README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1800-1899/1844.Replace%20All%20Digits%20with%20Characters/README_EN.md	1300	Biweekly Contest 51 Q1	String

1844. Replace All Digits with Characters

中文文档

Description

You are given a 0-indexed string s that has lowercase English letters in its even indices and digits in its odd indices.

There is a function shift(c, x), where c is a character and x is a digit, that returns the xth character after c.

• For example, shift('a', 5) = 'f' and shift('x', 0) = 'x'.

For every odd index i, you want to replace the digit s[i] with shift(s[i-1], s[i]).

Return s after replacing all digits. It is guaranteed that shift(s[i-1], s[i]) will never exceed 'z'.

 

Example 1:

Input: s = "a1c1e1"
Output: "abcdef"
Explanation: The digits are replaced as follows:
- s[1] -> shift('a',1) = 'b'
- s[3] -> shift('c',1) = 'd'
- s[5] -> shift('e',1) = 'f'

Example 2:

Input: s = "a1b2c3d4e"
Output: "abbdcfdhe"
Explanation: The digits are replaced as follows:
- s[1] -> shift('a',1) = 'b'
- s[3] -> shift('b',2) = 'd'
- s[5] -> shift('c',3) = 'f'
- s[7] -> shift('d',4) = 'h'

 

Constraints:

• 1 <= s.length <= 100
• s consists only of lowercase English letters and digits.
• shift(s[i-1], s[i]) <= 'z' for all odd indices i.

Solutions

Solution 1: Simulation

Traverse the string, for characters at odd indices, replace them with the character that is a certain number of positions after the previous character.

Finally, return the replaced string.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

Python3

class Solution:
    def replaceDigits(self, s: str) -> str:
        s = list(s)
        for i in range(1, len(s), 2):
            s[i] = chr(ord(s[i - 1]) + int(s[i]))
        return ''.join(s)

Java

class Solution {
    public String replaceDigits(String s) {
        char[] cs = s.toCharArray();
        for (int i = 1; i < cs.length; i += 2) {
            cs[i] = (char) (cs[i - 1] + (cs[i] - '0'));
        }
        return String.valueOf(cs);
    }
}

C++

class Solution {
public:
    string replaceDigits(string s) {
        int n = s.size();
        for (int i = 1; i < n; i += 2) {
            s[i] = s[i - 1] + s[i] - '0';
        }
        return s;
    }
};

Go

func replaceDigits(s string) string {
	cs := []byte(s)
	for i := 1; i < len(s); i += 2 {
		cs[i] = cs[i-1] + cs[i] - '0'
	}
	return string(cs)
}

TypeScript

function replaceDigits(s: string): string {
    const n = s.length;
    const ans = [...s];
    for (let i = 1; i < n; i += 2) {
        ans[i] = String.fromCharCode(ans[i - 1].charCodeAt(0) + Number(ans[i]));
    }
    return ans.join('');
}

Rust

impl Solution {
    pub fn replace_digits(s: String) -> String {
        let n = s.len();
        let mut ans = s.into_bytes();
        let mut i = 1;
        while i < n {
            ans[i] = ans[i - 1] + (ans[i] - b'0');
            i += 2;
        }
        ans.into_iter().map(char::from).collect()
    }
}

C

char* replaceDigits(char* s) {
    int n = strlen(s);
    for (int i = 1; i < n; i += 2) {
        s[i] = s[i - 1] + s[i] - '0';
    }
    return s;
}