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// 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
}
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