Tejas9535

which question are talking about ?

why would you need avarage word lenght and all that. solution-1 You can just => check if the array is odd or even const even= countries % 2 === 0...

My soliution which was gitving undefined i have fixed it => const even= countries.length % 2 === 0 const len= Math.round(countries.length/2) if (even) { console.log(countries[len - 1], countries[len]) } else...

> solution-2 I think the correct solution would be: // ---------- Number of letters in country name ---------- // for (let i = 0; i < countries.length; i++) { countries[i]...

> I have a problem with Exercise level1 number 14: I don't know how to filter the array to print the companies that have more than 'o'?! if you don't...

> > > I have a problem with Exercise level1 number 14: I don't know how to filter the array to print the companies that have more than 'o'?! >...

> > itCompanies.forEach(element => { > > console.log(element); > > console.log(!/(.)._\1/.test(element)); if (/(o)._\1/.test(element)) { > > console.log(`${element} contains more than one 'O'`); > > } else { > > console.log(`${element}...

> > You can try to enable module "iptable_magle" with "modprod" in your host with `/sbin/modprobe iptable_mangle` > > For me it's work :) > > After days of trying,...