Create 02 - Top-Down | DP | Approach.cpp

Author: JawadSher

24 - Dynamic Programming Problems/05 - Coin Change/02 - Top-Down | DP | Approach.cpp

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+class Solution {
+public:
+    // Helper function to solve the coin change problem using recursion with memoization
+    int solve(vector<int>& coins, int x, vector<int>& dp) {
+        // Base case: If the target amount is 0, no coins are needed
+        if(x == 0) return 0;
+
+        // Base case: If the target amount is negative, return an invalid result
+        if(x < 0) return INT_MAX;
+
+        // If the result for the current amount is already computed, return it
+        if(dp[x] != -1) return dp[x];
+
+        // Initialize the minimum number of coins required to a large value
+        int mini = INT_MAX;
+
+        // Iterate through each coin
+        for(int i = 0; i < coins.size(); i++) {
+            // Recursive call: Try using the current coin and solve for the remaining amount
+            int ans = solve(coins, x - coins[i], dp);
+
+            // If the recursive result is valid, update the minimum number of coins
+            if(ans != INT_MAX) 
+                mini = min(mini, 1 + ans); // Add 1 to include the current coin
+        }
+
+        // Store the computed result in the memoization array
+        dp[x] = mini;
+
+        // Return the computed result
+        return mini;
+    }
+
+    // Main function to find the minimum number of coins needed to make up the given amount
+    int coinChange(vector<int>& coins, int amount) {
+        // Create a memoization array initialized with -1
+        vector<int> dp(amount + 1, -1);
+
+        // Call the recursive helper function with memoization
+        int ans = solve(coins, amount, dp);
+
+        // If no valid combination of coins can form the amount, return -1
+        if(ans == INT_MAX) return -1;
+
+        // Otherwise, return the computed result
+        return ans;
+    }
+};