{"id":8766,"date":"2021-11-25T12:57:49","date_gmt":"2021-11-25T20:57:49","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=8766"},"modified":"2021-11-25T13:18:47","modified_gmt":"2021-11-25T21:18:47","slug":"leetcode-2007-find-original-array-from-doubled-array","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/hashtable\/leetcode-2007-find-original-array-from-doubled-array\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 2007. Find Original Array From Doubled Array"},"content":{"rendered":"\n

An integer array original<\/code> is transformed into a doubled<\/strong> array changed<\/code> by appending twice the value<\/strong> of every element in original<\/code>, and then randomly shuffling<\/strong> the resulting array.<\/p>\n\n\n\n

Given an array changed<\/code>, return original<\/code> if <\/em>changed<\/code> is a doubled<\/strong> array. If <\/em>changed<\/code> is not a doubled<\/strong> array, return an empty array. The elements in<\/em> original<\/code> may be returned in any<\/strong> order<\/em>.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input:<\/strong> changed = [1,3,4,2,6,8]\nOutput:<\/strong> [1,3,4]\nExplanation:<\/strong> One possible original array could be [1,3,4]:\n- Twice the value of 1 is 1 * 2 = 2.\n- Twice the value of 3 is 3 * 2 = 6.\n- Twice the value of 4 is 4 * 2 = 8.\nOther original arrays could be [4,3,1] or [3,1,4].\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> changed = [6,3,0,1]\nOutput:<\/strong> []\nExplanation:<\/strong> changed is not a doubled array.\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> changed = [1]\nOutput:<\/strong> []\nExplanation:<\/strong> changed is not a doubled array.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= changed.length <= 105<\/sup><\/code><\/li>
  • 0 <= changed[i] <= 105<\/sup><\/code><\/li><\/ul>\n\n\n\n
    Solution 1: Multiset<\/strong><\/h2>\n\n\n\n
    Start from the smallest number x, erase one x * 2 from the set.<\/p>\n\n\n\n
    Time complexity: O(nlogn)
    Space complexity: O(n)<\/p>\n\n\n\n
    \n
    C++\/Multiset<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  vector findOriginalArray(vector& changed) {\n    if (changed.size() & 1) return {};\n    const int n = changed.size() \/ 2;\n    multiset s(begin(changed), end(changed));    \n    while (ans.size() != n) {\n      int o = *begin(s);\n      s.erase(begin(s));\n      ans.push_back(o);\n      auto it = s.find(o * 2);\n      if (it == end(s)) return {};\n      s.erase(it);\n    }\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n\n\n\n
    Solution 2: Hashtable<\/strong><\/h2>\n\n\n\n
    Time complexity: O(max(nums) + n)
    Space complexity: O(max(nums))<\/p>\n\n\n\n
    \n
    C++\/Hashtable<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  vector findOriginalArray(vector& changed) {\n    if (changed.size() & 1) return {};\n    const int n = changed.size() \/ 2;\n    const int kMax = *max_element(begin(changed), end(changed));\n    vector m(kMax + 1);\n    for (int x : changed) ++m[x];\n    if (m[0] & 1) return {};\n    vector ans(m[0] \/ 2, 0);\n    for (int x = 1; ans.size() != n; ++x) {\n      if (x * 2 > kMax || m[x * 2] < m[x]) return {};      \n      ans.insert(end(ans), m[x], x);\n      m[x * 2] -= m[x];   \n    }\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n\n\n\n
    <\/p>\n","protected":false},"excerpt":{"rendered":"
    An integer array original is transformed into a doubled array changed by appending twice the value of every element in original, and then randomly shuffling the resulting array. Given an array changed, return original if changed is a doubled array. If changed is not a doubled array,…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[70],"tags":[82,177,335,656],"class_list":["post-8766","post","type-post","status-publish","format-standard","hentry","category-hashtable","tag-hashtable","tag-medium","tag-multiset","tag-treeset","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8766","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=8766"}],"version-history":[{"count":6,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8766\/revisions"}],"predecessor-version":[{"id":8774,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8766\/revisions\/8774"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=8766"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=8766"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=8766"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}