<prompt> Ha u fuoa likhoele tse peli, fumana me u khutlisetse khoele e telele ka ho fetisisa e tloaelehileng.str1 abcdef str2 abcxyz  <response>: Instruction in English: Given two strings, find and return the longest common substring. . Response in English: The longest common substring of the strings abcdef and abcxyz is abc.To find this, we can use dynamic programming to keep track of the length of the longest common substring ending at each position in both strings. Heres one way to do it:def longest_common_substringstr1, str2: m, n lenstr1, lenstr2 dp 0 n 1 for _ in rangem 1 max_len, end_idx 0, 0 for i in range1, m 1: for j in range1, n 1: if str1i - 1 str2j - 1: dpij dpi - 1j - 1 1 if dpij max_len: max_len dpij end_idx i - 1 return str1end_idx - max_len 1: end_idx 1 longest_common_substringabcdef, abcxyzThis code returns abc which is the longest common substring between the two strings. . Response in Sesotho: Mongolo o molelele o tloaelehileng oa likhoele abcdef le abcxyz ke abc. Ho fumana sena, re ka sebelisa dynamic programming ho boloka tlaleho ea bolelele ba substring e telele ka ho fetisisa e qetellang sebakeng se seng le se seng sa likhoele ka bobeli. Tsela e ngoe ea ho e etsa ke ena: def longest_common_substringstr1, str2: m, n lenstr1, lenstr2 dp 0 n 1 bakeng sa _ moelingm 1 max_len, end_idx 0, 0 bakeng sa i ka lethathamong1, m 1: bakeng sa j moeling1, n 1: haeba str1i - 1 str2j - 1: dpij dpi - 1j - 1 1 haeba dpij max_len: max_len dpij end_idx i - 1 khutlisetsa str1end_idx - max_len 1: end_idx 1 longest_common_substringabcdef, abcxyz Khoutu ena e khutlisa abc e leng eona khoele e telele ka ho fetisisa e tloaelehileng lipakeng tsa likhoele tse peli. 