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1309. Decrypt String from Alphabet to Integer Mapping

Description

You are given a string s formed by digits and '#'. We want to map s to English lowercase characters as follows:

  • Characters ('a' to 'i') are represented by ('1' to '9') respectively.
  • Characters ('j' to 'z') are represented by ('10#' to '26#') respectively.

Return the string formed after mapping.

The test cases are generated so that a unique mapping will always exist.

 

Example 1:

Input: s = "10#11#12"
Output: "jkab"
Explanation: "j" -> "10#" , "k" -> "11#" , "a" -> "1" , "b" -> "2".

Example 2:

Input: s = "1326#"
Output: "acz"

 

Constraints:

  • 1 <= s.length <= 1000
  • s consists of digits and the '#' letter.
  • s will be a valid string such that mapping is always possible.

Solutions

  • class Solution {
    public String freqAlphabets(String s) {
        int i = 0, n = s.length();
        StringBuilder res = new StringBuilder();
        while (i < n) {
            if (i + 2 < n && s.charAt(i + 2) == '#') {
                res.append(get(s.substring(i, i + 2)));
                i += 3;
            } else {
                res.append(get(s.substring(i, i + 1)));
                i += 1;
            }
        }
        return res.toString();
    }

    private char get(String s) {
        return (char) ('a' + Integer.parseInt(s) - 1);
    }
}

  • class Solution:
    def freqAlphabets(self, s: str) -> str:
        def get(s):
            return chr(ord('a') + int(s) - 1)

        i, n = 0, len(s)
        res = []
        while i < n:
            if i + 2 < n and s[i + 2] == '#':
                res.append(get(s[i : i + 2]))
                i += 3
            else:
                res.append(get(s[i]))
                i += 1
        return ''.join(res)


  • function freqAlphabets(s: string): string {
    const n = s.length;
    const ans = [];
    let i = 0;
    while (i < n) {
        if (s[i + 2] == '#') {
            ans.push(s.slice(i, i + 2));
            i += 3;
        } else {
            ans.push(s[i]);
            i += 1;
        }
    }
    return ans.map(c => String.fromCharCode('a'.charCodeAt(0) + Number(c) - 1)).join('');
}


  • impl Solution {
    pub fn freq_alphabets(s: String) -> String {
        let s = s.as_bytes();
        let n = s.len();
        let mut res = String::new();
        let mut i = 0;
        while i < n {
            let code: u8;
            if s.get(i + 2).is_some() && s[i + 2] == b'#' {
                code = (s[i] - b'0') * 10 + s[i + 1];
                i += 3;
            } else {
                code = s[i];
                i += 1;
            }
            res.push(char::from(('a' as u8) + code - b'1'));
        }
        res
    }
}

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