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.
Given a string s containing only lowercase English letters and the '?' character, convert all the '?' characters into lowercase letters such that the final string does not contain any consecutive repeating characters. You cannot modify the non '?' characters.
It is guaranteed that there are no consecutive repeating characters in the given string except for '?'.
Return the final string after all the conversions (possibly zero) have been made. If there is more than one solution, return any of them. It can be shown that an answer is always possible with the given constraints.
Example 1:
Input: s = "?zs" Output: "azs" Explanation: There are 25 solutions for this problem. From "azs" to "yzs", all are valid. Only "z" is an invalid modification as the string will consist of consecutive repeating characters in "zzs".
Example 2:
Input: s = "ubv?w" Output: "ubvaw" Explanation: There are 24 solutions for this problem. Only "v" and "w" are invalid modifications as the strings will consist of consecutive repeating characters in "ubvvw" and "ubvww".
Constraints:
1 <= s.length <= 100s consist of lowercase English letters and '?'.Problem summary: Given a string s containing only lowercase English letters and the '?' character, convert all the '?' characters into lowercase letters such that the final string does not contain any consecutive repeating characters. You cannot modify the non '?' characters. It is guaranteed that there are no consecutive repeating characters in the given string except for '?'. Return the final string after all the conversions (possibly zero) have been made. If there is more than one solution, return any of them. It can be shown that an answer is always possible with the given constraints.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
"?zs"
"ubv?w"
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1576: Replace All ?'s to Avoid Consecutive Repeating Characters
class Solution {
public String modifyString(String s) {
char[] cs = s.toCharArray();
int n = cs.length;
for (int i = 0; i < n; ++i) {
if (cs[i] == '?') {
for (char c = 'a'; c <= 'c'; ++c) {
if ((i > 0 && cs[i - 1] == c) || (i + 1 < n && cs[i + 1] == c)) {
continue;
}
cs[i] = c;
break;
}
}
}
return String.valueOf(cs);
}
}
// Accepted solution for LeetCode #1576: Replace All ?'s to Avoid Consecutive Repeating Characters
func modifyString(s string) string {
n := len(s)
cs := []byte(s)
for i := range s {
if cs[i] == '?' {
for c := byte('a'); c <= byte('c'); c++ {
if (i > 0 && cs[i-1] == c) || (i+1 < n && cs[i+1] == c) {
continue
}
cs[i] = c
break
}
}
}
return string(cs)
}
# Accepted solution for LeetCode #1576: Replace All ?'s to Avoid Consecutive Repeating Characters
class Solution:
def modifyString(self, s: str) -> str:
s = list(s)
n = len(s)
for i in range(n):
if s[i] == "?":
for c in "abc":
if (i and s[i - 1] == c) or (i + 1 < n and s[i + 1] == c):
continue
s[i] = c
break
return "".join(s)
// Accepted solution for LeetCode #1576: Replace All ?'s to Avoid Consecutive Repeating Characters
struct Solution;
impl Solution {
fn modify_string(s: String) -> String {
let mut s: Vec<char> = s.chars().collect();
let n = s.len();
for i in 0..n {
if s[i] == '?' {
s[i] = 'a';
while (i > 0 && s[i] == s[i - 1]) || (i + 1 < n && s[i] == s[i + 1]) {
s[i] = (s[i] as u8 + 1) as char;
}
}
}
s.into_iter().collect()
}
}
#[test]
fn test() {
let s = "?zs".to_string();
let res = "azs".to_string();
assert_eq!(Solution::modify_string(s), res);
let s = "ubv?w".to_string();
let res = "ubvaw".to_string();
assert_eq!(Solution::modify_string(s), res);
let s = "j?qg??b".to_string();
let res = "jaqgacb".to_string();
assert_eq!(Solution::modify_string(s), res);
let s = "??yw?ipkj?".to_string();
let res = "abywaipkja".to_string();
assert_eq!(Solution::modify_string(s), res);
}
// Accepted solution for LeetCode #1576: Replace All ?'s to Avoid Consecutive Repeating Characters
function modifyString(s: string): string {
const cs = s.split('');
const n = s.length;
for (let i = 0; i < n; ++i) {
if (cs[i] === '?') {
for (const c of 'abc') {
if ((i > 0 && cs[i - 1] === c) || (i + 1 < n && cs[i + 1] === c)) {
continue;
}
cs[i] = c;
break;
}
}
}
return cs.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.