{ "data": { "question": { "questionId": "3249", "questionFrontendId": "2997", "categoryTitle": "Algorithms", "boundTopicId": 2591565, "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": "
You are given a 0-indexed integer array nums and a positive integer k.
You can apply the following operation on the array any number of times:
\n\n0 to 1 or vice versa.Return the minimum number of operations required to make the bitwise XOR of all elements of the final array equal to k.
Note that you can flip leading zero bits in the binary representation of elements. For example, for the number (101)2 you can flip the fourth bit and obtain (1101)2.
\n
Example 1:
\n\n\nInput: nums = [2,1,3,4], k = 1\nOutput: 2\nExplanation: We can do the following operations:\n- Choose element 2 which is 3 == (011)2, we flip the first bit and we obtain (010)2 == 2. nums becomes [2,1,2,4].\n- Choose element 0 which is 2 == (010)2, we flip the third bit and we obtain (110)2 = 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\n\n
Example 2:
\n\n\nInput: nums = [2,0,2,0], k = 0\nOutput: 0\nExplanation: The XOR of elements of the array is (2 XOR 0 XOR 2 XOR 0) == 0 == k. So no operation is needed.\n\n\n
\n
Constraints:
\n\n1 <= nums.length <= 1050 <= nums[i] <= 1060 <= k <= 106给你一个下标从 0 开始的整数数组 nums 和一个正整数 k 。
你可以对数组执行以下操作 任意次 :
\n\n0 变成 1 或者将 1 变成 0 。你的目标是让数组里 所有 元素的按位异或和得到 k ,请你返回达成这一目标的 最少 操作次数。
注意,你也可以将一个数的前导 0 翻转。比方说,数字 (101)2 翻转第四个数位,得到 (1101)2 。
\n\n
示例 1:
\n\n\n输入:nums = [2,1,3,4], k = 1\n输出:2\n解释: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)2 ,我们翻转第一个数位得到 (010)2 == 2 。数组变为 [2,1,2,4] 。\n- 选择下标为 0 的元素,也就是 2 == (010)2 ,我们翻转第三个数位得到 (110)2 == 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\n\n
示例 2:
\n\n\n输入:nums = [2,0,2,0], k = 0\n输出:0\n解释: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\n\n
\n\n
提示:
\n\n1 <= nums.length <= 1050 <= nums[i] <= 1060 <= k <= 106XOR of all elements of the original array and compare it to k in their binary representation.",
"For each different bit between the bitwise XOR of elements of the original array and k we have to flip exactly one bit of an element in nums to make that bit equal."
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