{ "data": { "question": { "questionId": "2255", "questionFrontendId": "2134", "categoryTitle": "Algorithms", "boundTopicId": 1196321, "title": "Minimum Swaps to Group All 1's Together II", "titleSlug": "minimum-swaps-to-group-all-1s-together-ii", "content": "
A swap is defined as taking two distinct positions in an array and swapping the values in them.
\n\nA circular array is defined as an array where we consider the first element and the last element to be adjacent.
\n\nGiven a binary circular array nums
, return the minimum number of swaps required to group all 1
's present in the array together at any location.
\n
Example 1:
\n\n\nInput: nums = [0,1,0,1,1,0,0]\nOutput: 1\nExplanation: Here are a few of the ways to group all the 1's together:\n[0,0,1,1,1,0,0] using 1 swap.\n[0,1,1,1,0,0,0] using 1 swap.\n[1,1,0,0,0,0,1] using 2 swaps (using the circular property of the array).\nThere is no way to group all 1's together with 0 swaps.\nThus, the minimum number of swaps required is 1.\n\n\n
Example 2:
\n\n\nInput: nums = [0,1,1,1,0,0,1,1,0]\nOutput: 2\nExplanation: Here are a few of the ways to group all the 1's together:\n[1,1,1,0,0,0,0,1,1] using 2 swaps (using the circular property of the array).\n[1,1,1,1,1,0,0,0,0] using 2 swaps.\nThere is no way to group all 1's together with 0 or 1 swaps.\nThus, the minimum number of swaps required is 2.\n\n\n
Example 3:
\n\n\nInput: nums = [1,1,0,0,1]\nOutput: 0\nExplanation: All the 1's are already grouped together due to the circular property of the array.\nThus, the minimum number of swaps required is 0.\n\n\n
\n
Constraints:
\n\n1 <= nums.length <= 105
nums[i]
is either 0
or 1
.交换 定义为选中一个数组中的两个 互不相同 的位置并交换二者的值。
\n\n环形 数组是一个数组,可以认为 第一个 元素和 最后一个 元素 相邻 。
\n\n给你一个 二进制环形 数组 nums
,返回在 任意位置 将数组中的所有 1
聚集在一起需要的最少交换次数。
\n\n
示例 1:
\n\n输入:nums = [0,1,0,1,1,0,0]\n输出:1\n解释:这里列出一些能够将所有 1 聚集在一起的方案:\n[0,0,1,1,1,0,0] 交换 1 次。\n[0,1,1,1,0,0,0] 交换 1 次。\n[1,1,0,0,0,0,1] 交换 2 次(利用数组的环形特性)。\n无法在交换 0 次的情况下将数组中的所有 1 聚集在一起。\n因此,需要的最少交换次数为 1 。\n\n\n
示例 2:
\n\n输入:nums = [0,1,1,1,0,0,1,1,0]\n输出:2\n解释:这里列出一些能够将所有 1 聚集在一起的方案:\n[1,1,1,0,0,0,0,1,1] 交换 2 次(利用数组的环形特性)。\n[1,1,1,1,1,0,0,0,0] 交换 2 次。\n无法在交换 0 次或 1 次的情况下将数组中的所有 1 聚集在一起。\n因此,需要的最少交换次数为 2 。\n\n\n
示例 3:
\n\n输入:nums = [1,1,0,0,1]\n输出:0\n解释:得益于数组的环形特性,所有的 1 已经聚集在一起。\n因此,需要的最少交换次数为 0 。\n\n
\n\n
提示:
\n\n1 <= nums.length <= 105
nums[i]
为 0
或者 1
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