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1029. Two City Scheduling 👍

  • Time: $O(\texttt{sort})$
  • Space: $O(\texttt{sort})$
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class Solution {
 public:
  int twoCitySchedCost(vector<vector<int>>& costs) {
    const int n = costs.size() / 2;
    int ans = 0;

    // How much money can we save if we fly a person to A instead of B?
    // To save money, we should
    //   1. Fly the person with the maximum saving to A.
    //   2. Fly the person with the minimum saving to B.

    // Sort `costs` in ascending order by the money saved if we fly a person to
    // B instead of A.
    ranges::sort(costs, ranges::less{},
                 [](const vector<int>& cost) { return cost[0] - cost[1]; });

    for (int i = 0; i < n; ++i)
      ans += costs[i][0] + costs[i + n][1];

    return ans;
  }
};

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class Solution {
  public int twoCitySchedCost(int[][] costs) {
    final int n = costs.length / 2;
    int ans = 0;

    // How much money can we save if we fly a person to A instead of B?
    // To save money, we should
    //   1. Fly the person with the maximum saving to A.
    //   2. Fly the person with the minimum saving to B.

    // Sort `costs` in ascending order by the money saved if we fly a person to
    // B instead of A.
    Arrays.sort(costs, (a, b) -> Integer.compare(a[0] - a[1], b[0] - b[1]));

    for (int i = 0; i < n; ++i)
      ans += costs[i][0] + costs[i + n][1];

    return ans;
  }
}

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class Solution:
  def twoCitySchedCost(self, costs: list[list[int]]) -> int:
    n = len(costs) // 2

    # How much money can we save if we fly a person to A instead of B?
    # To save money, we should
    #   1. Fly the person with the maximum saving to A.
    #   2. Fly the person with the minimum saving to B.

    # Sort `costs` in ascending order by the money saved if we fly a person to
    # B instead of A.
    costs.sort(key=lambda x: x[0] - x[1])
    return sum(costs[i][0] + costs[i + n][1] for i in range(n))