Group Shifted Strings

This problem requires grouping strings that can be shifted to become each other. Shifting means moving each letter to its next letter in the alphabet (wrapping around from 'z' to 'a').

Naive Approach

A simple but inefficient way to solve this is to iterate through each string and, for every other string, check if they belong to the same shifting sequence. This involves writing a helper function to determine if one string can be shifted to become another.

Algorithm

1. Create a list to store the groups of shifted strings.
2. Iterate through the input list of strings.
3. For each string, create a new group or find an existing group to add it to.
4. To find a group, iterate through the existing groups.
5. For each group, check if the current string can be shifted to any string in that group using a helper function.
6. If a match is found, add the current string to that group.
7. If no match is found, create a new group with the current string.
8. Return the list of groups.

Big O Analysis

• Time Complexity: O(n² * k), where n is the number of strings and k is the average length of the strings. The nested loops dominate the runtime, and the helper function can take O(k) time.
• Space Complexity: O(n * k) in the worst case, where n is the number of strings and k is the average length of the strings. This is because, in the worst-case scenario, each string could be in its own group.

Optimal Approach

The optimal approach involves normalizing each string to a standard form and then grouping them based on this normalized form. The normalized form can be generated by calculating the difference between adjacent characters.

Algorithm

1. Create a hash map to store the normalized strings as keys and lists of original strings as values.
2. Iterate through the input list of strings.
3. For each string, normalize it by calculating the difference between adjacent characters and wrapping around if necessary.
4. Use the normalized string as a key in the hash map. If the key already exists, add the original string to the list of values. Otherwise, create a new entry in the hash map with the normalized string as the key and a new list containing the original string as the value.
5. After processing all strings, convert the values of the hash map into a list of lists, which represents the grouped strings.
6. Return the list of lists.

Edge Cases

• Empty input list: Return an empty list of lists.
• List containing empty strings: Empty strings can be grouped together.
• List containing only one string: Return a list containing a list with that string.
• Strings with different lengths cannot be shifted to become each other.

Code Example (Python)

def groupStrings(strings):
    groups = {}
    for s in strings:
        key = ""
        for i in range(1, len(s)):
            diff = (ord(s[i]) - ord(s[i - 1])) % 26
            key += str(diff) + ","
        if len(s) > 0:
          if key not in groups:
              groups[key] = [s]
          else:
              groups[key].append(s)
        else:
          if '' not in groups:
            groups[''] = [s]
          else:
            groups[''].append(s)

    return list(groups.values())

Big O Analysis

• Time Complexity: O(n * k), where n is the number of strings and k is the average length of the strings. This is because we iterate through each string once and normalize it in O(k) time.
• Space Complexity: O(n * k) in the worst case, where n is the number of strings and k is the average length of the strings. This is because we store each string in the hash map.