Create 04 - Space Optimized | DP | Approach.cpp

Author: JawadSher

24 - Dynamic Programming Problems/32 - Best Time To Buy and Sell Stock II/04 - Space Optimized | DP | Approach.cpp

@@ -0,0 +1,50 @@
+class Solution {
+public:
+    // Function to calculate the maximum profit using space-optimized dynamic programming
+    int solve(vector<int>& prices) {
+        int n = prices.size(); // Get the number of days
+
+        // Space-optimized DP approach:
+        // Instead of a 2D table, use two 1D arrays to store current and next states
+        vector<int> curr(2, 0); // Array to store the current day's profit for both states (buy/sell)
+        vector<int> next(2, 0); // Array to store the next day's profit for both states (buy/sell)
+
+        // Iterate backward from the second last day to the first day
+        for (int index = n - 1; index >= 0; index--) {
+            // Iterate over the two states: buy (1) and sell (0)
+            for (int buy = 0; buy <= 1; buy++) {
+                int profit = 0; // Initialize profit for the current state
+
+                if (buy) {
+                    // If we are allowed to buy:
+                    // - Option 1: Buy the stock today (-prices[index]) and move to the next day with buy=0
+                    // - Option 2: Skip buying today and move to the next day with buy=1
+                    int buyStock = -prices[index] + next[0];
+                    int notBuyStock = 0 + next[1];
+                    profit = max(buyStock, notBuyStock);
+                } else {
+                    // If we are allowed to sell:
+                    // - Option 1: Sell the stock today (+prices[index]) and move to the next day with buy=1
+                    // - Option 2: Skip selling today and move to the next day with buy=0
+                    int sellStock = +prices[index] + next[1];
+                    int notSellStock = 0 + next[0];
+                    profit = max(sellStock, notSellStock);
+                }
+
+                // Store the profit for the current state in the current day's array
+                curr[buy] = profit;
+            }
+
+            // Update the next day's array with the current day's results
+            next = curr;
+        }
+
+        // The maximum profit starting from day 0 with the ability to buy (buy=1)
+        return curr[1];
+    }
+
+    // Main function to calculate the maximum profit
+    int maxProfit(vector<int>& prices) {
+        return solve(prices); // Call the solve function
+    }
+};