Two City Scheduling

Author: ishpreet-singh

3.cpp

@@ -0,0 +1,52 @@
+/*
+
+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 <= costs.length <= 100
+It is guaranteed that costs.length is even.
+1 <= costs[i][0], costs[i][1] <= 1000
+*/
+
+auto cmp = [](const vector<int>& lhs, const vector<int>& rhs) {
+    return lhs[0] - lhs[1] < rhs[0] - rhs[1];
+};
+
+class Solution {
+public:
+    
+    int twoCitySchedCost(vector<vector<int>>& costs) {
+        long long sum = 0;
+        sort(costs.begin(), costs.end(), cmp);
+        int n = costs.size()/2;
+        for(int i=0; i<2*n; i++) {
+            if(i < n) {
+                sum += costs[i][0];
+            } else {
+                sum += costs[i][1];
+            }
+        }
+        return sum;
+    }
+};