💥Beats 100% on runtime [EXPLAINED]
IntuitionWe need to figure out a string that can match a given LCP (Longest Common Prefix) matrix. The LCP matrix tells us how much of the prefix between any tw
# Intuition
We need to figure out a string that can match a given LCP (Longest Common Prefix) matrix. The LCP matrix tells us how much of the prefix between any two characters in the string matches. The key observation is that if the LCP between two characters is 0, their values in the string must be different. This allows us to construct a string step by step, adjusting characters and ensuring the string can match the given LCP matrix.

# Approach
Start with a string initialized to all 'a's and adjust the characters in the string by checking the LCP values between pairs of characters; if the LCP is 0 and the characters are the same, we increment one of them; then, construct a DP table to verify the string’s correctness by comparing the LCP matrix generated from this string with the given matrix.

# Complexity
- Time complexity:
O(n^2) because we iterate over pairs of characters twice: once to adjust the string and once to build and compare the DP table.

- Space complexity:
Also O(n^2) due to the DP table that stores the longest common prefix lengths between all pairs of characters, along with the string array.

# Code
```rust []
impl Solution {
    pub fn find_the_string(lcp: Vec<Vec<i32>>) -> String {
        let n = lcp.len();
        let mut cand = vec!['a'; n];

        // Update cand based on the LCP conditions
        for r in 0..n {
            for c in r..n {
                if lcp[r][c] == 0 && cand[c] == cand[r] {
                    cand[c] = (cand[c] as u8 + 1) as char; // Bump character
                }
            }
        }

        // Check if any character exceeds 'z'
        if cand.iter().any(|&ch| ch > 'z') {
            return String::new();
        }

        // Use DP to build the would-be LCP matrix
        let mut would = vec![vec![0; n + 1]; n + 1];

        for r in (0..n).rev() {
            for c in (0..n).rev() {
                if cand[r] == cand[c] {
                    would[r][c] = 1 + would[r + 1][c + 1];
                }
            }
        }

        // Compare the constructed LCP matrix with the given one
        for r in 0..n {
            for c in 0..n {
                if would[r][c] != lcp[r][c] {
                    return String::new();
                }
            }
        }

        cand.iter().collect() // Convert the vector of chars into a string
    }
}

```
['Array', 'String', 'Dynamic Programming', 'Greedy', 'Union Find', 'Graph', 'Rolling Hash', 'Matrix', 'Rust']