to the 'odd' word.
Return the word that generated that 'odd' sequence.

Code Implementation

def odd_string(words):
    difference_sequences = {}
    
    for word in words:
        numerical_sequence = []
        for i in range(len(word) - 1):
            numerical_sequence.append(ord(word[i+1]) - ord(word[i]))
        numerical_sequence = tuple(numerical_sequence)
        
        # Count occurrences of each difference sequence.
        if numerical_sequence not in difference_sequences:
            difference_sequences[numerical_sequence] = [word]
        else:
            difference_sequences[numerical_sequence].append(word)
            
    # Find the sequence that appears only once.
    for sequence, word_list in difference_sequences.items():
        if len(word_list) == 1:
            # Return the word associated with the odd sequence.
            return word_list[0]

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n*m) – Let n be the number of words in the input list and m be the maximum length of any word. The dominant operation involves iterating through each word in the list (n words) and, for each word, calculating the difference sequence which requires traversing the word (maximum length m). The comparison of sequences created from the difference is assumed to take constant time since the problem statement doesn't indicate this is the bottleneck. Thus, the time complexity is approximately n * m, which gives us O(n*m).

Space Complexity

O(N) – The algorithm stores the difference sequences in a hash map to count their occurrences. In the worst case, all the words have different difference sequences, resulting in the hash map storing N sequence-count pairs, where N is the number of words. Additionally, creating each difference sequence uses a temporary list whose length is at most M-1 where M is the length of the longest word. Therefore, the auxiliary space is dominated by the storage of the difference sequences within the hash map, leading to O(N) space complexity, considering each sequence has length at most M-1, and the number of words is N.

Edge Cases

Case	How to Handle
Empty input array `words`	Return an empty string since there are no words to process.
Input array `words` contains an empty string	Handle the empty string gracefully, either skipping it or assigning it an empty difference sequence (e.g., []).
Input array `words` contains strings with only one character	Strings with only one character have no difference sequence, so return an empty list for them.
All strings in `words` have the same difference sequence	Return an empty string or null, indicating that no string has a unique difference sequence.
Strings with very long lengths, leading to large difference sequences	The solution should efficiently compute and store the difference sequence without exceeding memory limits.
Strings with characters having extreme ASCII values (e.g., near 0 or 255)	The integer subtraction in the difference sequence calculation should not overflow or underflow.
Input `words` is very large, impacting performance	The solution should use an efficient data structure (e.g., hash map) to store and compare difference sequences to maintain acceptable performance.
Multiple strings have the same unique difference sequence	The problem statement specifies that only one string has a unique difference sequence, but handle the edge case by perhaps returning the first encountered string with a unique sequence.