There are two types of soup: type A and type B. Initially we have\u00a0 When we serve some soup, we give it to someone and we no longer have it.\u00a0 Each turn,\u00a0we will choose from the four operations with equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve\u00a0as much as we can.\u00a0 We stop once we no longer have some quantity of both types of soup.<\/p>\n Note that we do not have the operation where all 100 ml’s of soup B are used first.<\/p>\n Return the probability that soup A will be empty\u00a0first, plus half the probability that A and B become empty at the same time.<\/p>\n Notes:<\/strong><\/p>\n Time complexity: O(N^2) N ~ 5000 \/ 25 = 200<\/p>\n Space complexity: O(N^2)<\/p>\n C++<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" Problem There are two types of soup: type A and type B. Initially we have\u00a0N\u00a0ml of each type of soup. There are four kinds of…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[46],"tags":[18,31,177,120,17],"class_list":["post-3407","post","type-post","status-publish","format-standard","hentry","category-dynamic-programming","tag-dp","tag-math","tag-medium","tag-probability","tag-recursion","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/3407","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=3407"}],"version-history":[{"count":5,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/3407\/revisions"}],"predecessor-version":[{"id":3412,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/3407\/revisions\/3412"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=3407"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=3407"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=3407"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}N<\/code>\u00a0ml of each type of soup. There are four kinds of operations:<\/p>\n\n
Example:<\/strong>\r\nInput:<\/strong> N = 50\r\nOutput:<\/strong> 0.625\r\nExplanation:<\/strong> \r\nIf we choose the first two operations, A will become empty first. For the third operation, A and B will become empty at the same time. For the fourth operation, B will become empty first. So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.25 * (1 + 1 + 0.5 + 0) = 0.625.\r\n\r\n<\/pre>\n
\n
0 <= N <= 10^9<\/code>.<\/li>\n10^-6<\/code>\u00a0of the true value will be accepted as correct.<\/li>\n<\/ul>\nSolution 1: Recursion with Memorization<\/strong><\/h1>\n
\/\/ Author: Huahua\r\n\/\/ Running time: 4 ms\r\nclass Solution {\r\npublic:\r\n double soupServings(int N) {\r\n if (N >= 5000) return 1.0;\r\n return prob(N, N);\r\n }\r\nprivate:\r\n unordered_map<int, unordered_map<int, double>> m_;\r\n double prob(int A, int B) {\r\n static int ops[][2] = {{100, 0}, {75, 25}, {50, 50}, {25, 75}};\r\n if (A <= 0 && B <= 0) return 0.5;\r\n if (A <= 0) return 1.0; \r\n if (B <= 0) return 0.0;\r\n if (m_.count(A) && m_[A].count(B)) return m_[A][B]; \r\n m_[A][B] = 0.0;\r\n for (int i = 0; i < 4; ++i)\r\n m_[A][B] += 0.25 * prob(A - ops[i][0], B - ops[i][1]);\r\n return m_[A][B];\r\n }\r\n};<\/pre>\n