Create 04 - Space Optimized | DP | Approach.cpp

Author: JawadSher

24 - Dynamic Programming Problems/14 - Minimum Cost for Tickets/04 - Space Optimized | DP | Approach.cpp

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+class Solution {
+public:
+    int mincostTickets(vector<int>& days, vector<int>& costs) {
+        int ans = 0; // Variable to keep track of the minimum cost so far
+
+        // Queues to track the minimum cost for 7-day and 30-day passes
+        queue<pair<int, int>> month; // Stores {day, cost} for the 30-day pass
+        queue<pair<int, int>> week;  // Stores {day, cost} for the 7-day pass
+
+        // Iterate through all the travel days
+        for (int day : days) {
+            // Remove outdated entries from the month queue
+            // If the current day is greater than or equal to the day of the oldest entry + 30,
+            // it means the 30-day pass for that day is no longer valid
+            while (!month.empty() && month.front().first + 30 <= day) {
+                month.pop();
+            }
+
+            // Remove outdated entries from the week queue
+            // If the current day is greater than or equal to the day of the oldest entry + 7,
+            // it means the 7-day pass for that day is no longer valid
+            while (!week.empty() && week.front().first + 7 <= day) {
+                week.pop();
+            }
+
+            // Add a new entry for the current day to the week queue
+            // This represents the cost of buying a 7-day pass starting today
+            week.push(make_pair(day, ans + costs[1]));
+
+            // Add a new entry for the current day to the month queue
+            // This represents the cost of buying a 30-day pass starting today
+            month.push(make_pair(day, ans + costs[2]));
+
+            // Update the minimum cost (ans) for the current day
+            // Consider three options:
+            // 1. Buying a 1-day pass for the current day
+            // 2. Using the cheapest cost from the week queue
+            // 3. Using the cheapest cost from the month queue
+            ans = min(ans + costs[0], min(week.front().second, month.front().second));
+        }
+
+        // Return the total minimum cost to cover all the travel days
+        return ans;
+    }
+};