elif not and not and are even return gcd( > > < elif not and is even is odd return gcd( > yelif and not is odd is even return gcd(xy > return gcd(xy xboth and are odd note that the last step leads to recursive call with one even and one odd number consequentlyin every two callswe reduce the combined bit length of the two numbers by at least onemeaning that the time complexity is proportional to the sum of the number of bits in and yi (log log ) compute the maximum product of all entries but one suppose you are given an array of integersand are asked to find the largest product that can be made by multiplying all but one of the entries in (you cannot use an entry more than once for exampleif ithe result is if - ithe result is and if - - ithe result is - - one approach is to form the product of all the elementsand then find the maximum of / [iover all this takes multiplications (to form pand divisions (to compute each / [ ]suppose because of finite precision considerations we cannot use division-based approachwe can only use multiplications the brute-force solution entails computing all products of elementseach such product takes multiplicationsi ( time complexity given an array of length whose entries are integerscompute the largest product that can be made using entries in you cannot use an entry more than once array entries may be positivenegativeor your algorithm cannot use the division operatorexplicitly or implicitly hintconsider the products of the first and the last elements alternativelycount the number of negative entries and zero entries solutionthe brute-force approach to compute / [iis to multiplying the entries appearing before with those that appear after this leads to ( multiplicationssince for each term / [iwe need multiplications note that there is substantial overlap in computation when computing / [iand / [ in particularthe product of the entries appearing before is [itimes the product of the entries appearing before we can compute all products of entries before with multiplicationsand all products of entries after with multiplicationsand store these values in an array of prefix productsand an array of suffix products the desired result then is the maximum prefix-suffix product slightly more space efficient approach is to build the suffix product array and then iterate forward through the arrayusing single variable to compute the prefix products this prefix product is multiplied with the corresponding suffix product and the maximum prefix-suffix product is updated if required def find_biggest_n_minus_one_product ( )builds suffix products suffix_products list