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]. -
rowis guaranteed to be a permutation of0...len(row)-1.
Related Topics:
Greedy, Union Find, Graph
Similar Questions:
// OJ: https://leetcode.com/problems/couples-holding-hands/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
// Ref: https://leetcode.com/problems/couples-holding-hands/discuss/117520/Java-union-find-easy-to-understand-5-ms
class UnionFind {
vector<int> id, rank;
int cnt;
public:
UnionFind(int n) : id(n), rank(n, 1), cnt(n) {
for(int i = 0; i < n; ++i) id[i] = i;
}
void connect(int a, int b) {
int x = find(a), y = find(b);
if (x == y) return;
if (rank[x] <= rank[y]) {
id[x] = y;
if (rank[x] == rank[y]) rank[y]++;
} else id[y] = x;
--cnt;
}
int find(int a) {
return id[a] == a ? a : (id[a] = find(id[a]));
}
int getCount() { return cnt; }
};
class Solution {
public:
int minSwapsCouples(vector<int>& row) {
int N = row.size() / 2;
UnionFind uf(N);
for (int i = 0; i < N; ++i) uf.connect(row[2 * i] / 2, row[2 * i + 1] / 2);
return N - uf.getCount();
}
};