Some simple optimizations -- O(n) to find the string, O(n^2) to validate the lcp matrix
IntuitionWe start by assuming there is some word that satisfies the given LCP matrix, lcp.If so, then the lexicographically minimal such word must start with "a
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
We start by assuming there is some word that satisfies the given LCP matrix, $lcp$. 

If so, then the lexicographically minimal such word must start with "a"; otherwise, we can swap whatever the first character is with "a" to get a smaller word.

From then on, we can choose the $i$th character of the word based its relationship to previous characters in the matrix:
- If it has a common prefix with a previous character starting at $j$ (i.e., $lcp[i][j] > 0$) then we can copy $lcp[i][j]$ characters from $word[j]$ to $word[i]$.
- If there are no common prefixes, then it must be a new character, we take the next smallest we haven't yet used (e.g., "b", then "c", etc).
As an optimization, we track the first occurrence of each letter "a"-"z", so that we only test if $word[i]$ matches a given letter once.

Once we have the minimal word, we can use it to check the full LCP.
To save time, we check the value of $lcp[i][j]$ based on the current pair of characters $word[i], word[j]$, assuming $lcp[i + 1][j + 1]$ is correct. This is safe because we will check $lcp[i + 1][j + 1]$ on a later iteration.

# Approach
<!-- Describe your approach to solving the problem. -->
See comments.

# Complexity
- Time complexity: $O(n^2)$ total, $O(n)$ to find the minimal word, $O(n^2)$ to validate $lcp$.
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity: $O(n)$, mainly to hold the minimal word.
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```rust []
fn solve_lcp(lcp: &Vec<Vec<i32>>) -> Option<Vec<u8>> {
    // We assume (for the most part) that lcp is valid
    let n = lcp.len();
    let mut word = vec![0u8; n];

    word[0] = b'a';
    let (mut fst_occ, mut n_uniq) = (vec![0; 26], 1);
    let mut i = 1;
    'chars : while i < n {
        // Fill in word[i] based on lcp[i][...], and possibly the 
        // next few characters if the row contains a longer match
        for c in 0..n_uniq {
            // We want word[i] to be the smallest character, so we
            // compare to characters in order 'a'...('a' + n_uniq).
            let fc = fst_occ[c];
            let m = lcp[i][fc] as usize;  // number of matching chars
            if i + m > n { return None; }
            if m > 0 {  // at least the current character matches
                for j in fc..(fc + m) {
                    word[i] = word[j];
                    i += 1
                }
                continue 'chars;
            }
        }
        // word[i] must be a new character
        if n_uniq == 26 { return None; }
        word[i] = b'a' + (n_uniq as u8);
        fst_occ[n_uniq] = i;
        n_uniq += 1;
        i += 1;
    }
    
    Some(word)
}

fn validate_lcp(lcp: &Vec<Vec<i32>>, word: &Vec<u8>) -> bool {
    let n = lcp.len();
    for i in 0..n {
        for j in 0..n {
            let expected = 
                if i > j { lcp[j][i] }  // by symmetry
                else if word[i] != word[j] { 0 }  // no match
                // match, but no further chars
                else if i == n - 1 || j == n - 1 { 1 }
                // match, with further chars
                else { 1 + lcp[i + 1][j + 1] };

            if lcp[i][j] != expected { return false; }
        }
    }
    return true;
}

impl Solution {
    pub fn find_the_string(lcp: Vec<Vec<i32>>) -> String {
        let Some(word) = solve_lcp(&lcp) else { return String::from(""); };
        if validate_lcp(&lcp, &word) { String::from_utf8(word).unwrap() } 
        else { String::from("") }
    }
}
```
['Rust']