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Find the String with LCP

Difficulty: Hard

Problem Description

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. The lcp matrix is defined 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].

Key Insights

  • The lcp matrix must satisfy certain constraints for a valid string to exist.
  • The value lcp[i][j] should be consistent with the characters of the string at positions i and j.
  • Each position in the resulting string can be deduced based on the longest common prefix properties.
  • The solution involves a greedy approach to construct the lexicographically smallest string.

Space and Time Complexity

Time Complexity: O(n^2)
Space Complexity: O(n)

Solution

To solve this problem, we can use an iterative approach to fill the characters of the string based on the lcp matrix. The algorithm checks the values in the lcp matrix to determine what characters can occupy each position while ensuring the resulting string is lexicographically smallest.

  1. Start by initializing an array to hold the characters of the string.
  2. For each pair of indices (i, j), if lcp[i][j] > 0, it indicates that the characters at these indices must be the same.
  3. Use a simple greedy strategy to assign characters starting from the smallest possible character ('a') and ensuring that all conditions derived from the lcp matrix are satisfied.
  4. Finally, ensure that the constructed string does not violate any constraints defined by the lcp matrix.

Code Solutions

def find_string_with_lcp(lcp):
    n = len(lcp)
    word = [''] * n
    
    for i in range(n):
        for j in range(i):
            if lcp[i][j] > 0:
                if word[i] == '':
                    word[i] = word[j] if word[j] != '' else 'a'
                elif word[i] != word[j] and lcp[i][j] > 1:
                    return ""
    
    for i in range(n):
        if word[i] == '':
            word[i] = 'a'

    for i in range(n):
        for j in range(n):
            if lcp[i][j] != common_prefix_length(word[i:], word[j:]):
                return ""
    
    return ''.join(word)

def common_prefix_length(s1, s2):
    length = 0
    while length < min(len(s1), len(s2)) and s1[length] == s2[length]:
        length += 1
    return length
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