{"id":7388,"date":"2020-09-19T23:11:13","date_gmt":"2020-09-20T06:11:13","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=7388"},"modified":"2020-09-19T23:13:01","modified_gmt":"2020-09-20T06:13:01","slug":"leetcode-1589-maximum-sum-obtained-of-any-permutation","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/greedy\/leetcode-1589-maximum-sum-obtained-of-any-permutation\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1589. Maximum Sum Obtained of Any Permutation"},"content":{"rendered":"\n

We have an array of integers, nums<\/code>, and an array of requests<\/code> where requests[i] = [starti<\/sub>, endi<\/sub>]<\/code>. The ith<\/sup><\/code> request asks for the sum of nums[starti<\/sub>] + nums[starti<\/sub> + 1] + ... + nums[endi<\/sub> - 1] + nums[endi<\/sub>]<\/code>. Both starti<\/sub><\/code> and endi<\/sub><\/code> are 0-indexed<\/em>.<\/p>\n\n\n\n

Return the maximum total sum of all requests among all permutations<\/strong> of<\/em> nums<\/code>.<\/p>\n\n\n\n

Since the answer may be too large, return it modulo<\/strong> 109<\/sup> + 7<\/code>.<\/p>\n\n\n\n

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

Input:<\/strong> nums = [1,2,3,4,5], requests = [[1,3],[0,1]]\nOutput:<\/strong> 19\nExplanation:<\/strong> One permutation of nums is [2,1,3,4,5] with the following result: \nrequests[0] -> nums[1] + nums[2] + nums[3] = 1 + 3 + 4 = 8\nrequests[1] -> nums[0] + nums[1] = 2 + 1 = 3\nTotal sum: 8 + 3 = 11.\nA permutation with a higher total sum is [3,5,4,2,1] with the following result:\nrequests[0] -> nums[1] + nums[2] + nums[3] = 5 + 4 + 2 = 11\nrequests[1] -> nums[0] + nums[1] = 3 + 5  = 8\nTotal sum: 11 + 8 = 19, which is the best that you can do.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> nums = [1,2,3,4,5,6], requests = [[0,1]]\nOutput:<\/strong> 11\nExplanation:<\/strong> A permutation with the max total sum is [6,5,4,3,2,1] with request sums [11].<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> nums = [1,2,3,4,5,10], requests = [[0,2],[1,3],[1,1]]\nOutput:<\/strong> 47\nExplanation:<\/strong> A permutation with the max total sum is [4,10,5,3,2,1] with request sums [19,18,10].<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • n == nums.length<\/code><\/li>
  • 1 <= n <= 105<\/sup><\/code><\/li>
  • 0 <= nums[i] <= 105<\/sup><\/code><\/li>
  • 1 <= requests.length <= 105<\/sup><\/code><\/li>
  • requests[i].length == 2<\/code><\/li>
  • 0 <= starti<\/sub> <= endi<\/sub> < n<\/code><\/li><\/ul>\n\n\n\n
    Solution: Greedy + Sweep line<\/strong><\/h2>\n\n\n\n
    Sort the numbers, and sort the frequency of each index, it’s easy to show largest number with largest frequency gives us max sum.<\/p>\n\n\n\n
    ans = sum(nums[i] * freq[i])<\/p>\n\n\n\n
    We can use sweep line to compute the frequency of each index in O(n) time and space.<\/p>\n\n\n\n
    For each request [start, end] : ++freq[start], –freq[end + 1]<\/p>\n\n\n\n
    Then the prefix sum of freq array is the frequency for each index.<\/p>\n\n\n\n
    Time complexity: O(nlogn)
    Space complexity: O(n)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \nclass Solution {\npublic:\n  int maxSumRangeQuery(vector& nums, vector>& requests) {\n    constexpr int kMod = 1e9 + 7;\n    const int n = nums.size();    \n    vector freq(n);\n    for (const auto& r : requests) {\n      ++freq[r[0]];\n      if (r[1] + 1 < n) --freq[r[1] + 1];\n    }\n    for (int i = 1; i < n; ++i)\n      freq[i] += freq[i - 1];\n    \n    sort(begin(freq), end(freq));\n    sort(begin(nums), end(nums));\n    \n    long ans = 0;\n    for (int i = 0; i < n; ++i)\n      ans += freq[i] * nums[i];\n    \n    return ans % kMod;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    We have an array of integers, nums, and an array of requests where requests[i] = [starti, endi]. The ith request asks for the sum of nums[starti] + nums[starti + 1] + … +…<\/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":[88,177,104],"class_list":["post-7388","post","type-post","status-publish","format-standard","hentry","category-greedy","tag-greedy","tag-medium","tag-sweep-line","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7388","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=7388"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7388\/revisions"}],"predecessor-version":[{"id":7390,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7388\/revisions\/7390"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=7388"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=7388"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=7388"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}