1309. Decrypt String from Alphabet to Integer Mapping

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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

Solution 1: Simulation

We can directly simulate the process.

Traverse the string $s$. For the current index $i$, if $i + 2 < n$ and $s[i + 2]$ is #, then convert the substring formed by $s[i]$ and $s[i + 1]$ to an integer, add the ASCII value of a minus 1, convert it to a character, add it to the result array, and increment $i$ by 3. Otherwise, convert $s[i]$ to an integer, add the ASCII value of a minus 1, convert it to a character, add it to the result array, and increment $i$ by 1.

Finally, convert the result array to a string and return it.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.

Python3

class Solution:
    def freqAlphabets(self, s: str) -> str:
        ans = []
        i, n = 0, len(s)
        while i < n:
            if i + 2 < n and s[i + 2] == "#":
                ans.append(chr(int(s[i : i + 2]) + ord("a") - 1))
                i += 3
            else:
                ans.append(chr(int(s[i]) + ord("a") - 1))
                i += 1
        return "".join(ans)

Java

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

C++

class Solution {
public:
    string freqAlphabets(string s) {
        string ans = "";
        int i = 0, n = s.size();
        while (i < n) {
            if (i + 2 < n && s[i + 2] == '#') {
                ans += char(stoi(s.substr(i, 2)) + 'a' - 1);
                i += 3;
            } else {
                ans += char(s[i] - '0' + 'a' - 1);
                i += 1;
            }
        }
        return ans;
    }
};

Go

func freqAlphabets(s string) string {
	var ans []byte
	for i, n := 0, len(s); i < n; {
		if i+2 < n && s[i+2] == '#' {
			num := (int(s[i])-'0')*10 + int(s[i+1]) - '0'
			ans = append(ans, byte(num+int('a')-1))
			i += 3
		} else {
			num := int(s[i]) - '0'
			ans = append(ans, byte(num+int('a')-1))
			i += 1
		}
	}
	return string(ans)
}

TypeScript

function freqAlphabets(s: string): string {
    const ans: string[] = [];
    for (let i = 0, n = s.length; i < n; ) {
        if (i + 2 < n && s[i + 2] === '#') {
            ans.push(String.fromCharCode(96 + +s.slice(i, i + 2)));
            i += 3;
        } else {
            ans.push(String.fromCharCode(96 + +s[i]));
            i++;
        }
    }
    return ans.join('');
}

Rust

impl Solution {
    pub fn freq_alphabets(s: String) -> String {
        let s = s.as_bytes();
        let mut ans = String::new();
        let mut i = 0;
        let n = s.len();
        while i < n {
            if i + 2 < n && s[i + 2] == b'#' {
                let num = (s[i] - b'0') * 10 + (s[i + 1] - b'0');
                ans.push((96 + num) as char);
                i += 3;
            } else {
                let num = s[i] - b'0';
                ans.push((96 + num) as char);
                i += 1;
            }
        }
        ans
    }
}

C

char* freqAlphabets(char* s) {
    int n = strlen(s);
    int i = 0;
    int j = 0;
    char* ans = malloc(sizeof(s) * n);
    while (i < n) {
        int t;
        if (i + 2 < n && s[i + 2] == '#') {
            t = (s[i] - '0') * 10 + s[i + 1];
            i += 3;
        } else {
            t = s[i];
            i += 1;
        }
        ans[j++] = 'a' + t - '1';
    }
    ans[j] = '\0';
    return ans;
}