and maps them to objects which are comparable--this function defines the sort order for exampleif =[ then sort(key=lambda xstr( )maps integers to stringsand is updated to [ ] the entries are sorted according to the lexicographical ordering of their string representation if reverse is set to truethe sort is in descending orderotherwise it is in ascending order the sorted function takes an iterable and return new list containing all items from the iterable in ascending order the original list is unchanged for exampleb sorted(akey=lambda xstr( )leaves unchangedand assigns to [ the optional arguments key and reverse work identically to sort compute the intersection of two sorted arrays natural implementation for search engine is to retrieve documents that match the set of words in query by maintaining an inverted index each page is assigned an integer identifierits documentid an inverted index is mapping that takes word and returns sorted array of page-ids which contain --the sort order could befor examplethe page rank in descending order when query contains multiple wordsthe search engine finds the sorted array for each word and then computes the intersection of these arrays--these are the pages containing all the words in the query the most computationally intensive step of doing this is finding the intersection of the sorted arrays write program which takes as input two sorted arraysand returns new array containing elements that are present in both of the input arrays the input arrays may have duplicate entriesbut the returned array should be free of duplicates for examplethe input is and iyour output should be hintsolve the problem if the input array lengths differ by orders of magnitude what if they are approximately equalsolutionthe brute-force algorithm is "loop join" traversing through all the elements of one array and comparing them to the elements of the other array let and be the lengths of the two input arrays def intersect_two_sorted_arrays (ab)return [ for ia in enumerate (aif ( = or ! [ ]and in bthe brute-force algorithm has (mntime complexity since both the arrays are sortedwe can make some optimizations firstwe can iterate through the first array and use binary search in array to test if the element is present in the second array def intersect_two_sorted_arrays (ab)def is_present ( ) bisect bisect_left (bkreturn len(band [ = return for ia in enumerate (aif ( = or ! [ ]and is_present (