{"id":9234,"date":"2021-12-25T18:20:42","date_gmt":"2021-12-26T02:20:42","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=9234"},"modified":"2021-12-25T18:21:52","modified_gmt":"2021-12-26T02:21:52","slug":"leetcode-1920-build-array-from-permutation","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/algorithms\/array\/leetcode-1920-build-array-from-permutation\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1920. Build Array from Permutation"},"content":{"rendered":"\n

Given a zero-based permutation<\/strong> nums<\/code> (0-indexed<\/strong>), build an array ans<\/code> of the same length<\/strong> where ans[i] = nums[nums[i]]<\/code> for each 0 <= i < nums.length<\/code> and return it.<\/p>\n\n\n\n

A zero-based permutation<\/strong> nums<\/code> is an array of distinct<\/strong> integers from 0<\/code> to nums.length - 1<\/code> (inclusive<\/strong>).<\/p>\n\n\n\n

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

Input:<\/strong> nums = [0,2,1,5,3,4]\nOutput:<\/strong> [0,1,2,4,5,3]\nExplanation:<\/strong> The array ans is built as follows: \nans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]]\n    = [nums[0], nums[2], nums[1], nums[5], nums[3], nums[4]]\n    = [0,1,2,4,5,3]<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> nums = [5,0,1,2,3,4]\nOutput:<\/strong> [4,5,0,1,2,3]\nExplanation:<\/strong> The array ans is built as follows:\nans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]]\n    = [nums[5], nums[0], nums[1], nums[2], nums[3], nums[4]]\n    = [4,5,0,1,2,3]<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= nums.length <= 1000<\/code><\/li>
  • 0 <= nums[i] < nums.length<\/code><\/li>
  • The elements in nums<\/code> are distinct<\/strong>.<\/li><\/ul>\n\n\n\n
    Follow-up:<\/strong> Can you solve it without using an extra space (i.e., O(1)<\/code> memory)?<\/p>\n\n\n\n
    Solution 1: Straight forward<\/strong><\/h2>\n\n\n\n
    Time complexity: O(n)
    Space complexity: O(n)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  vector buildArray(vector& nums) {\n    vector ans(nums);\n    for (size_t i = 0; i < nums.size(); ++i)\n      ans[i] = nums[nums[i]];\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n\n\n\n
    Solution 2: Follow up: Inplace Encoding<\/strong><\/h2>\n\n\n\n
    Since nums[i] <= 1000, we can use low 16 bit to store the original value and high 16 bit for new value.<\/p>\n\n\n\n
    Time complexity: O(n)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  vector buildArray(vector& nums) {\n    const int n = nums.size();\n    for (int i = 0; i < n; ++i)\n      nums[i] |= (nums[nums[i]] & 0xffff) << 16;\n    for (int i = 0; i < n; ++i)\n      nums[i] >>= 16;\n    return nums;\n  }\n};\n<\/pre>\n<\/div><\/div>\n\n\n\n
    <\/p>\n","protected":false},"excerpt":{"rendered":"
    Given a zero-based permutation nums (0-indexed), build an array ans of the same length where ans[i] = nums[nums[i]] for each 0 <= i < nums.length and return it. A zero-based permutation nums is an array of distinct integers from 0 to nums.length – 1 (inclusive).…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[184],"tags":[20,16,222],"class_list":["post-9234","post","type-post","status-publish","format-standard","hentry","category-array","tag-array","tag-bit","tag-easy","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9234","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=9234"}],"version-history":[{"count":3,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9234\/revisions"}],"predecessor-version":[{"id":9237,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9234\/revisions\/9237"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=9234"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=9234"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=9234"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}