LeetCode 1029. Two City Scheduling

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.

 

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

 

Constraints:

• 2 * n == costs.length
• 2 <= costs.length <= 100
• costs.length is even.
• 1 <= aCosti, bCosti <= 1000

	class Solution {
	public:
	int twoCitySchedCost(vector<vector<int>>& costs) {
	// sort by difference
	sort(costs.begin(), costs.end(), [&](vector<int> a, vector<int> b){
	return a[0] - a[1] < b[0] - b[1];
	});
	int sum = 0;
	for(int i = 0; i < costs.size(); i++){
	// left half: it is best to use a
	if(i < costs.size() / 2)
	sum += costs[i][0];
	// right half: it is best to use b
	else
	sum += costs[i][1];
	}
	return sum;
	}
	};

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