Build confidence with an intuition-first walkthrough focused on two pointers fundamentals.
You are given a string s consisting of lowercase English letters and special characters.
Your task is to perform these in order:
Return the resulting string after performing the reversals.
Example 1:
Input: s = ")ebc#da@f("
Output: "(fad@cb#e)"
Explanation:
['e', 'b', 'c', 'd', 'a', 'f']:
['f', 'a', 'd', 'c', 'b', 'e']s becomes ")fad#cb@e("[')', '#', '@', '(']:
['(', '@', '#', ')']s becomes "(fad@cb#e)"Example 2:
Input: s = "z"
Output: "z"
Explanation:
The string contains only one letter, and reversing it does not change the string. There are no special characters.
Example 3:
Input: s = "!@#$%^&*()"
Output: ")(*&^%$#@!"
Explanation:
The string contains no letters. The string contains all special characters, so reversing the special characters reverses the whole string.
Constraints:
1 <= s.length <= 100s consists only of lowercase English letters and the special characters in "!@#$%^&*()".Problem summary: You are given a string s consisting of lowercase English letters and special characters. Your task is to perform these in order: Reverse the lowercase letters and place them back into the positions originally occupied by letters. Reverse the special characters and place them back into the positions originally occupied by special characters. Return the resulting string after performing the reversals.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers
")ebc#da@f(""z"
"!@#$%^&*()"
reverse-only-letters)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3823: Reverse Letters Then Special Characters in a String
class Solution {
public String reverseByType(String s) {
StringBuilder a = new StringBuilder();
StringBuilder b = new StringBuilder();
char[] t = s.toCharArray();
for (char c : t) {
if (Character.isLetter(c)) {
a.append(c);
} else {
b.append(c);
}
}
int j = a.length(), k = b.length();
for (int i = 0; i < t.length; ++i) {
if (Character.isLetter(t[i])) {
t[i] = a.charAt(--j);
} else {
t[i] = b.charAt(--k);
}
}
return new String(t);
}
}
// Accepted solution for LeetCode #3823: Reverse Letters Then Special Characters in a String
func reverseByType(s string) string {
a := make([]byte, 0)
b := make([]byte, 0)
t := []byte(s)
for _, c := range t {
if (c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z') {
a = append(a, c)
} else {
b = append(b, c)
}
}
j, k := len(a), len(b)
for i := 0; i < len(t); i++ {
if (t[i] >= 'A' && t[i] <= 'Z') || (t[i] >= 'a' && t[i] <= 'z') {
j--
t[i] = a[j]
} else {
k--
t[i] = b[k]
}
}
return string(t)
}
# Accepted solution for LeetCode #3823: Reverse Letters Then Special Characters in a String
class Solution:
def reverseByType(self, s: str) -> str:
a = []
b = []
for c in s:
if c.isalpha():
a.append(c)
else:
b.append(c)
return ''.join(a.pop() if c.isalpha() else b.pop() for c in s)
// Accepted solution for LeetCode #3823: Reverse Letters Then Special Characters in a String
fn reverse_by_type(s: String) -> String {
let len = s.len();
let cs: Vec<_> = s.chars().collect();
let mut ret: Vec<char> = cs.clone();
let mut left = 0;
let mut right = len - 1;
while left < right {
while left < len && !cs[left].is_ascii_alphabetic() {
left += 1;
}
while right > 0 && !cs[right].is_ascii_alphabetic() {
right -= 1;
}
if left >= right {
break;
}
ret[left] = cs[right];
ret[right] = cs[left];
left += 1;
right -= 1;
}
left = 0;
right = len - 1;
while left < right {
while left < len && cs[left].is_ascii_alphabetic() {
left += 1;
}
while right > 0 && cs[right].is_ascii_alphabetic() {
right -= 1;
}
if left >= right {
break;
}
ret[left] = cs[right];
ret[right] = cs[left];
left += 1;
right -= 1;
}
ret.into_iter().collect()
}
fn main() {
let ret = reverse_by_type(")ebc#da@f(".to_string());
println!("ret={ret}");
}
#[test]
fn test() {
assert_eq!(reverse_by_type("(q".to_string()), "(q");
assert_eq!(reverse_by_type(")ebc#da@f(".to_string()), "(fad@cb#e)");
assert_eq!(reverse_by_type("z".to_string()), "z");
assert_eq!(reverse_by_type("!@#$%^&*()".to_string()), ")(*&^%$#@!");
}
// Accepted solution for LeetCode #3823: Reverse Letters Then Special Characters in a String
function reverseByType(s: string): string {
const a: string[] = [];
const b: string[] = [];
const t = s.split('');
for (const c of t) {
if (/[a-zA-Z]/.test(c)) {
a.push(c);
} else {
b.push(c);
}
}
let j = a.length,
k = b.length;
for (let i = 0; i < t.length; i++) {
if (/[a-zA-Z]/.test(t[i])) {
t[i] = a[--j];
} else {
t[i] = b[--k];
}
}
return t.join('');
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.