Find the String with LCP

Hard

Google

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Topics:

StringsArrays

We define the lcp matrix of any 0-indexed string word of n lowercase English letters as an n x n grid such that:

lcp[i][j] is equal to the length of the longest common prefix between the substrings word[i,n-1] and word[j,n-1].

Given an n x n matrix lcp, return the alphabetically smallest string word that corresponds to lcp. If there is no such string, return an empty string.

A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b. For example, "aabd" is lexicographically smaller than "aaca" because the first position they differ is at the third letter, and 'b' comes before 'c'.

Example 1:

Input: lcp = [[4,0,2,0],[0,3,0,1],[2,0,2,0],[0,1,0,1]]
Output: "abab"
Explanation: lcp corresponds to any 4 letter string with two alternating letters. The lexicographically smallest of them is "abab".

Example 2:

Input: lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,1]]
Output: "aaaa"
Explanation: lcp corresponds to any 4 letter string with a single distinct letter. The lexicographically smallest of them is "aaaa".

Example 3:

Input: lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,3]]
Output: ""
Explanation: lcp[3][3] cannot be equal to 3 since word[3,...,3] consists of only a single letter; Thus, no answer exists.

Constraints:

1 <= n == lcp.length == lcp[i].length <= 1000
0 <= lcp[i][j] <= n

Find the String with LCP

Hard

Google

3 views

Topics:

StringsArrays

We define the lcp matrix of any 0-indexed string word of n lowercase English letters as an n x n grid such that:

lcp[i][j] is equal to the length of the longest common prefix between the substrings word[i,n-1] and word[j,n-1].

Given an n x n matrix lcp, return the alphabetically smallest string word that corresponds to lcp. If there is no such string, return an empty string.

A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b. For example, "aabd" is lexicographically smaller than "aaca" because the first position they differ is at the third letter, and 'b' comes before 'c'.

Example 1:

Input: lcp = [[4,0,2,0],[0,3,0,1],[2,0,2,0],[0,1,0,1]]
Output: "abab"
Explanation: lcp corresponds to any 4 letter string with two alternating letters. The lexicographically smallest of them is "abab".

Example 2:

Input: lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,1]]
Output: "aaaa"
Explanation: lcp corresponds to any 4 letter string with a single distinct letter. The lexicographically smallest of them is "aaaa".

Example 3:

Input: lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,3]]
Output: ""
Explanation: lcp[3][3] cannot be equal to 3 since word[3,...,3] consists of only a single letter; Thus, no answer exists.

Constraints:

1 <= n == lcp.length == lcp[i].length <= 1000
0 <= lcp[i][j] <= n

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 possible characters in the strings? Is it limited to lowercase English letters, or are there other possibilities like uppercase letters, numbers, or special characters?
Can the input array be empty, or contain null or empty strings? If so, what should the function return?
If multiple strings share the longest common prefix, can I return any one of them, or is there a specific string I should prioritize?
What is the maximum length of any given string in the input array?
Is there a guaranteed minimum number of strings in the input array (e.g., at least two), or is a single-string input a possibility?

Brute Force Solution

Approach

The brute force approach to finding a string with a specific Longest Common Prefix (LCP) tries all possible strings within certain constraints. We essentially create every possible string and then check if it meets the given LCP criteria with all the strings in the input list. If it does, we have found our string.

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

First, consider the shortest string from the input list. The maximum possible length of the LCP we need to create cannot be longer than this shortest string.
Now, begin constructing strings, character by character. Start with a single character at a time, then two characters, then three, up to the length of the shortest string in the input list.
For each of the strings we construct, we will check if the new string is indeed a common prefix to ALL input strings. We do this by comparing it to the start of each input string.
If the string we made so far is a prefix of every input string, that means we found a valid LCP. Keep this string in mind.
Continue generating and checking all possible prefixes in the same way. The longest valid common prefix generated during this process is your answer.

Code Implementation

def find_string_with_lcp_brute_force(strings):
    shortest_string_length = min(len(string) for string in strings)
    longest_common_prefix = ""

    for prefix_length in range(1, shortest_string_length + 1):
        for i in range(26):
            character = chr(ord('a') + i)
            potential_prefix = longest_common_prefix + character

            is_common_prefix = True
            # Need to check if the potential prefix is common to all strings
            for string in strings:

                if not string.startswith(potential_prefix):

                    is_common_prefix = False
                    break

            if is_common_prefix:
                # If it is, then update the longest common prefix
                longest_common_prefix = potential_prefix

    return longest_common_prefix

Big(O) Analysis

Time Complexity

O(m * k * n) – Let n be the number of strings in the input list, m be the length of the shortest string in the input list, and k be the number of possible characters in the alphabet. The algorithm iterates through possible prefixes of length 1 to m. For each prefix length, it generates all possible strings of that length (O(k^length)). For each generated string, it checks if it's a prefix of all n input strings. This prefix check involves comparing the generated string (of length at most m) with the prefix of each input string. Thus, for each of approximately k^length generated strings, we perform m * n character comparisons. Summing this across prefix lengths 1 to m gives a complexity roughly proportional to (k + k^2 + ... + k^m) * n * m. This can be loosely approximated by O(k^m * m * n). However, since the problem description specifies that the prefixes are generated one character at a time, and only the longest valid prefix is needed, the actual number of prefix checks becomes m (the length of the potential prefix). Within this loop the code will compare each string with the n inputs, for a length of at most m. Therefore we get O(m*k*n).

Space Complexity

O(1) – The algorithm constructs and checks potential LCPs one at a time and keeps track of only the longest valid LCP found so far. This means we are only storing a single prefix string at a time, and updating it when a longer common prefix is found. The length of this string is bounded by the length of the shortest string in the input, but since we are only ever storing *one* such prefix, the auxiliary space required is constant with respect to the number of input strings N. Therefore, the space complexity is O(1).

Optimal Solution

Approach

The best way to solve this problem is by building the longest common prefix (LCP) string character by character. Start with a plausible answer and then refine it until a valid solution is found. The core idea is to iteratively extend the LCP while ensuring it remains consistent with the inputs.

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

Take the first string in the input list and consider it as the starting point for our potential LCP.
Compare this potential LCP with the second string. Find the longest sequence of characters that both strings share from the very beginning.
Update our potential LCP to be just that shared sequence, discarding the rest.
Repeat this comparison process, comparing our current LCP with each of the remaining strings in the input list, one at a time.
Each time we find a shorter shared sequence at the beginning, we shorten our LCP to match.
After comparing with all the strings, what's left of our potential LCP is guaranteed to be the longest common prefix of all the input strings.
If at any point the potential LCP becomes empty (no shared characters), then there is no common prefix and you can stop.

Code Implementation

def find_longest_common_prefix(list_of_strings):
    if not list_of_strings:
        return ""

    longest_common_prefix = list_of_strings[0]

    for current_string in list_of_strings[1:]:
        # Iterate through all strings to find common prefix.
        index = 0
        while index < len(longest_common_prefix) and index < len(current_string) and \
                longest_common_prefix[index] == current_string[index]:
            index += 1

        # Update the longest common prefix based on the comparison.
        longest_common_prefix = longest_common_prefix[:index]

        # If the prefix becomes empty, there's no common prefix.
        if not longest_common_prefix:
            return ""

    return longest_common_prefix

Big(O) Analysis

Time Complexity

O(N * M) – The algorithm iterates through each of the N strings in the input list. Within each iteration, it compares the current potential LCP of length M with the next string. The comparison involves iterating through the characters of both strings up to the length of the shorter string (at most M). Therefore, the time complexity is proportional to N * M, where N is the number of strings and M is the length of the longest common prefix, or length of the shortest string. Thus, the overall time complexity is O(N * M).

Space Complexity

O(1) – The algorithm's space complexity is O(1) because it primarily uses a single variable to store the potential LCP string, which is repeatedly updated in place. The length of this LCP string is bounded by the length of the shortest string in the input list, but the space occupied by LCP does not scale with the number of input strings. Therefore, the auxiliary space used is constant regardless of the number and size of input strings, where N represents the total number of characters across all input strings.

Edge Cases

Case	How to Handle
Empty input array of strings	Return an empty string as the LCP of an empty array is defined as empty.
Input array with only one string	Return the single string in the array as the LCP with itself is the string itself.
Input array where one of the strings is an empty string	Return an empty string because any string compared to an empty string will result in an empty LCP.
Input array with strings of significantly different lengths	Iterate only up to the length of the shortest string to avoid index out of bounds errors.
Input array with strings that have no common prefix	The algorithm should return an empty string in this case after the first character comparison fails.
Input array with all strings identical	The algorithm should return any of the strings since they are all the same and thus, the LCP.
Input array with strings containing non-ASCII characters	Ensure the comparison logic is Unicode-aware and handles multi-byte characters correctly; consider encoding issues.
Very large input array of very long strings	Consider the space complexity of storing intermediate results, and optimize for minimal memory usage to avoid memory exhaustion.