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-rw-r--r--min_swap_group_1.c53
1 files changed, 53 insertions, 0 deletions
diff --git a/min_swap_group_1.c b/min_swap_group_1.c
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+// 2134. Minimum Swaps to Group All 1's Together II
+#include <math.h>
+#include <stdio.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
+}