Build confidence with an intuition-first walkthrough focused on two pointers fundamentals.
Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string.
If there are fewer than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and leave the other as original.
Example 1:
Input: s = "abcdefg", k = 2 Output: "bacdfeg"
Example 2:
Input: s = "abcd", k = 2 Output: "bacd"
Constraints:
1 <= s.length <= 104s consists of only lowercase English letters.1 <= k <= 104Problem summary: Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string. If there are fewer than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and leave the other as original.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers
"abcdefg" 2
"abcd" 2
reverse-string)reverse-words-in-a-string-iii)faulty-keyboard)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #541: Reverse String II
class Solution {
public String reverseStr(String s, int k) {
char[] cs = s.toCharArray();
int n = cs.length;
for (int i = 0; i < n; i += k * 2) {
for (int l = i, r = Math.min(i + k - 1, n - 1); l < r; ++l, --r) {
char t = cs[l];
cs[l] = cs[r];
cs[r] = t;
}
}
return new String(cs);
}
}
// Accepted solution for LeetCode #541: Reverse String II
func reverseStr(s string, k int) string {
cs := []byte(s)
n := len(cs)
for i := 0; i < n; i += 2 * k {
for l, r := i, min(i+k-1, n-1); l < r; l, r = l+1, r-1 {
cs[l], cs[r] = cs[r], cs[l]
}
}
return string(cs)
}
# Accepted solution for LeetCode #541: Reverse String II
class Solution:
def reverseStr(self, s: str, k: int) -> str:
cs = list(s)
for i in range(0, len(cs), 2 * k):
cs[i : i + k] = reversed(cs[i : i + k])
return "".join(cs)
// Accepted solution for LeetCode #541: Reverse String II
struct Solution;
impl Solution {
fn rev_half(s: &mut [char], k: usize) -> &[char] {
if s.len() <= k {
s.reverse();
} else {
let half = &mut s[0..k];
half.reverse();
}
s
}
fn reverse_str(s: String, k: i32) -> String {
let k: usize = k as usize;
let mut s: Vec<char> = s.chars().collect();
let n = s.len();
let mut i = 0;
while i * 2 * k < n {
let r = (i + 1) * 2 * k;
if r < n {
let ss = &mut s[i * 2 * k..r];
Self::rev_half(ss, k);
} else {
let ss = &mut s[i * 2 * k..n];
Self::rev_half(ss, k);
}
i += 1;
}
s.iter().collect()
}
}
#[test]
fn test() {
let s = "abcdefg".to_string();
let k = 2;
let t = "bacdfeg".to_string();
assert_eq!(Solution::reverse_str(s, k), t);
}
// Accepted solution for LeetCode #541: Reverse String II
function reverseStr(s: string, k: number): string {
const n = s.length;
const cs = s.split('');
for (let i = 0; i < n; i += 2 * k) {
for (let l = i, r = Math.min(i + k - 1, n - 1); l < r; l++, r--) {
[cs[l], cs[r]] = [cs[r], cs[l]];
}
}
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.