Alice and Bob take turns playing a game, with Alice starting first.<\/p>\n\n\n\n
Initially, there are Also, if a player cannot make a move, he\/she loses the game.<\/p>\n\n\n\n Given a positive integer Example 1:<\/strong><\/p>\n\n\n\n Example 2:<\/strong><\/p>\n\n\n\n Example 3:<\/strong><\/p>\n\n\n\n Example 4:<\/strong><\/p>\n\n\n\n Example 5:<\/strong><\/p>\n\n\n\n Constraints:<\/strong><\/p>\n\n\n\n Let win(n) denotes whether the current play will win or not. Time complexity: O(nsqrt(n)) Alice and Bob take turns playing a game, with Alice starting first. Initially, there are n stones in a pile. On each player’s turn, that player makes…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[581],"tags":[18,27,217],"class_list":["post-7107","post","type-post","status-publish","format-standard","hentry","category-game-theory","tag-dp","tag-game","tag-hard","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7107","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=7107"}],"version-history":[{"count":3,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7107\/revisions"}],"predecessor-version":[{"id":7110,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7107\/revisions\/7110"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=7107"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=7107"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=7107"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}n<\/code> stones in a pile. On each player’s turn, that player makes a move<\/em> consisting of removing any<\/strong> non-zero square number<\/strong> of stones in the pile.<\/p>\n\n\n\nn<\/code>. Return True<\/code> if and only if Alice wins the game otherwise return False<\/code>, assuming both players play optimally.<\/p>\n\n\n\nInput:<\/strong> n = 1\nOutput:<\/strong> true\nExplanation: <\/strong>Alice can remove 1 stone winning the game because Bob doesn't have any moves.<\/pre>\n\n\n\n
Input:<\/strong> n = 2\nOutput:<\/strong> false\nExplanation: <\/strong>Alice can only remove 1 stone, after that Bob removes the last one winning the game (2 -> 1 -> 0).<\/pre>\n\n\n\n
Input:<\/strong> n = 4\nOutput:<\/strong> true\nExplanation:<\/strong> n is already a perfect square, Alice can win with one move, removing 4 stones (4 -> 0).\n<\/pre>\n\n\n\n
Input:<\/strong> n = 7\nOutput:<\/strong> false\nExplanation: <\/strong>Alice can't win the game if Bob plays optimally.\nIf Alice starts removing 4 stones, Bob will remove 1 stone then Alice should remove only 1 stone and finally Bob removes the last one (7 -> 3 -> 2 -> 1 -> 0). \nIf Alice starts removing 1 stone, Bob will remove 4 stones then Alice only can remove 1 stone and finally Bob removes the last one (7 -> 6 -> 2 -> 1 -> 0).<\/pre>\n\n\n\n
Input:<\/strong> n = 17\nOutput:<\/strong> false\nExplanation: <\/strong>Alice can't win the game if Bob plays optimally.\n<\/pre>\n\n\n\n
1 <= n <= 10^5<\/code><\/li><\/ul>\n\n\n\nSolution: Recursion w\/ Memoization \/ DP<\/strong><\/h2>\n\n\n\n
Try all possible square numbers and see whether the other player will lose or not.
win(n) = any(win(n – i*i) == False) ? True : False
base case: win(0) = False<\/p>\n\n\n\n
Space complexity: O(n)<\/p>\n\n\n\nC++<\/h2>\n
\nclass Solution {\npublic:\n bool winnerSquareGame(int n) { \n vectorJava<\/h2>\n
\nclass Solution {\n private int[] cache;\n public boolean winnerSquareGame(int n) {\n this.cache = new int[n + 1];\n return this.win(n) > 0;\n }\n \n private int win(int n) {\n if (n == 0) return -1;\n if (this.cache[n] != 0) return this.cache[n];\n for (int i = (int)Math.sqrt(n); i >= 1; --i)\n if (win(n - i * i) < 0) \n return this.cache[n] = 1;\n return this.cache[n] = -1;\n }\n}\n<\/pre>\n\n<\/div>Python3<\/h2>\n
\nclass Solution:\n def winnerSquareGame(self, n: int) -> bool:\n dp = [None] * (n + 1)\n dp[0] = False\n for i in range(0, n): \n if dp[i]: continue\n for j in range(1, n + 1): \n if i + j * j > n: break\n dp[i + j * j] = True\n return dp[n]\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"