Build confidence with an intuition-first walkthrough focused on two pointers fundamentals.
You are given a string s and an integer k.
Reverse the first k characters of s and return the resulting string.
Example 1:
Input: s = "abcd", k = 2
Output: "bacd"
Explanation:
The first k = 2 characters "ab" are reversed to "ba". The final resulting string is "bacd".
Example 2:
Input: s = "xyz", k = 3
Output: "zyx"
Explanation:
The first k = 3 characters "xyz" are reversed to "zyx". The final resulting string is "zyx".
Example 3:
Input: s = "hey", k = 1
Output: "hey"
Explanation:
The first k = 1 character "h" remains unchanged on reversal. The final resulting string is "hey".
Constraints:
1 <= s.length <= 100s consists of lowercase English letters.1 <= k <= s.lengthProblem summary: You are given a string s and an integer k. Reverse the first k characters of s and return the resulting string.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers
"abcd" 2
"xyz" 3
"hey" 1
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3794: Reverse String Prefix
class Solution {
public String reversePrefix(String s, int k) {
StringBuilder sb = new StringBuilder(s.substring(0, k));
return sb.reverse().toString() + s.substring(k);
}
}
// Accepted solution for LeetCode #3794: Reverse String Prefix
func reversePrefix(s string, k int) string {
t := []byte(s[:k])
slices.Reverse(t)
return string(t) + s[k:]
}
# Accepted solution for LeetCode #3794: Reverse String Prefix
class Solution:
def reversePrefix(self, s: str, k: int) -> str:
return s[:k][::-1] + s[k:]
// Accepted solution for LeetCode #3794: Reverse String Prefix
fn reverse_prefix(s: String, k: i32) -> String {
let k = k as usize;
let mut pre: Vec<_> = s.chars().take(k).collect();
pre.reverse();
let mut post: Vec<_> = s.chars().skip(k).collect();
pre.append(&mut post);
pre.into_iter().collect()
}
fn main() {
let ret = reverse_prefix("abcd".to_string(), 2);
println!("ret={ret}");
}
#[test]
fn test() {
assert_eq!(reverse_prefix("abcd".to_string(), 2), "bacd");
assert_eq!(reverse_prefix("xyz".to_string(), 3), "zyx");
assert_eq!(reverse_prefix("hey".to_string(), 1), "hey");
}
// Accepted solution for LeetCode #3794: Reverse String Prefix
function reversePrefix(s: string, k: number): string {
return s.slice(0, k).split('').reverse().join('') + s.slice(k);
}
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.