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leetcode-problemset/leetcode-cn/originData/find-subarray-with-bitwise-and-closest-to-k.json
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"question": {
"questionId": "3436",
"questionFrontendId": "3171",
"categoryTitle": "Algorithms",
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"title": "Find Subarray With Bitwise AND Closest to K",
"titleSlug": "find-subarray-with-bitwise-and-closest-to-k",
"content": "<p>You are given an array <code>nums</code> and an integer <code>k</code>. You need to find a <span data-keyword=\"subarray-nonempty\">subarray</span> of <code>nums</code> such that the <strong>absolute difference</strong> between <code>k</code> and the bitwise <code>AND</code> of the subarray elements is as<strong> small</strong> as possible. In other words, select a subarray <code>nums[l..r]</code> such that <code>|k - (nums[l] AND nums[l + 1] ... AND nums[r])|</code> is minimum.</p>\n\n<p>Return the <strong>minimum</strong> possible value of the absolute difference.</p>\n\n<p>A <strong>subarray</strong> is a contiguous <b>non-empty</b> sequence of elements within an array.</p>\n\n<p>&nbsp;</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 = [1,2,4,5], k = 3</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">1</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The subarray <code>nums[2..3]</code> has <code>AND</code> value 4, which gives the minimum absolute difference <code>|3 - 4| = 1</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 = [1,2,1,2], k = 2</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">0</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The subarray <code>nums[1..1]</code> has <code>AND</code> value 2, which gives the minimum absolute difference <code>|2 - 2| = 0</code>.</p>\n</div>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [1], k = 10</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">9</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>There is a single subarray with <code>AND</code> value 1, which gives the minimum absolute difference <code>|10 - 1| = 9</code>.</p>\n</div>\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>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
"translatedTitle": "找到按位与最接近 K 的子数组",
"translatedContent": "<p>给你一个数组&nbsp;<code>nums</code>&nbsp;和一个整数&nbsp;<code>k</code>&nbsp;。你需要找到&nbsp;<code>nums</code>&nbsp;的一个&nbsp;<span data-keyword=\"subarray-nonempty\">子数组</span>&nbsp;,满足子数组中所有元素按位与运算&nbsp;<code>AND</code>&nbsp;的值与 <code>k</code>&nbsp;的 <strong>绝对差</strong>&nbsp;尽可能 <strong>小</strong>&nbsp;。换言之,你需要选择一个子数组&nbsp;<code>nums[l..r]</code>&nbsp;满足&nbsp;<code>|k - (nums[l] AND nums[l + 1] ... AND nums[r])|</code>&nbsp;最小。</p>\n\n<p>请你返回 <strong>最小</strong>&nbsp;的绝对差值。</p>\n\n<p><strong>子数组</strong>是数组中连续的&nbsp;<strong>非空</strong>&nbsp;元素序列。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入:</b>nums = [1,2,4,5], k = 3</span></p>\n\n<p><span class=\"example-io\"><b>输出:</b>1</span></p>\n\n<p><strong>解释:</strong></p>\n\n<p>子数组&nbsp;<code>nums[2..3]</code> 的按位&nbsp;<code>AND</code>&nbsp;运算值为 4 ,得到最小差值&nbsp;<code>|3 - 4| = 1</code>&nbsp;。</p>\n</div>\n\n<p><strong class=\"example\">示例 2</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入:</b>nums = [1,2,1,2], k = 2</span></p>\n\n<p><span class=\"example-io\"><b>输出:</b>0</span></p>\n\n<p><strong>解释:</strong></p>\n\n<p>子数组&nbsp;<code>nums[1..1]</code> 的按位&nbsp;<code>AND</code>&nbsp;运算值为 2 ,得到最小差值&nbsp;<code>|2 - 2| = 0</code>&nbsp;。</p>\n</div>\n\n<p><strong class=\"example\">示例 3</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入:</b>nums = [1], k = 10</span></p>\n\n<p><span class=\"example-io\"><b>输出:</b>9</span></p>\n\n<p><strong>解释:</strong></p>\n\n<p>只有一个子数组,按位&nbsp;<code>AND</code>&nbsp;运算值为 1 ,得到最小差值&nbsp;<code>|10 - 1| = 9</code>&nbsp;。</p>\n</div>\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>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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"Let <code>dp[i]</code> be the set of all the bitwise <code>AND</code> of all the subarrays ending at index <code>i</code>.",
"We start from <code>nums[i]</code>, taking the bitwise <code>AND</code> result by including elements one by one from <code>i</code> towards left. Notice that only set bits can become unset on adding a element, and unset bits never become set again.",
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