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 # -*- coding: utf-8 -*- # # @lc app=leetcode id=765 lang=ruby # # [765] Couples Holding Hands # # https://leetcode.com/problems/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: # ⁠ # ⁠len(row) is even and in the range of [4, 60]. # ⁠row is guaranteed to be a permutation of 0...len(row)-1. # This problem seems interesting. Easy to write hard to prove. # I'll look into this if I got interviewed. # @param {Integer[]} row # @return {Integer} def min_swaps_couples(row) i, j = 0, 1 count = 0 while j < row.length if !is_cp(row, i, j) cp = other(row[i]) swap(row, row.find_index(cp), j) count += 1 end i += 2 j += 2 end count end def is_cp(row, i, j) row[j] == other(row[i]) end def swap(row, i, j) row[i], row[j] = row[j], row[i] end def other(p) if p.odd? p-1 else p+1 end end
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