<prompt> Hlahisa mekhoa ea ho ntlafatsa khoutu ka hara boloko e fanoeng.bakeng saint i0;in;i haebai20 System.out.printlni;  <response>: Instruction in English: Suggest ways to optimize the code in the given block. . Response in English: There are several ways you could optimize the code block youve provided.1. Increment i by 2 instead of 1. As youre only printing out the even values, you could change the increment in the for loop to i 2 instead of i. This way, you dont need the if statement, so the code will execute faster:forint i 0; i n; i 2 System.out.printlni;2. Preallocate the size of the buffer. If you know the maximum size of integers, you could pre-allocate the size of the buffer to reduce the time spent on memory allocation. Additionally, you could use StringBuilder instead of System.out.println to be more efficient when dealing with string manipulations.StringBuilder sb new StringBuildern;forint i 0; i n; i 2 sb.appendi; sb.appendn;System.out.printlnsb.toString; . Response in Sesotho: Ho na le mekhoa e mengata eo u ka ntlafatsang code block eo u faneng ka eona. 1. Keketseho ea i ka 2 sebakeng sa 1. Ha u ntse u hatisa feela boleng bo lekanang, u ka fetola increment ho loop ho i 2 sebakeng sa i. Ka tsela ena, ha u hloke polelo ea if, kahoo khoutu e tla sebetsa kapele: bakeng saint i 0; i n; i 2 System.out.printlni; 2. Beha pele boholo ba buffer. Haeba u tseba boholo ba boholo ba linomoro, u ka arola esale pele boholo ba buffer ho fokotsa nako e sebelisoang kabong ea memori. Ho feta moo, o ka sebelisa StringBuilder sebakeng sa System.out.println hore e sebetse hantle haholoanyane ha e sebetsana le likhoele tsa manipulations. StringBuilder sb StringBuilder e nchan; bakeng saint i 0; ke n; ke 2 sb.appendi; sb.appendn; System.out.printlnsb.toString; 