Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.
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.
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 <= 100s consists only of lowercase English letters and digits.shift(s[i-1], s[i]) <= 'z' for all odd indices i.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
"a1c1e1"
"a1b2c3d4e"
shifting-letters)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1844: Replace All Digits with Characters
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);
}
}
// Accepted solution for LeetCode #1844: Replace All Digits with Characters
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)
}
# Accepted solution for LeetCode #1844: Replace All Digits with Characters
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)
// Accepted solution for LeetCode #1844: Replace All Digits with Characters
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()
}
}
// Accepted solution for LeetCode #1844: Replace All Digits with Characters
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('');
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.