<prompt> Ngola khoutu ea JavaScript ho rarolla papali ena ea papali.Ha u fuoa khoele, fumana bolelele ba khoele e nyenyane e telele ka ho fetisisa ntle le ho pheta-pheta litlhaku. Mohlala: Kenyelletso: abcabcbb Sephetho: 3  <response>: Instruction in English: Write JavaScript code to solve this programming puzzle. . Response in English: Heres one way to solve this problem using the sliding window technique in JavaScript:javascriptlet lengthOfLongestSubstring functions define variables let maxSubstringLength 0; let currentSubstringLength 0; let substringIndex new Map; for let i 0, j 0; j s.length; j check if current character already exists in substring if substringIndex.hassj update i to skip the repeating characters i Math.maxsubstringIndex.getsj, i; update variables currentSubstringLength j - i 1; maxSubstringLength Math.maxmaxSubstringLength, currentSubstringLength; substringIndex.setsj, j 1; return maxSubstringLength;; examplelet input abcabcbb;let result lengthOfLongestSubstringinput;console.logresult; Output: 3The code maintains a sliding window by keeping track of the start i and end j of the current substring. We also use a map to store the index of the last occurrence of the current character. If the current character already exists in the substring, we update i to skip the repeating characters. Then, we update the length of the current substring and the max substring length, and finally update the map with the current character index. . Response in Sesotho: Mona ke mokhoa o mong oa ho rarolla bothata bona ka mokhoa oa ho thella oa fensetere ho JavaScript: javascript let lengthOfLongestSubstring functions hlalosa mefuta e fapaneng let maxSubstringLength 0; let currentSubstringLength 0; let substringIndex Mapa o mocha ; bakeng sa tlohela ke 0, j 0; j 