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 of 0...len(row)-1.
Solution:
class Solution(object):
def minSwapsCouples(self, row):
"""
:type row: List[int]
:rtype: int
"""
seats = {} # store the two people on couch i
cur_pos = {} # store the couch person x is sitting now
for i in range(0, len(row), 2):
seats[i] = [row[i], row[i+1]]
cur_pos[row[i]] = i
cur_pos[row[i+1]] = i
# check each couch, if the two people are already couple, skip
# otherwise iteratively find the partner for the small one
ans = 0
for i in range(0, len(row), 2):
x, y = seats[i]
if x > y:
x, y = y, x
if x % 2 == 0 and y % 2 == 1:
continue
pre = i
while not (x % 2 == 0 and y % 2 == 1): # not couple yet
ans += 1
partner = (x+1) if x % 2 == 0 else (x-1)#x couple id
partner_cur_pos = cur_pos[partner]#locate x parter's position
seats[pre] = [x, partner]#update current seats to x+x's couple
seats[partner_cur_pos].remove(partner)
seats[partner_cur_pos].append(y)#update the x's couplt's previous seats, executing the swapping
cur_pos[y] = partner_cur_pos
cur_pos[partner] = pre
pre = partner_cur_pos
x, y = seats[pre]
if x > y:
x, y = y, x
return ansSolution: https://leetcode.com/problems/couples-holding-hands/