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leetcode-problemset/leetcode/originData/sum-of-distances.json
2023-12-09 19:57:46 +08:00

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{
"data": {
"question": {
"questionId": "2721",
"questionFrontendId": "2615",
"boundTopicId": null,
"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",
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"code": "class Solution {\npublic:\n vector<long long> distance(vector<int>& nums) {\n \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.",
"For each occurrence of x, the indices to the right will be regular subtraction while the indices to the left will be reversed subtraction."
],
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