mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
190 lines
21 KiB
JSON
190 lines
21 KiB
JSON
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"questionId": "1798",
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"questionFrontendId": "1679",
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"categoryTitle": "Algorithms",
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"boundTopicId": 508585,
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"title": "Max Number of K-Sum Pairs",
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"content": "<p>You are given an integer array <code>nums</code> and an integer <code>k</code>.</p>\n\n<p>In one operation, you can pick two numbers from the array whose sum equals <code>k</code> and remove them from the array.</p>\n\n<p>Return <em>the maximum number of operations you can perform on the array</em>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3,4], k = 5\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> Starting with nums = [1,2,3,4]:\n- Remove numbers 1 and 4, then nums = [2,3]\n- Remove numbers 2 and 3, then nums = []\nThere are no more pairs that sum up to 5, hence a total of 2 operations.</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3,1,3,4,3], k = 6\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> Starting with nums = [3,1,3,4,3]:\n- Remove the first two 3's, then nums = [1,4,3]\nThere are no more pairs that sum up to 6, hence a total of 1 operation.</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= nums[i] <= 10<sup>9</sup></code></li>\n\t<li><code>1 <= k <= 10<sup>9</sup></code></li>\n</ul>\n",
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"translatedTitle": "K 和数对的最大数目",
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"translatedContent": "<p>给你一个整数数组 <code>nums</code> 和一个整数 <code>k</code> 。</p>\n\n<p>每一步操作中，你需要从数组中选出和为 <code>k</code> 的两个整数，并将它们移出数组。</p>\n\n<p>返回你可以对数组执行的最大操作数。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,2,3,4], k = 5\n<strong>输出：</strong>2\n<strong>解释：</strong>开始时 nums = [1,2,3,4]：\n- 移出 1 和 4 ，之后 nums = [2,3]\n- 移出 2 和 3 ，之后 nums = []\n不再有和为 5 的数对，因此最多执行 2 次操作。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [3,1,3,4,3], k = 6\n<strong>输出：</strong>1\n<strong>解释：</strong>开始时 nums = [3,1,3,4,3]：\n- 移出前两个 3 ，之后nums = [1,4,3]\n不再有和为 6 的数对，因此最多执行 1 次操作。</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= nums[i] <= 10<sup>9</sup></code></li>\n\t<li><code>1 <= k <= 10<sup>9</sup></code></li>\n</ul>\n",
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"The abstract problem asks to count the number of disjoint pairs with a given sum k.",
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"For each possible value x, it can be paired up with k - x.",
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"The number of such pairs equals to min(count(x), count(k-x)), unless that x = k / 2, where the number of such pairs will be floor(count(x) / 2)."
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