Two city scheduling

Input: [[10,20],[30,200],[400,50],[30,20]]
Output: 110
Explanation: 
The first person goes to city A for a cost of 10.
The second person goes to city A for a cost of 30.
The third person goes to city B for a cost of 50.
The fourth person goes to city B for a cost of 20.
The total minimum cost is 10 + 30 + 50 + 20 = 110 to have half the people interviewing in each city.

Code: 
var twoCitySchedCost = function(costs) {
  //Here we are sorting based on the absolute difference between 0 and 1st elemet
  //this extra line compared to previous one
   costs.sort((a,b) =>Math.abs(b[0]-b[1])-Math.abs(a[0]-a[1]));
  
    let minCost = 0;
    let countA = 0;
    let countB = 0;
    
    for(let i=0;i<costs.length;i++){
        if(costs[i][0] <costs[i][1]){
           if(countA <costs.length/2){
               countA++;
               minCost += costs[i][0];
           }
            else{
                countB++;
                minCost += costs[i][1];
            }
        }else{
            if(countB <costs.length/2){
                countB++;
                minCost +=costs[i][1];
            }
            else{
                countA++;
                minCost += costs[i][0];
            }
        }
    }
    return minCost;
};
Status: Couldn't solve, found question ambiguous. Switched to javascript because comparator class here is much simpler.