Find the String with LCP

• Define a recursive function, say solve(index, currentWord), that attempts to build the string character by character.
• The base case is index == n. When reached, a full candidate string is formed. Validate this string against the lcp matrix. If it matches, a solution is found.
• In the recursive step for index, iterate through characters c from 'a' to 'z'.
• For each character c, assign currentWord[index] = c.
• Perform a partial validation of the prefix currentWord[0...index] against the lcp matrix. Check for any immediate contradictions with lcp[i][j] for i, j <= index.
• If the prefix is consistent, make a recursive call solve(index + 1, currentWord).
• If the recursive call finds a solution, return it. Otherwise, backtrack and try the next character.
• Since characters are tried in alphabetical order ('a' to 'z'), the first complete valid string found will be the lexicographically smallest.

The backtracking algorithm explores the search space of all possible strings of length n. It does so by constructing the string one character at a time. To ensure the result is lexicographically smallest, it tries placing 'a', then 'b', and so on, at each position.

After adding a character at index k, the algorithm checks if the prefix of length k+1 violates any conditions imposed by the lcp matrix. For example, if we have word[i] and word[j] (with i, j <= k), and word[i] != word[j], then lcp[i][j] must be 0. If the input lcp[i][j] is greater than 0, this path is invalid, and the algorithm must backtrack.

This pruning of invalid paths helps reduce the search space, but the worst-case complexity remains exponential. A full validation is required once a complete string is formed.

Due to its high time complexity, this approach is not practical for the given problem constraints but serves as a theoretical brute-force method.

• Get the size of the matrix, n.
• Greedy Construction:
  • Initialize a character array word of size n and a character nextChar = 'a'.
  • Iterate i from 0 to n-1:
    • If word[i] has not been assigned a character yet:
      • If nextChar > 'z', return "" as more than 26 unique characters are needed.
      • Assign word[i] = nextChar.
      • For all j > i, if lcp[i][j] > 0, it implies word[j] must be the same as word[i]. Assign word[j] = nextChar.
      • Increment nextChar.
• Verification:
  • The constructed word is the only candidate for the lexicographically smallest solution. Now, we must verify it.
  • Create a new n x n matrix, computedLcp.
  • Fill computedLcp using dynamic programming, iterating i and j from n-1 down to 0:
    • If word[i] == word[j], computedLcp[i][j] = 1 + (i+1 < n && j+1 < n ? computedLcp[i+1][j+1] : 0).
    • Otherwise, computedLcp[i][j] = 0.
  • During the computation, if at any point computedLcp[i][j] does not match the input lcp[i][j], the candidate is invalid. Return "".
• If the entire verification process completes without returning, the candidate string is valid. Return it.

This approach is broken down into two main parts: construction and verification. It's also good practice to perform some initial sanity checks on the input matrix.

Initial Sanity Checks (Optional but Recommended): Before starting, we can check for basic properties that any valid lcp matrix must have:

• lcp[i][i] == n - i for all i.
• The matrix must be symmetric: lcp[i][j] == lcp[j][i].
• The LCP value cannot exceed the length of the suffixes: lcp[i][j] <= n - max(i, j). If any of these fail, we can immediately return an empty string.

Phase 1: Greedy Construction To find the lexicographically smallest string, we should use the smallest characters ('a', 'b', 'c', ...) as early as possible. We iterate through the string positions from left to right (i = 0 to n-1). If the character at the current position word[i] hasn't been determined yet, we assign it the next available character from the alphabet. This assignment creates a new character group. The lcp matrix tells us which other indices must belong to the same group. Specifically, if lcp[i][j] > 0, then word[j] must equal word[i]. We enforce this by assigning the same character to all such j's.

1public String findString(int[][] lcp) {2    int n = lcp.length;3    char[] word = new char[n];4    char nextChar = 'a';5 6    for (int i = 0; i < n; i++) {7        if (word[i] == 0) { // Character not yet assigned8            if (nextChar > 'z') {9                return ""; // Need more than 26 distinct characters10            }11            word[i] = nextChar;12            for (int j = i + 1; j < n; j++) {13                if (lcp[i][j] > 0) {14                    word[j] = nextChar;15                }16            }17            nextChar++;18        }19    }

Phase 2: Verification The candidate string is built by only considering if lcp[i][j] is zero or non-zero. We must now verify the exact values. We do this by computing the LCP matrix for our generated word and comparing it to the input lcp matrix. This can be done efficiently with dynamic programming.

We iterate backwards from the end of the string. The LCP of two suffixes word[i...] and word[j...] is 0 if word[i] != word[j]. If word[i] == word[j], it is 1 + the LCP of the rest of the suffixes, word[i+1...] and word[j+1...].

1    // Verification Phase2    for (int i = n - 1; i >= 0; i--) {3        for (int j = n - 1; j >= 0; j--) {4            int currentLcp;5            if (word[i] != word[j]) {6                currentLcp = 0;7            } else {8                if (i == n - 1 || j == n - 1) {9                    currentLcp = 1;10                } else {11                    currentLcp = 1 + lcp[i + 1][j + 1];12                }13            }14            if (lcp[i][j] != currentLcp) {15                return "";16            }17        }18    }19    // A more robust way is to recompute from scratch20    int[][] computedLcp = new int[n][n];21    for (int i = n - 1; i >= 0; i--) {22        for (int j = n - 1; j >= 0; j--) {23            if (word[i] == word[j]) {24                computedLcp[i][j] = 1;25                if (i + 1 < n && j + 1 < n) {26                    computedLcp[i][j] += computedLcp[i + 1][j + 1];27                }28            }29            if (computedLcp[i][j] != lcp[i][j]) {30                return "";31            }32        }33    }34 35    return new String(word);36}

This combined approach correctly and efficiently finds the solution.