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// 2134. Minimum Swaps to Group All 1's Together II
#include "leetcode.h"
/*
* a swap is defined as taking two distinct position 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 adjacent. given a binary circular
* array 'nums', return the minimum number of swaps required to group all 1's
* preent in the array together at any location.
*/
class Solution {
public:
int minSwaps(vector<int> &nums) {
int n = nums.size(), k = 0;
for (int i = 0; i < n; i++)
k += (nums[i] == 1);
nums.insert(nums.end(), nums.begin(), nums.end());
int ans = n, i = 0, j = 0, l = 0;
while (j < 2 * n) {
l += (nums[j] == 1);
if (j - i + 1 == k) {
ans = min(ans, k - l);
l -= (nums[i] == 1);
i++;
}
j++;
}
return ans == n ? 0 : ans;
}
};
int main() {
Solution obj;
vector<int> n1 = {0, 1, 0, 1, 1, 0, 0}, n2 = {0, 1, 1, 1, 0, 0, 1, 1, 0},
n3 = {1, 1, 0, 0, 1};
printf("%d\n", obj.minSwaps(n1)); // expect: 1
printf("%d\n", obj.minSwaps(n2)); // expect: 2
printf("%d\n", obj.minSwaps(n3)); // expect: 0
}
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