Skip to content

1844. Replace All Digits with Characters

Description

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

You must perform an operation 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 the result of the shift(s[i-1], s[i]) operation.

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

Note that shift(c, x) is not a preloaded function, but an operation to be implemented as part of the solution.

 

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 \(O(1)\).

1
2
3
4
5
6
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)

1
2
3
4
5
6
7
8
9
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);
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
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;
    }
};

1
2
3
4
5
6
7
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)
}

1
2
3
4
5
6
7
8
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('');
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
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()
    }
}

1
2
3
4
5
6
7
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;
}

Comments