Couples Holding Hands

Description

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.

Solution(javascript)

/**
 * @param {number[]} row
 * @return {number}
 */
const minSwapsCouples = function (row) {
  const map = row.reduce((acc, num, index) => {
    acc[num] = index
    return acc
  }, {})
  const swap = (a, b) => {
    const indexA = map[a]
    const indexB = map[b]
    let temp = map[a]
    map[a] = map[b]
    map[b] = map[a]
    temp = row[indexA]
    row[indexA] = row[indexB]
    row[indexB] = temp
  }
  let count = 0
  for (let i = 0; i < row.length; i += 2) {
    if (row[i] % 2 === 0) {
      if (row[i + 1] !== row[i] + 1) {
        swap(row[i + 1], row[i] + 1)
        count += 1
      }
    } else if (row[i + 1] !== row[i] - 1) {
      swap(row[i + 1], row[i] - 1)
      count += 1
    }
  }
  return count
}