2134. Minimum Swaps to Group All 1's Together II
Description
A swap is defined as taking two distinct positions in an array and swapping the values in them.
A circular array is defined as an array where we consider the first element and the last element to be adjacent.
Given 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.
Example 1:
Input: nums = [0,1,0,1,1,0,0] Output: 1 Explanation: Here are a few of the ways to group all the 1's together: [0,0,1,1,1,0,0] using 1 swap. [0,1,1,1,0,0,0] using 1 swap. [1,1,0,0,0,0,1] using 2 swaps (using the circular property of the array). There is no way to group all 1's together with 0 swaps. Thus, the minimum number of swaps required is 1.
Example 2:
Input: nums = [0,1,1,1,0,0,1,1,0] Output: 2 Explanation: Here are a few of the ways to group all the 1's together: [1,1,1,0,0,0,0,1,1] using 2 swaps (using the circular property of the array). [1,1,1,1,1,0,0,0,0] using 2 swaps. There is no way to group all 1's together with 0 or 1 swaps. Thus, the minimum number of swaps required is 2.
Example 3:
Input: nums = [1,1,0,0,1] Output: 0 Explanation: All the 1's are already grouped together due to the circular property of the array. Thus, the minimum number of swaps required is 0.
Constraints:
1 <= nums.length <= 105
nums[i]
is either0
or1
.
Solutions
Solution 1: Sliding Window
First, we count the number of $1$s in the array, denoted as $k$. The problem is actually asking for a circular subarray of length $k$ that contains the maximum number of $1$s. Therefore, the minimum number of swaps is $k$ minus the maximum number of $1$s in that subarray.
We can solve this problem using a sliding window. First, we count the number of $1$s in the first $k$ elements of the array, denoted as $cnt$. Then, we maintain a sliding window of length $k$. Each time we move the window one position to the right, we update $cnt$ and simultaneously update the maximum $cnt$ value, i.e., $mx = \max(mx, cnt)$. Finally, the answer is $k - mx$.
The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.
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