mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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187 lines
19 KiB
JSON
187 lines
19 KiB
JSON
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"questionId": "3363",
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"questionFrontendId": "3092",
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"boundTopicId": null,
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"title": "Most Frequent IDs",
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"titleSlug": "most-frequent-ids",
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"content": "<p>The problem involves tracking the frequency of IDs in a collection that changes over time. You have two integer arrays, <code>nums</code> and <code>freq</code>, of equal length <code>n</code>. Each element in <code>nums</code> represents an ID, and the corresponding element in <code>freq</code> indicates how many times that ID should be added to or removed from the collection at each step.</p>\n\n<ul>\n\t<li><strong>Addition of IDs:</strong> If <code>freq[i]</code> is positive, it means <code>freq[i]</code> IDs with the value <code>nums[i]</code> are added to the collection at step <code>i</code>.</li>\n\t<li><strong>Removal of IDs:</strong> If <code>freq[i]</code> is negative, it means <code>-freq[i]</code> IDs with the value <code>nums[i]</code> are removed from the collection at step <code>i</code>.</li>\n</ul>\n\n<p>Return an array <code>ans</code> of length <code>n</code>, where <code>ans[i]</code> represents the <strong>count</strong> of the <em>most frequent ID</em> in the collection after the <code>i<sup>th</sup></code> step. If the collection is empty at any step, <code>ans[i]</code> should be 0 for that step.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [2,3,2,1], freq = [3,2,-3,1]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[3,3,2,2]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>After step 0, we have 3 IDs with the value of 2. So <code>ans[0] = 3</code>.<br />\nAfter step 1, we have 3 IDs with the value of 2 and 2 IDs with the value of 3. So <code>ans[1] = 3</code>.<br />\nAfter step 2, we have 2 IDs with the value of 3. So <code>ans[2] = 2</code>.<br />\nAfter step 3, we have 2 IDs with the value of 3 and 1 ID with the value of 1. So <code>ans[3] = 2</code>.</p>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [5,5,3], freq = [2,-2,1]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[2,0,1]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>After step 0, we have 2 IDs with the value of 5. So <code>ans[0] = 2</code>.<br />\nAfter step 1, there are no IDs. So <code>ans[1] = 0</code>.<br />\nAfter step 2, we have 1 ID with the value of 3. So <code>ans[2] = 1</code>.</p>\n</div>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length == freq.length <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= nums[i] <= 10<sup>5</sup></code></li>\n\t<li><code>-10<sup>5</sup> <= freq[i] <= 10<sup>5</sup></code></li>\n\t<li><code>freq[i] != 0</code></li>\n\t<li>The input is generated<!-- notionvc: a136b55a-f319-4fa6-9247-11be9f3b1db8 --> such that the occurrences of an ID will not be negative in any step.</li>\n</ul>\n",
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"difficulty": "Medium",
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"exampleTestcases": "[2,3,2,1]\n[3,2,-3,1]\n[5,5,3]\n[2,-2,1]",
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"categoryTitle": "Algorithms",
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"name": "Array",
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"name": "Heap (Priority Queue)",
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"code": "# @param {Integer[]} nums\n# @param {Integer[]} freq\n# @return {Integer[]}\ndef most_frequent_i_ds(nums, freq)\n \nend",
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"code": "impl Solution {\n pub fn most_frequent_i_ds(nums: Vec<i32>, freq: Vec<i32>) -> Vec<i64> {\n \n }\n}",
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"Use an ordered set for maintaining the occurrences of each ID.",
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"After step <code>i</code> find the occurrences of <code>nums[i]</code>.",
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"Change the occurrences of <code>nums[i]</code> in the ordered set."
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