8. Two City Scheduling

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.

Example 1:
Input: costs = [[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.

Example 2:
Input: costs = [[259,770],[448,54],[926,667],[184,139],[840,118],[577,469]]
Output: 1859


Example 3:
Input: costs = [[515,563],[451,713],[537,709],[343,819],[855,779],[457,60],[650,359],[631,42]]
Output: 3086

Solution:

Approach: Find the profit in travelling from city A to city B Sort according to the profit and find first n values

class Solution
{
public:
    static bool comparator(pair<int, int> &a, pair<int, int> &b)
    {
        return a.first > b.first;
    }
    
    int twoCitySchedCost(vector<vector<int>> &costs)
    {

        int n = costs.size() / 2;
        vector<pair<int, int>> v;
        for (int i = 0; i < costs.size(); i++)
        {

            int a = costs[i][0];
            int b = costs[i][1];

            int dif = b - a;
            v.push_back({dif, i});
        }

        sort(v.begin(), v.end(), comparator);
        int s = 0;
        for (int i = 0; i < n; i++)
        {
            int index = v[i].second;
            s += costs[index][0];
        }

        for (int i = n; i < costs.size(); i++)
        {
            int index = v[i].second;
            s += costs[index][1];
        }

        return s;
    }
};

Time Complexity: O(n log n)