Update s0014_longest_common_prefix.rs

[pro100skm]

Solution with one loop

[shivaraaman]

Old code only finds the longest common prefix and doesn't support . or *, so it fails on regex cases. New code uses 2D DP with nested loops to handle full pattern matching, passes all test cases, and runs efficiently in O(m × n) time for strings of length m and n. This code Suggests Solution using 2D Dynamic Programming with nested loops to handle . and * patterns for full string match.

// submission codes start here

impl Solution {
pub fn is_match(s: String, p: String) -> bool {
let s = s.as_bytes(); // Convert strings to byte arrays for fast indexing
let p = p.as_bytes();
let m = s.len();
let n = p.len();
let mut dp = vec![vec![false; n + 1]; m + 1];
dp[0][0] = true; // Empty pattern matches empty string
for j in 2..=n {
if p[j - 1] == b'' {
dp[0][j] = dp[0][j - 2];
}
}
for i in 1..=m {
for j in 1..=n {
if p[j - 1] == b'.' || p[j - 1] == s[i - 1] {
dp[i][j] = dp[i - 1][j - 1];
} else if p[j - 1] == b'' {
dp[i][j] = dp[i][j - 2];
if p[j - 2] == b'.' || p[j - 2] == s[i - 1] {
dp[i][j] |= dp[i - 1][j];
}
}
}
}
dp[m][n]
}
}

// submission codes end