Create 03 - Bottom-Up | DP | Approach.cpp

Author: JawadSher

24 - Dynamic Programming Problems/13 - Perfect Squares/03 - Bottom-Up | DP | Approach.cpp

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+class Solution {
+public:
+    // Function to calculate the minimum number of perfect squares
+    int solve(int n){
+        // Create a DP array of size n+1 initialized with INT_MAX (a large value representing infinity)
+        // dp[i] will store the minimum number of perfect squares required to sum to i.
+        vector<int> dp(n+1, INT_MAX);
+
+        // Base case: dp[0] = 0, as no squares are needed to sum to 0.
+        dp[0] = 0;
+
+        // Iterate over all values of i from 1 to n
+        for(int i = 1; i <= n; i++){
+            // For each i, check all perfect squares (j * j) that are less than or equal to i
+            for(int j = 1; j * j <= i; j++){
+                int SQRT = j * j;  // Square of j
+
+                // If subtracting SQRT from i doesn't make it negative, update dp[i]
+                // Take the minimum of the current value of dp[i] and dp[i - SQRT] + 1
+                // (the +1 is because we are using one more square, i.e., SQRT)
+                dp[i] = min(dp[i], dp[i - SQRT] + 1);
+            }
+        }
+
+        // Return the minimum number of perfect squares for the number n
+        return dp[n];
+    }
+
+    // Main function to return the minimum number of perfect squares for n
+    int numSquares(int n) {
+        return solve(n);  // Call the solve function to calculate the result
+    }
+};