Solution 1 - Optimized

For every block of 2k characters, reverse only its first k characters.

TimeComplexity: O(n)

SpaceComplexity: O(n)

class Solution:
    def reverseStr(self,s,k):
        a=list(s)
        for i in range(0,len(a),2*k): a[i:i+k]=reversed(a[i:i+k])
        return "".join(a)
if __name__ == "__main__": assert Solution().reverseStr("abcdefg",2)=="bacdfeg"

class Solution{reverseStr(s,k){const a=[...s];for(let i=0;i<a.length;i+=2*k){const x=a.slice(i,i+k).reverse();a.splice(i,x.length,...x);}return a.join("");}}console.assert(new Solution().reverseStr("abcdefg",2)==="bacdfeg");

class Solution{reverseStr(s:string,k:number):string{const a=[...s];for(let i=0;i<a.length;i+=2*k){const x=a.slice(i,i+k).reverse();a.splice(i,x.length,...x);}return a.join("");}}console.assert(new Solution().reverseStr("abcdefg",2)==="bacdfeg");

package main
import "fmt"
func rev(s string,k int)string{a:=[]byte(s);for i:=0;i<len(a);i+=2*k{j:=i+k-1;if j>=len(a){j=len(a)-1};for l:=i;l<j;l,j=l+1,j-1{a[l],a[j]=a[j],a[l]}};return string(a)}
func main(){fmt.Println(rev("abcdefg",2))}

fn rev(s:String,k:usize)->String{let mut a=s.into_bytes();for i in(0..a.len()).step_by(2*k){let e=(i+k).min(a.len());a[i..e].reverse()}String::from_utf8(a).unwrap()}fn main(){assert_eq!(rev("abcdefg".into(),2),"bacdfeg");}

0541 - Reverse String II

Solution 1 - Optimized