README.md

2301. Match Substring After Replacement

You are given two strings s and sub. You are also given a 2D character array mappings where mappings[i] = [oldi, newi] indicates that you may perform the following operation any number of times:

• Replace a character oldi of sub with newi.

Each character in sub cannot be replaced more than once.

Return true if it is possible to make sub a substring of s by replacing zero or more characters according to mappings. Otherwise, return false.

A substring is a contiguous non-empty sequence of characters within a

Example 1:

Input: s = "fool3e7bar", sub = "leet", mappings = [["e","3"],["t","7"],["t","8"]]
Output: true
Explanation: Replace the first 'e' in sub with '3' and 't' in sub with '7'.
Now sub = "l3e7" is a substring of s, so we return true.

Example 2:

Input: s = "fooleetbar", sub = "f00l", mappings = [["o","0"]]
Output: false
Explanation: The string "f00l" is not a substring of s and no replacements can be made.
Note that we cannot replace '0' with 'o'.

Example 3:

Input: s = "Fool33tbaR", sub = "leetd", mappings = [["e","3"],["t","7"],["t","8"],["d","b"],["p","b"]]
Output: true
Explanation: Replace the first and second 'e' in sub with '3' and 'd' in sub with 'b'.
Now sub = "l33tb" is a substring of s, so we return true.

Constraints:

• 1 <= sub.length <= s.length <= 5000
• 0 <= mappings.length <= 1000
• mappings[i].length == 2
• oldi != newi
• s and sub consist of uppercase and lowercase English letters and digits.
• oldi and newi are either uppercase or lowercase English letters or digits.

Solutions (Rust)

1. Solution

impl Solution {
    pub fn match_replacement(s: String, sub: String, mappings: Vec<Vec<char>>) -> bool {
        let s = s.as_bytes();
        let sub = sub.as_bytes();
        let mut bitmappings = [0_i128; 128];

        for i in 0..mappings.len() {
            let (old, new) = (mappings[i][0] as usize, mappings[i][1] as usize);
            bitmappings[old] |= 1 << new;
        }

        for i in 0..=s.len() - sub.len() {
            let mut flag = true;

            for j in 0..sub.len() {
                if sub[j] != s[i + j] && bitmappings[sub[j] as usize] & (1 << s[i + j]) == 0 {
                    flag = false;
                    break;
                }
            }

            if flag {
                return true;
            }
        }

        false
    }
}