### Description:

A company is planning to interview 2n people. Given the array costs where costs[i] = [aCosti, bCosti], the cost of flying the ith person to city a is aCosti, and the cost of flying the ith person to city b is bCosti.

Return the minimum cost to fly every person to a city such that exactly n people arrive in each city.

### Solution:

Time Complexity : O(nlog(n))

Space Complexity: O(1)

```
var twoCitySchedCost = function(costs) {
// Calculate the amount of people we need per city
let n = costs.length/2;
// Pointers and total cost
let a = 0, b = 0, total = 0;
// Sort costs by greatest difference
costs.sort((a,b) => Math.abs(b[0]-b[1])-Math.abs(a[0]-a[1]));
// Add costs of flights to the total
for (let cost of costs) {
if (cost[0] <= cost[1] && a < n) {
total += cost[0];
a++;
} else if (cost[0] >= cost[1] && b < n) {
total += cost[1];
b++;
} else total += a < n ? cost[0] : cost[1];
}
return total;
};
```

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