mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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185 lines
23 KiB
JSON
185 lines
23 KiB
JSON
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"questionId": "2240",
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"questionFrontendId": "2121",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1174201,
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"title": "Intervals Between Identical Elements",
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"titleSlug": "intervals-between-identical-elements",
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"content": "<p>You are given a <strong>0-indexed</strong> array of <code>n</code> integers <code>arr</code>.</p>\n\n<p>The <strong>interval</strong> between two elements in <code>arr</code> is defined as the <strong>absolute difference</strong> between their indices. More formally, the <strong>interval</strong> between <code>arr[i]</code> and <code>arr[j]</code> is <code>|i - j|</code>.</p>\n\n<p>Return <em>an array</em> <code>intervals</code> <em>of length</em> <code>n</code> <em>where</em> <code>intervals[i]</code> <em>is <strong>the sum of intervals</strong> between </em><code>arr[i]</code><em> and each element in </em><code>arr</code><em> with the same value as </em><code>arr[i]</code><em>.</em></p>\n\n<p><strong>Note:</strong> <code>|x|</code> is the absolute value of <code>x</code>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [2,1,3,1,2,3,3]\n<strong>Output:</strong> [4,2,7,2,4,4,5]\n<strong>Explanation:</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<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> arr = [10,5,10,10]\n<strong>Output:</strong> [5,0,3,4]\n<strong>Explanation:</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<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == arr.length</code></li>\n\t<li><code>1 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= arr[i] <= 10<sup>5</sup></code></li>\n</ul>\n",
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"translatedTitle": "相同元素的间隔之和",
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"translatedContent": "<p>给你一个下标从 <strong>0</strong> 开始、由 <code>n</code> 个整数组成的数组 <code>arr</code> 。</p>\n\n<p><code>arr</code> 中两个元素的 <strong>间隔</strong> 定义为它们下标之间的 <strong>绝对差</strong> 。更正式地,<code>arr[i]</code> 和 <code>arr[j]</code> 之间的间隔是 <code>|i - j|</code> 。</p>\n\n<p>返回一个长度为 <code>n</code> 的数组 <code>intervals</code> ,其中 <code>intervals[i]</code> 是<em> </em><code>arr[i]</code><em> </em>和<em> </em><code>arr</code><em> </em>中每个相同元素(与 <code>arr[i]</code> 的值相同)的 <strong>间隔之和</strong> <em>。</em></p>\n\n<p><strong>注意:</strong><code>|x|</code> 是 <code>x</code> 的绝对值。</p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre><strong>输入:</strong>arr = [2,1,3,1,2,3,3]\n<strong>输出:</strong>[4,2,7,2,4,4,5]\n<strong>解释:</strong>\n- 下标 0 :另一个 2 在下标 4 ,|0 - 4| = 4\n- 下标 1 :另一个 1 在下标 3 ,|1 - 3| = 2\n- 下标 2 :另两个 3 在下标 5 和 6 ,|2 - 5| + |2 - 6| = 7\n- 下标 3 :另一个 1 在下标 1 ,|3 - 1| = 2\n- 下标 4 :另一个 2 在下标 0 ,|4 - 0| = 4\n- 下标 5 :另两个 3 在下标 2 和 6 ,|5 - 2| + |5 - 6| = 4\n- 下标 6 :另两个 3 在下标 2 和 5 ,|6 - 2| + |6 - 5| = 5\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre><strong>输入:</strong>arr = [10,5,10,10]\n<strong>输出:</strong>[5,0,3,4]\n<strong>解释:</strong>\n- 下标 0 :另两个 10 在下标 2 和 3 ,|0 - 2| + |0 - 3| = 5\n- 下标 1 :只有这一个 5 在数组中,所以到相同元素的间隔之和是 0\n- 下标 2 :另两个 10 在下标 0 和 3 ,|2 - 0| + |2 - 3| = 3\n- 下标 3 :另两个 10 在下标 0 和 2 ,|3 - 0| + |3 - 2| = 4\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>n == arr.length</code></li>\n\t<li><code>1 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= arr[i] <= 10<sup>5</sup></code></li>\n</ul>\n",
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"For each unique value found in the array, store a sorted list of indices of elements that have this value in the array.",
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"Process each list of indices separately and get the sum of intervals for the elements of that value by utilizing prefix sums.",
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