Create: 0983-minimum-cost-for-tickets.cpp

Author: ashutosh2706

cpp/0983-minimum-cost-for-tickets.cpp

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+/*
+The days of the year in which you will travel are given as an integer array days. Each day is an integer from 1 to 365.
+
+Train tickets are sold in three different ways:
+
+a 1-day pass is sold for costs[0] dollars,
+a 7-day pass is sold for costs[1] dollars, and
+a 30-day pass is sold for costs[2] dollars.
+The passes allow that many days of consecutive travel.
+
+Return the minimum number of dollars you need to travel every day in the given list of days.
+
+
+Example. For days = [1,4,6,7,8,20] and costs = [2,7,15] we can buy a 1-day pass
+	 for costs[0] = $2, which covers day 1. On day 3 we can buy a 7-day pass
+	 for costs[1] = $7, which covers days 3,4....9. On day 20 we can again buy a
+	 1-day pass for costs[0] = $2 that will cover the 20th day. So in total we spent
+	 2 + 7 + 2 = $11 which is the minimum dollars needed for travelling in this case.
+
+
+Time: O(n)
+Space: O(n)
+
+*/
+
+
+class Solution {
+public:
+    int mincostTickets(vector<int>& days, vector<int>& costs) {
+
+    vector<int> dp(days.size() + 1, 1e9);
+    dp[days.size()] = 0;
+    for(int i=days.size()-1; i>=0; i--) {
+        for(int j=0; j<3; j++) {
+            if(j == 0) {
+                auto it = lower_bound(days.begin()+i, days.end(), days[i]+1);
+                dp[i] = min(dp[i], costs[j] + dp[it-days.begin()]);
+            }
+            else if(j == 1) {
+                auto it = lower_bound(days.begin()+i, days.end(), days[i]+7);
+                dp[i] = min(dp[i], costs[j] + dp[it-days.begin()]);
+            } else {
+                auto it = lower_bound(days.begin()+i, days.end(), days[i]+30);
+                dp[i] = min(dp[i], costs[j] + dp[it-days.begin()]);
+            }
+        }
+    }
+    return dp[0];
+
+    }
+};