{"id":4956,"date":"2019-03-09T21:58:33","date_gmt":"2019-03-10T05:58:33","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=4956"},"modified":"2019-03-09T21:59:27","modified_gmt":"2019-03-10T05:59:27","slug":"leetcode-1005-maximize-sum-of-array-after-k-negations","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/greedy\/leetcode-1005-maximize-sum-of-array-after-k-negations\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1005. Maximize Sum Of Array After K Negations"},"content":{"rendered":"\n

Given an array A<\/code> of integers, we must<\/strong> modify the array in the following way: we choose an i<\/code> and replace A[i]<\/code> with -A[i]<\/code>, and we repeat this process K<\/code> times in total.  (We may choose the same index i<\/code> multiple times.)<\/p>\n\n\n\n

Return the largest possible sum of the array after modifying it in this way.<\/p>\n\n\n\n

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

Input: <\/strong>A = [4,2,3], K = 1\nOutput: <\/strong>5\nExplanation: <\/strong>Choose indices (1,) and A becomes [4,-2,3].\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input: <\/strong>A = [3,-1,0,2], K = 3\nOutput: <\/strong>6\nExplanation: <\/strong>Choose indices (1, 2, 2) and A becomes [3,1,0,2].\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input: <\/strong>A = [2,-3,-1,5,-4], K = 2\nOutput: <\/strong>13\nExplanation: <\/strong>Choose indices (1, 4) and A becomes [2,3,-1,5,4].\n<\/pre>\n\n\n\n
Note:<\/strong><\/p>\n\n\n\n
  1. 1 <= A.length <= 10000<\/code><\/li>
  2. 1 <= K <= 10000<\/code><\/li>
  3. -100 <= A[i] <= 100<\/code><\/li><\/ol>\n\n\n\n
    Solution: Sorting<\/strong><\/h2>\n\n\n\n
    Flip as many as negative numbers as possible, if K > # of negative numbers and K is odd, flip the smallest element in the array.<\/p>\n\n\n\n
    Time complexity: O(nlogn)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua 8 ms 8.6MB\nclass Solution {\npublic:\n  int largestSumAfterKNegations(vector& A, int K) {    \n    std::sort(begin(A), end(A));\n    for (int& a : A) {\n      if (!K) break;\n      if (a < 0) {\n        a = -a;\n        --K;\n      }\n    }\n    return accumulate(begin(A), end(A), 0) - (K % 2 ? *min_element(begin(A), end(A)) * 2 : 0);    \n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    Given an array A of integers, we must modify the array in the following way: we choose an i and replace A[i] with -A[i], and we repeat this process K times in total.  (We may choose…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[51],"tags":[222,88],"class_list":["post-4956","post","type-post","status-publish","format-standard","hentry","category-greedy","tag-easy","tag-greedy","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/4956","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=4956"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/4956\/revisions"}],"predecessor-version":[{"id":4958,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/4956\/revisions\/4958"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=4956"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=4956"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=4956"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}