Two City - Implementation

Here is the solution:

function twoCitySchedCost(costs)
 sort costs by (cost[0] - cost[1]) ascending
 total := 0
 n := length / 2
 for i from 0 to n - 1
 total := total + costs[i][0]
 for i from n to length - 1
 total := total + costs[i][1]
 return total

Time: $O(n \log n)$. Space: $O(1)$.

People with the smallest $cost_A - cost_B$ should go to A. That means A is relatively cheaper for them.