// 2134. Minimum Swaps to Group All 1's Together II #include "leetcode.h" #pragma GCC optimize("-O3") /* * 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. */ int minSwaps(int *nums, int nums_size) { int ones = 0, n = nums_size; for (int i = 0; i < n; i++) if (nums[i] == 1) ones++; int cnt = 0; for (int i = 0; i < ones; i++) if (nums[i] == 1) cnt++; if (cnt == ones) return 0; int max = cnt; for (int i = ones; i < n; i++) { if (nums[i] == 1) cnt++; if (nums[i - ones] == 1) cnt--; if (cnt == ones) return 0; max = fmax(max, cnt); } for (int i = n; i < n + ones; i++) { if (nums[i % n] == 1) cnt++; if (nums[i - ones] == 1) cnt--; if (cnt == ones) return 0; max = fmax(max, cnt); } return ones - max; } int main() { 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", minSwaps(n1, 7)); // expect: 1 printf("%d\n", minSwaps(n2, 9)); // expect: 2 printf("%d\n", minSwaps(n3, 5)); // expect: 0 }