<prompt> Hlahisa tharollo bakeng sa bothata bona ba leetcodeHo fanoe ka khoele e nang le mefuta e meraro feela ea litlhaku: , le , ngola tšebetso ho hlahloba hore na khoele ena e nepahetse. Re hlalosa bonnete ba khoele ka melao ena: Mashaka afe kapa afe a ka ho le letšehali a tlameha ho ba le mashaka a nepahetseng . Mashakana afe kapa afe a ka ho le letona a tlameha ho ba le mashaka a ka ho le letšehali . Mashaka a ka ho le letšehali a tlameha ho ea pele ho mashaka a ka ho le letona . e ka nkoa e le mashaka a le letona kapa le letšehali le le leng. masakana kapa khoele e se nang letho. Khoele e se nang letho le eona e ea sebetsa.  <response>: Instruction in English: Suggest a solution for this leetcode problem . Response in English: One solution to this problem is to use two stacks. Heres a step by step guide:- Initialize two stacks, one to store the indices of left parentheses and the other to store the indices of characters.- Iterate over the characters in the string.- If a character is , push its index to the left parentheses stack.- If the character is , push its index to the stack.- If the character is , pop from the left parentheses stack if its not empty, otherwise pop from the stack. If both stacks are empty, then return False as there is no matching left parenthesis or to balance this right parenthesis.- After the iteration, we are left with indices of some s and s in their respective stacks. The s can be used to balance the s from their right. So, for every remaining in the stack, we need to ensure there is a to its right. We can do this by checking that indices of s are greater than indices of s while popping from both stacks. If any is left without a to its right, return False.- Return TrueHere is a sample code in Python:pythondef checkValidStrings: left_parentheses stack to store indices of asterisk stack to store indices of for i, char in enumerates: if char : left_parentheses.appendi elif char : asterisk.appendi else: char if lenleft_parentheses 0: left_parentheses.pop elif lenasterisk 0: asterisk.pop else: return False while lenleft_parentheses 0 and lenasterisk 0: if left_parentheses.pop asterisk.pop: return False return lenleft_parentheses 0This solution has a time complexity of On and space complexity of On, where n is the length of the input string. . Response in Sesotho: Tharollo e ngoe ea bothata bona ke ho sebelisa li-stacks tse peli. Mona ke tataiso ea mohato ka mohato: - Qala ka mekoloko e meli, e ngoe e boloke li-indices tsa masakaneng a ka ho le letšehali me e ngoe e boloke li-indices tsa litlhaku tsa - Pheta litlhaku tse khoeleng. , sutumelletsa index ea eona ho stack ea mashaka a ka ho le letšehali. - Haeba tlhaku e le , sutumelletsa index ea eona ho stack. - Haeba tlhaku e le , pop ho tloha ka letsohong le letšehali la masakakana haeba ho se joalo. letho, ho seng joalo pop ho tsoa ho . Haeba mekotla ka bobeli e se na letho, khutlisetsa False kaha ha ho na mashaka a ka ho le letšehali kapa ho leka-lekanya mashaka ana a ka ho le letona. s le s ka mekoloko ea tsona. The s e ka sebelisoa ho leka-lekanya ho tloha ka ho le letona. Kahoo, bakeng sa e ngoe le e ngoe e setseng, re hloka ho netefatsa hore ho na le ke ka ho le letona. Re ka etsa sena ka ho lekola hore li-indices tsa s li kholo ho feta li-indices tsa s ha li ntse li hlaha ka bobeli. Haeba e sala e se na ho e nepahetseng, khutlisetsa Bohata - Khutlisa Nete Mona ke mohlala oa khoutu ho Python: python def checkValidStrings: left_parentheses stack ho boloka indices ea asterisk stack to store indices ea bakeng sa i, char in enumerateli: if char : left_parentheses.appendi elif char : asterisk.appendi e ngoe: char haeba lenleft_parentheses 0: left_parentheses.pop elif lenasterisk 0: asterisk.pop else: return False while lenleft_parentheses 0 le lenasterisk 0: if left_parentheses.pop asterisk.pop: return False return lenleft_parentheses 0 Tharollo ena e na le nako e rarahaneng ya On le sebaka sa ho rarahana sa On, moo n e leng bolelele ba se kentsweng. khoele. 