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"categoryTitle": "Algorithms",
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"title": "Minimum Number of Operations to Make Array XOR Equal to K",
"titleSlug": "minimum-number-of-operations-to-make-array-xor-equal-to-k",
"content": "<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> and a positive integer <code>k</code>.</p>\n\n<p>You can apply the following operation on the array <strong>any</strong> number of times:</p>\n\n<ul>\n\t<li>Choose <strong>any</strong> element of the array and <strong>flip</strong> a bit in its <strong>binary</strong> representation. Flipping a bit means changing a <code>0</code> to <code>1</code> or vice versa.</li>\n</ul>\n\n<p>Return <em>the <strong>minimum</strong> number of operations required to make the bitwise </em><code>XOR</code><em> of <strong>all</strong> elements of the final array equal to </em><code>k</code>.</p>\n\n<p><strong>Note</strong> that you can flip leading zero bits in the binary representation of elements. For example, for the number <code>(101)<sub>2</sub></code> you can flip the fourth bit and obtain <code>(1101)<sub>2</sub></code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,1,3,4], k = 1\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> We can do the following operations:\n- Choose element 2 which is 3 == (011)<sub>2</sub>, we flip the first bit and we obtain (010)<sub>2</sub> == 2. nums becomes [2,1,2,4].\n- Choose element 0 which is 2 == (010)<sub>2</sub>, we flip the third bit and we obtain (110)<sub>2</sub> = 6. nums becomes [6,1,2,4].\nThe XOR of elements of the final array is (6 XOR 1 XOR 2 XOR 4) == 1 == k.\nIt can be shown that we cannot make the XOR equal to k in less than 2 operations.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,0,2,0], k = 0\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> The XOR of elements of the array is (2 XOR 0 XOR 2 XOR 0) == 0 == k. So no operation is needed.\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>6</sup></code></li>\n\t<li><code>0 &lt;= k &lt;= 10<sup>6</sup></code></li>\n</ul>\n",
"translatedTitle": "使数组异或和等于 K 的最少操作次数",
"translatedContent": "<p>给你一个下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;和一个正整数&nbsp;<code>k</code>&nbsp;。</p>\n\n<p>你可以对数组执行以下操作 <strong>任意次</strong>&nbsp;</p>\n\n<ul>\n\t<li>选择数组里的 <strong>任意</strong>&nbsp;一个元素,并将它的&nbsp;<strong>二进制</strong>&nbsp;表示&nbsp;<strong>翻转</strong>&nbsp;一个数位,翻转数位表示将&nbsp;<code>0</code> 变成&nbsp;<code>1</code>&nbsp;或者将 <code>1</code>&nbsp;变成 <code>0</code>&nbsp;。</li>\n</ul>\n\n<p>你的目标是让数组里 <strong>所有</strong>&nbsp;元素的按位异或和得到 <code>k</code>&nbsp;,请你返回达成这一目标的 <strong>最少&nbsp;</strong>操作次数。</p>\n\n<p><strong>注意</strong>,你也可以将一个数的前导 0 翻转。比方说,数字&nbsp;<code>(101)<sub>2</sub></code>&nbsp;翻转第四个数位,得到&nbsp;<code>(1101)<sub>2</sub></code>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1</strong></p>\n\n<pre>\n<b>输入:</b>nums = [2,1,3,4], k = 1\n<b>输出:</b>2\n<b>解释:</b>我们可以执行以下操作:\n- 选择下标为 2 的元素,也就是 3 == (011)<sub>2</sub>&nbsp;,我们翻转第一个数位得到 (010)<sub>2</sub> == 2 。数组变为 [2,1,2,4] 。\n- 选择下标为 0 的元素,也就是 2 == (010)<sub>2</sub> ,我们翻转第三个数位得到 (110)<sub>2</sub> == 6 。数组变为 [6,1,2,4] 。\n最终数组的所有元素异或和为 (6 XOR 1 XOR 2 XOR 4) == 1 == k 。\n无法用少于 2 次操作得到异或和等于 k 。\n</pre>\n\n<p><strong class=\"example\">示例 2</strong></p>\n\n<pre>\n<strong>输入:</strong>nums = [2,0,2,0], k = 0\n<b>输出:</b>0\n<strong>解释:</strong>数组所有元素的异或和为 (2 XOR 0 XOR 2 XOR 0) == 0 == k 。所以不需要进行任何操作。\n</pre>\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>0 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>\n\t<li><code>0 &lt;= k &lt;= 10<sup>6</sup></code></li>\n</ul>\n",
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"For each different bit between the bitwise <code>XOR</code> of elements of the original array and <code>k</code> we have to flip <strong>exactly</strong> one bit of an element in <code>nums</code> to make that bit equal."
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