JuneChallenge (64) | Two City Scheduling

  Two City Scheduling

There are 2N people a company is planning to interview. The cost of flying the i-th person to city A is costs[i][0], and the cost of flying the i-th person to city B is costs[i][1].

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

Example 1:

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.

Note:

1. 1 <= costs.length <= 100
2. It is guaranteed that costs.length is even.
3. 1 <= costs[i][0], costs[i][1] <= 1000

If you random chooses, you can't control the best solution. Instead of find the best of minimum value for each, we delete the value will not choose.

Ex: [10,20] [30][200] [400,50] [30,20]. Not choose: 400, 200, 20, 30. So, we implement it with Greedy algorithm.

My code:

	class Solution {
	public int twoCitySchedCost(int[][] costs) {
	PriorityQueue<int[]> pq = new PriorityQueue<>((a,b)->(Math.abs(b[0]-b[1])-Math.abs(a[0]-a[1])));
	int res = 0;
	int numsA = 0;
	int numsB = 0;
	for (int arr[]: costs){
	pq.add(arr);
	}
	System.out.println(costs.length/2);
	while (!pq.isEmpty()){
	if ((pq.peek()[0]<=pq.peek()[1] && numsA <costs.length/2) || numsB == costs.length/2){
	numsA++;
	res+=pq.peek()[0];
	}
	else{
	numsB++;
	res+=pq.peek()[1];
	}
	pq.remove();
	}
	return res;
	}
	}

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