Build confidence with an intuition-first walkthrough focused on two pointers fundamentals.
Given a string s, reverse the string according to the following rules:
Return s after reversing it.
Example 1:
Input: s = "ab-cd" Output: "dc-ba"
Example 2:
Input: s = "a-bC-dEf-ghIj" Output: "j-Ih-gfE-dCba"
Example 3:
Input: s = "Test1ng-Leet=code-Q!" Output: "Qedo1ct-eeLg=ntse-T!"
Constraints:
1 <= s.length <= 100s consists of characters with ASCII values in the range [33, 122].s does not contain '\"' or '\\'.Problem summary: Given a string s, reverse the string according to the following rules: All the characters that are not English letters remain in the same position. All the English letters (lowercase or uppercase) should be reversed. Return s after reversing it.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers
"ab-cd"
"a-bC-dEf-ghIj"
"Test1ng-Leet=code-Q!"
faulty-keyboard)reverse-letters-then-special-characters-in-a-string)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #917: Reverse Only Letters
class Solution {
public String reverseOnlyLetters(String s) {
char[] cs = s.toCharArray();
int i = 0, j = cs.length - 1;
while (i < j) {
while (i < j && !Character.isLetter(cs[i])) {
++i;
}
while (i < j && !Character.isLetter(cs[j])) {
--j;
}
if (i < j) {
char t = cs[i];
cs[i] = cs[j];
cs[j] = t;
++i;
--j;
}
}
return new String(cs);
}
}
// Accepted solution for LeetCode #917: Reverse Only Letters
func reverseOnlyLetters(s string) string {
cs := []rune(s)
i, j := 0, len(s)-1
for i < j {
for i < j && !unicode.IsLetter(cs[i]) {
i++
}
for i < j && !unicode.IsLetter(cs[j]) {
j--
}
if i < j {
cs[i], cs[j] = cs[j], cs[i]
i++
j--
}
}
return string(cs)
}
# Accepted solution for LeetCode #917: Reverse Only Letters
class Solution:
def reverseOnlyLetters(self, s: str) -> str:
cs = list(s)
i, j = 0, len(cs) - 1
while i < j:
while i < j and not cs[i].isalpha():
i += 1
while i < j and not cs[j].isalpha():
j -= 1
if i < j:
cs[i], cs[j] = cs[j], cs[i]
i, j = i + 1, j - 1
return "".join(cs)
// Accepted solution for LeetCode #917: Reverse Only Letters
impl Solution {
pub fn reverse_only_letters(s: String) -> String {
let mut cs: Vec<char> = s.chars().collect();
let n = cs.len();
let mut l = 0;
let mut r = n - 1;
while l < r {
if !cs[l].is_ascii_alphabetic() {
l += 1;
} else if !cs[r].is_ascii_alphabetic() {
r -= 1;
} else {
cs.swap(l, r);
l += 1;
r -= 1;
}
}
cs.iter().collect()
}
}
// Accepted solution for LeetCode #917: Reverse Only Letters
function reverseOnlyLetters(s: string): string {
const cs = [...s];
let [i, j] = [0, cs.length - 1];
while (i < j) {
while (!/[a-zA-Z]/.test(cs[i]) && i < j) {
i++;
}
while (!/[a-zA-Z]/.test(cs[j]) && i < j) {
j--;
}
[cs[i], cs[j]] = [cs[j], cs[i]];
i++;
j--;
}
return cs.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.