leetcode-problemset/leetcode-cn/originData/sum-of-distances.json

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        "question": {
            "questionId": "2721",
            "questionFrontendId": "2615",
            "categoryTitle": "Algorithms",
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            "title": "Sum of Distances",
            "titleSlug": "sum-of-distances",
            "content": "<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>. There exists an array <code>arr</code> of length <code>nums.length</code>, where <code>arr[i]</code> is the sum of <code>|i - j|</code> over all <code>j</code> such that <code>nums[j] == nums[i]</code> and <code>j != i</code>. If there is no such <code>j</code>, set <code>arr[i]</code> to be <code>0</code>.</p>\n\n<p>Return <em>the array </em><code>arr</code><em>.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,3,1,1,2]\n<strong>Output:</strong> [5,0,3,4,0]\n<strong>Explanation:</strong> \nWhen i = 0, nums[0] == nums[2] and nums[0] == nums[3]. Therefore, arr[0] = |0 - 2| + |0 - 3| = 5. \nWhen i = 1, arr[1] = 0 because there is no other index with value 3.\nWhen i = 2, nums[2] == nums[0] and nums[2] == nums[3]. Therefore, arr[2] = |2 - 0| + |2 - 3| = 3. \nWhen i = 3, nums[3] == nums[0] and nums[3] == nums[2]. Therefore, arr[3] = |3 - 0| + |3 - 2| = 4. \nWhen i = 4, arr[4] = 0 because there is no other index with value 2. \n\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [0,5,3]\n<strong>Output:</strong> [0,0,0]\n<strong>Explanation:</strong> Since each element in nums is distinct, arr[i] = 0 for all i.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "等值距离和",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> 。现有一个长度等于 <code>nums.length</code> 的数组 <code>arr</code> 。对于满足 <code>nums[j] == nums[i]</code> 且 <code>j != i</code> 的所有 <code>j</code> ，<code>arr[i]</code> 等于所有 <code>|i - j|</code> 之和。如果不存在这样的 <code>j</code> ，则令 <code>arr[i]</code> 等于 <code>0</code> 。</p>\n\n<p>返回数组<em> </em><code>arr</code><em> 。</em></p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,3,1,1,2]\n<strong>输出：</strong>[5,0,3,4,0]\n<strong>解释：</strong>\ni = 0 ，nums[0] == nums[2] 且 nums[0] == nums[3] 。因此，arr[0] = |0 - 2| + |0 - 3| = 5 。 \ni = 1 ，arr[1] = 0 因为不存在值等于 3 的其他下标。\ni = 2 ，nums[2] == nums[0] 且 nums[2] == nums[3] 。因此，arr[2] = |2 - 0| + |2 - 3| = 3 。\ni = 3 ，nums[3] == nums[0] 且 nums[3] == nums[2] 。因此，arr[3] = |3 - 0| + |3 - 2| = 4 。 \ni = 4 ，arr[4] = 0 因为不存在值等于 2 的其他下标。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [0,5,3]\n<strong>输出：</strong>[0,0,0]\n<strong>解释：</strong>因为 nums 中的元素互不相同，对于所有 i ，都有 arr[i] = 0 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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                {
                    "name": "Array",
                    "slug": "array",
                    "translatedName": "数组",
                    "__typename": "TopicTagNode"
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                {
                    "name": "Hash Table",
                    "slug": "hash-table",
                    "translatedName": "哈希表",
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                    "name": "Prefix Sum",
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                    "lang": "C++",
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                    "code": "class Solution {\npublic:\n    vector<long long> distance(vector<int>& nums) {\n        \n    }\n};",
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                    "code": "class Solution(object):\n    def distance(self, nums):\n        \"\"\"\n        :type nums: List[int]\n        :rtype: List[int]\n        \"\"\"",
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                    "code": "/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nlong long* distance(int* nums, int numsSize, int* returnSize){\n\n}",
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            "hints": [
                "Can we use the prefix sum here?",
                "For each number x, collect all the indices where x occurs, and calculate the prefix sum of the array.",
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