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Formatted question description: https://leetcode.ca/all/1844.html

1844. Replace All Digits with Characters

Level

Easy

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 x-th 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.

Solution

Loop over odd indices of s and convert the characters to the shifted characters. Then return the modified string.

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

############

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);
    }
}

  • // OJ: https://leetcode.com/problems/replace-all-digits-with-characters/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    string replaceDigits(string s) {
        for (int i = 1; i < s.size(); i += 2) {
            s[i] = s[i-1] + (s[i] - '0');
        }
        return s;
    }
};

  • 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)

############

# 1844. Replace All Digits with Characters
# https://leetcode.com/problems/replace-all-digits-with-characters/

class Solution:
    def replaceDigits(self, s: str) -> str:
        res = []
        
        for i, c in enumerate(s):
            if c.isnumeric():
                res.append(chr(ord(s[i-1]) + int(c)))
            else:
                res.append(c)
        
        return "".join(res)


  • 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()
    }
}

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