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+// 765. Couples Holding Hands
+#include "leetcode.h"
+
+/*
+ * there are 'n' couples sitting '2n' seats arranged in a row and want to hold
+ * hands. the people and seats are represented by an integer array 'row' where
+ * 'row[i]' is the id of the person sitting in the i'th seat. the couples are
+ * numbered in order, the first couple being '(0, 1)', the second couple being
+ * '(2, 3)', and so on with the last couple being '(2n -2, 2n -1)'. return the
+ * minimum number of swaps so that every couple is sitting side by side. a swap
+ * consists of choosing any two people then they stand up and switch seats.
+ */
+
+class Solution {
+public:
+ int minSwapsCouples(vector<int> &row) {
+ int ans = 0, n = row.size();
+ vector<int> ptn(n, 0), pos(n, 0);
+ for (int i = 0; i < n; i++) {
+ ptn[i] = (!(i % 2) ? i + 1 : i - 1);
+ pos[row[i]] = i;
+ }
+ for (int i = 0; i < n; i++)
+ for (int j = ptn[pos[ptn[row[i]]]]; i != j; j = ptn[pos[ptn[row[i]]]]) {
+ swap(row[i], row[j]);
+ swap(pos[row[i]], pos[row[j]]);
+ ans++;
+ }
+ return ans;
+ }
+};
+
+int main() {
+ Solution obj;
+ vector<int> r1 = {0, 2, 1, 3}, r2 = {3, 2, 0, 1};
+ printf("%d\n", obj.minSwapsCouples(r1)); // expect: 1
+ printf("%d\n", obj.minSwapsCouples(r2)); // expect: 0
+}