N couples sit in 2N seats arranged in a row and want to hold hands. We want to know 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.
The people and seats are represented by an integer from 0
to 2N-1
, 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)
.
The couples' initial seating is given by row[i]
being the value of the person who is initially sitting in the i-th seat.
Example 1:
Input: row = [0, 2, 1, 3] Output: 1 Explanation: We only need to swap the second (row[1]) and third (row[2]) person.
Example 2:
Input: row = [3, 2, 0, 1] Output: 0 Explanation: All couples are already seated side by side.
Note:
len(row)
is even and in the range of[4, 60]
.row
is guaranteed to be a permutation of0...len(row)-1
.
Approach #1: C++.
class Solution { public: int minSwapsCouples(vector<int>& row) { int res = 0, N = row.size(); vector<int> ptn(N, 0); vector<int> pos(N, 0); for (int i = 0; i < N; ++i) { ptn[i] = (i % 2 == 0 ? 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]]); res++; } } return res; } };
Analysis:
https://leetcode.com/problems/couples-holding-hands/discuss/113362/JavaC%2B%2B-O(N)-solution-using-cyclic-swapping