2022-03-25 Daily-Challenge

Posted on:
Posted modified:
In: DailyChallenge

Today I have done leetcode's March LeetCoding Challenge with cpp.

March LeetCoding Challenge 25

Description

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

Solution

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

// Accepted
// 50/50 cases passed (4 ms)
// Your runtime beats 78.71 % of cpp submissions
// Your memory usage beats 78.35 % of cpp submissions (7.9 MB)
Algorithm