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The goal is to find the minimum number of swaps needed to group all 1's together. First, count the total number of 1's, which determines the window size you are interested in. Use the sliding window technique to find the window with the maximum number of 1's. This will ensure that the remaining positions in this window must be filled with 1's by swapping.
Since the array is circular, concatenate the array with itself to simulate the wrap-around. Slide a window of the calculated size and keep track of the maximum number of 1's within this window. To minimize swaps, the window should contain the maximum possible 1's.
Time Complexity: O(n), where n is the size of the array, due to the sliding window.
Space Complexity: O(1), because the space used does not scale with input size.
1import java.util.*;
2
3public class Solution {
4 public int minSwaps(int[] nums) {
5
The Java solution adopts similar logic to the C/C++ implementations. Through a sliding window approach applied over a virtual version of the circular array, it calculates and verifies the maximum aggregation of 1's possible within any window. The difference from the overall count gives the swap count needed.