Question

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 $O(1)$.

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)

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);
  }
}

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;
  }
};

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)
}

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('');
}

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()
  }
}

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;
}