1029. 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.

sorting + greedy solution

• sort the costs by the difference between a and b.
• add A’s cost in first half and B’s cost in second half.
• return the ans.
• TC: O(nlogn)
• SC: O(1)

Code

class Solution {
public:
    int twoCitySchedCost(vector<vector<int>>& costs)
    {
        int n = costs.size();
        int half = n / 2;
        sort(costs.begin(), costs.end(), [](vector<int>& a, vector<int>& b) {
            return a[0] - a[1] < b[0] - b[1];
        });
        int ans = 0;
        for (int i = 0; i < half; i++) {
            ans += costs[i][0];
        }
        for (int i = half; i < n; i++) {
            ans += costs[i][1];
        }
        return ans;
    }
};