LeetCode 1844. Replace All Digits with Characters Solution in Java, C++, Python & More | Explanation + Code

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).

PythonJavaC++GoTypeScriptRustC

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)(code-box)

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); } }(code-box)

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; } };(code-box)

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) }(code-box)

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(''); }(code-box)

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() } }(code-box)

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; }(code-box)