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

Author: JawadSher

24 - Dynamic Programming Problems/13 - Perfect Squares/02 - Top-Down | DP | Approach.cpp

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+class Solution {
+public:
+    // Recursive function to calculate the minimum number of perfect squares 
+    // using memoization to avoid redundant calculations
+    int solve(int n, vector<int> &dp){
+        // Base case: dp[0] = 0, as 0 can be represented by 0 perfect squares
+        dp[0] = 0;
+
+        // If the result for the current value of n has already been computed, return it
+        if(dp[n] != -1) return dp[n];
+
+        // Initialize ans to the maximum possible value, which is n (since n can be represented
+        // by the sum of n 1's (1^2), which is the worst case)
+        int ans = n;
+
+        // Try all perfect squares less than or equal to n
+        for(int i = 1; i * i <= n; i++){
+            // The current square number
+            int SQRT = i * i;
+
+            // Recursively calculate the minimum squares needed for the remaining value (n - SQRT)
+            ans = min(ans, solve(n - SQRT, dp) + 1); // Add 1 for the current perfect square used
+        }
+
+        // Store the result for the current value of n to avoid recalculating it
+        dp[n] = ans;
+
+        // Return the minimum number of perfect squares for the current value of n
+        return dp[n];
+    }
+
+    // Main function to calculate the minimum number of perfect squares for the input n
+    int numSquares(int n) {
+        // Create a DP array of size n+1 initialized with -1 (indicating that no value has been computed yet)
+        vector<int> dp(n+1, -1);
+
+        // Call the recursive function with memoization to compute the result
+        return solve(n, dp);
+    }
+};