Formatted question description: https://leetcode.ca/all/765.html

765. Couples Holding Hands (Hard)

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:

  1. len(row) is even and in the range of [4, 60].
  2. row is guaranteed to be a permutation of 0...len(row)-1.

Related Topics:
Greedy, Union Find, Graph

Similar Questions:

Solution 1. Union Find

// OJ: https://leetcode.com/problems/couples-holding-hands/

// 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();
    }
};

Java

  • class Solution {
        public int minSwapsCouples(int[] row) {
            int swaps = 0;
            Map<Integer, Integer> numberIndexMap = new HashMap<Integer, Integer>();
            int length = row.length;
            for (int i = 0; i < length; i++)
                numberIndexMap.put(row[i], i);
            for (int i = 0; i < length; i += 2) {
                int num1 = row[i];
                int couple = getCouple(num1);
                int coupleIndex = numberIndexMap.get(couple);
                if (coupleIndex != i + 1) {
                    int num2 = row[i + 1];
                    row[i + 1] = couple;
                    row[coupleIndex] = num2;
                    numberIndexMap.put(couple, i + 1);
                    numberIndexMap.put(num2, coupleIndex);
                    swaps++;
                }
            }
            return swaps;
        }
    
        public int getCouple(int num) {
            return num % 2 == 0 ? num + 1 : num - 1;
        }
    }
    
  • // OJ: https://leetcode.com/problems/couples-holding-hands/
    // 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();
        }
    };
    
  • class Solution(object):
        def minSwapsCouples(self, row):
            """
            :type row: List[int]
            :rtype: int
            """
            res = 0
            n = len(row)
            for i in range(0, n - 1, 2):
                if row[i + 1] == (row[i] ^ 1):
                    continue
                for j in range(i + 1, n):
                    if row[j] == (row[i] ^ 1):
                        row[j], row[i + 1] = row[i + 1], row[j]
                res += 1
            return res
    

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