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authorjack <[email protected]>2024-06-14 13:13:15 -0600
committerjack <[email protected]>2024-06-14 13:13:15 -0600
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tree96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /min_swap_group_1.cpp
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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
+}