1029 Two City Scheduling.py

Author: rajeevranjancom

1029 Two City Scheduling.py

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+#!/usr/bin/python3
+"""
+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 <= costs.length <= 100
+It is guaranteed that costs.length is even.
+1 <= costs[i][0], costs[i][1] <= 1000
+"""
+
+
+class Solution:
+    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
+        """
+        sort by city A and greedy? [30, 20]?
+        sort by total?
+        sort by diff - either choose A or B, the difference matters
+
+        a - b: incremental cost of flying A instead of B
+        """
+        A = [(a - b, a, b) for a, b in costs]
+        A.sort()
+        ret = 0
+        remain = len(A) // 2
+        for _, a, b in A:
+            if remain > 0:
+                ret += a
+                remain -= 1
+            else:
+                ret += b
+
+        return ret
+
+    def twoCitySchedCost_error(self, costs: List[List[int]]) -> int:
+        """
+        sort by city A and greedy? [30, 20]?
+        sort by total?
+        sort by diff - either choose A or B, the difference matters
+
+        Error in the abs of difference
+        """
+        A = [(abs(a - b), a, b) for a, b in costs]
+        A.sort(reverse=True)
+        ret = 0
+        remain = len(A) // 2
+        for _, a, b in A:
+            if a > b:
+                ret += b
+            elif remain > 0:
+                ret += a
+                remain -= 1
+            else:
+                ret += b
+
+        return ret