Welcome to Subscribe On Youtube

765. Couples Holding Hands

Description

There are n couples sitting in 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 ith 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.

 

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.

 

Constraints:

• 2n == row.length
• 2 <= n <= 30
• n is even.
• 0 <= row[i] < 2n
• All the elements of row are unique.

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• C#

• class Solution {
      private int[] p;
  
      public int minSwapsCouples(int[] row) {
          int n = row.length >> 1;
          p = new int[n];
          for (int i = 0; i < n; ++i) {
              p[i] = i;
          }
          for (int i = 0; i < n << 1; i += 2) {
              int a = row[i] >> 1, b = row[i + 1] >> 1;
              p[find(a)] = find(b);
          }
          int ans = n;
          for (int i = 0; i < n; ++i) {
              if (i == find(i)) {
                  --ans;
              }
          }
          return ans;
      }
  
      private int find(int x) {
          if (p[x] != x) {
              p[x] = find(p[x]);
          }
          return p[x];
      }
  }
  

• class Solution {
  public:
      int minSwapsCouples(vector<int>& row) {
          int n = row.size() / 2;
          int p[n];
          iota(p, p + n, 0);
          function<int(int)> find = [&](int x) -> int {
              if (p[x] != x) {
                  p[x] = find(p[x]);
              }
              return p[x];
          };
          for (int i = 0; i < n << 1; i += 2) {
              int a = row[i] >> 1, b = row[i + 1] >> 1;
              p[find(a)] = find(b);
          }
          int ans = n;
          for (int i = 0; i < n; ++i) {
              ans -= i == find(i);
          }
          return ans;
      }
  };
  

• class Solution:
      def minSwapsCouples(self, row: List[int]) -> int:
          def find(x: int) -> int:
              if p[x] != x:
                  p[x] = find(p[x])
              return p[x]
  
          n = len(row) >> 1
          p = list(range(n))
          for i in range(0, len(row), 2):
              a, b = row[i] >> 1, row[i + 1] >> 1
              p[find(a)] = find(b)
          return n - sum(i == find(i) for i in range(n))
  
  

• func minSwapsCouples(row []int) int {
  	n := len(row) >> 1
  	p := make([]int, n)
  	for i := range p {
  		p[i] = i
  	}
  	var find func(int) int
  	find = func(x int) int {
  		if p[x] != x {
  			p[x] = find(p[x])
  		}
  		return p[x]
  	}
  	for i := 0; i < n<<1; i += 2 {
  		a, b := row[i]>>1, row[i+1]>>1
  		p[find(a)] = find(b)
  	}
  	ans := n
  	for i := range p {
  		if find(i) == i {
  			ans--
  		}
  	}
  	return ans
  }
  

• function minSwapsCouples(row: number[]): number {
      const n = row.length >> 1;
      const p: number[] = Array(n)
          .fill(0)
          .map((_, i) => i);
      const find = (x: number): number => {
          if (p[x] !== x) {
              p[x] = find(p[x]);
          }
          return p[x];
      };
      for (let i = 0; i < n << 1; i += 2) {
          const a = row[i] >> 1;
          const b = row[i + 1] >> 1;
          p[find(a)] = find(b);
      }
      let ans = n;
      for (let i = 0; i < n; ++i) {
          if (i === find(i)) {
              --ans;
          }
      }
      return ans;
  }
  
  

• public class Solution {
      private int[] p;
  
      public int MinSwapsCouples(int[] row) {
          int n = row.Length >> 1;
          p = new int[n];
          for (int i = 0; i < n; ++i) {
              p[i] = i;
          }
          for (int i = 0; i < n << 1; i += 2) {
              int a = row[i] >> 1;
              int b = row[i + 1] >> 1;
              p[find(a)] = find(b);
          }
          int ans = n;
          for (int i = 0; i < n; ++i) {
              if (p[i] == i) {
                  --ans;
              }
          }
          return ans;
      }
  
      private int find(int x) {
          if (p[x] != x) {
              p[x] = find(p[x]);
          }
          return p[x];
      }
  }
  

All Problems

All Solutions