{"id":9253,"date":"2021-12-26T05:34:56","date_gmt":"2021-12-26T13:34:56","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=9253"},"modified":"2021-12-26T05:37:10","modified_gmt":"2021-12-26T13:37:10","slug":"leetcode-2121-intervals-between-identical-elements","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/hashtable\/leetcode-2121-intervals-between-identical-elements\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 2121. Intervals Between Identical Elements"},"content":{"rendered":"\n

You are given a 0-indexed<\/strong> array of n<\/code> integers arr<\/code>.<\/p>\n\n\n\n

The interval<\/strong> between two elements in arr<\/code> is defined as the absolute difference<\/strong> between their indices. More formally, the interval<\/strong> between arr[i]<\/code> and arr[j]<\/code> is |i - j|<\/code>.<\/p>\n\n\n\n

Return an array<\/em> intervals<\/code> of length<\/em> n<\/code> where<\/em> intervals[i]<\/code> is the sum of intervals<\/strong> between <\/em>arr[i]<\/code> and each element in <\/em>arr<\/code> with the same value as <\/em>arr[i]<\/code>.<\/em><\/p>\n\n\n\n

Note:<\/strong> |x|<\/code> is the absolute value of x<\/code>.<\/p>\n\n\n\n

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

Input:<\/strong> arr = [2,1,3,1,2,3,3]\nOutput:<\/strong> [4,2,7,2,4,4,5]\nExplanation:<\/strong>\n- Index 0: Another 2 is found at index 4. |0 - 4| = 4\n- Index 1: Another 1 is found at index 3. |1 - 3| = 2\n- Index 2: Two more 3s are found at indices 5 and 6. |2 - 5| + |2 - 6| = 7\n- Index 3: Another 1 is found at index 1. |3 - 1| = 2\n- Index 4: Another 2 is found at index 0. |4 - 0| = 4\n- Index 5: Two more 3s are found at indices 2 and 6. |5 - 2| + |5 - 6| = 4\n- Index 6: Two more 3s are found at indices 2 and 5. |6 - 2| + |6 - 5| = 5\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> arr = [10,5,10,10]\nOutput:<\/strong> [5,0,3,4]\nExplanation:<\/strong>\n- Index 0: Two more 10s are found at indices 2 and 3. |0 - 2| + |0 - 3| = 5\n- Index 1: There is only one 5 in the array, so its sum of intervals to identical elements is 0.\n- Index 2: Two more 10s are found at indices 0 and 3. |2 - 0| + |2 - 3| = 3\n- Index 3: Two more 10s are found at indices 0 and 2. |3 - 0| + |3 - 2| = 4\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • n == arr.length<\/code><\/li>
  • 1 <= n <= 105<\/sup><\/code><\/li>
  • 1 <= arr[i] <= 105<\/sup><\/code><\/li><\/ul>\n\n\n\n
    Solution: Math \/ Hashtable + Prefix Sum <\/strong><\/h2>\n\n\n\n
    For each arr[i], suppose it occurs in the array of total c times, among which k of them are in front of it and c – k – 1 of them are after it. Then the total sum intervals:
    (i – j1<\/sub>) + (i – j2<\/sub>) + … + (i – jk<\/sub>) + (jk+1<\/sub>-i) + (jk+2<\/sub>-i) + … + (jc<\/sub>-i)
    <=> k * i – sum(j1<\/sub>~jk<\/sub>)  + sum(jk+1<\/sub>~jc<\/sub>) – (c – k – 1) * i<\/p>\n\n\n\n
    Use a hashtable to store the indies of each unique number in the array and compute the prefix sum for fast range sum query.<\/p>\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 getDistances(vector& arr) {\n    const int n = arr.size();\n    unordered_map> m;\n    vector pos(n);\n    for (int i = 0; i < n; ++i) {\n      m[arr[i]].push_back(i);\n      pos[i] = m[arr[i]].size() - 1;\n    }\n    for (auto& [k, idx] : m)\n      partial_sum(begin(idx), end(idx), begin(idx));      \n    vector ans(n);\n    for (int i = 0; i < n; ++i) {\n      const auto& sums = m[arr[i]];\n      const long long k = pos[i];\n      const long long c = sums.size();      \n      if (k > 0) ans[i] += k * i - sums[k - 1];\n      if (k + 1 < c) ans[i] += (sums[c - 1] - sums[k]) - (c - k - 1) * i;\n    }\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    You are given a 0-indexed array of n integers arr. The interval between two elements in arr is defined as the absolute difference between their indices. More formally, the interval between arr[i] and arr[j] is |i – j|. Return an array intervals of length n where intervals[i] is the sum of…<\/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":[752,31,200],"class_list":["post-9253","post","type-post","status-publish","format-standard","hentry","category-hashtable","tag-hash-table","tag-math","tag-prefix-sum","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9253","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=9253"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9253\/revisions"}],"predecessor-version":[{"id":9255,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9253\/revisions\/9255"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=9253"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=9253"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=9253"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}