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

Author: JawadSher

24 - Dynamic Programming Problems/08 - Maximum The Cut Segments/03 - Bottom-Up | DP | Approach.cpp

@@ -0,0 +1,42 @@
+class Solution
+{
+    public:
+    // Function to calculate the maximum number of cuts using an iterative approach
+    int solve(int n, int x, int y, int z){
+        // Initialize a dp array with size (n+1) and set all values to INT_MIN
+        // INT_MIN is used to indicate invalid states where no cuts are possible
+        vector<int> dp(n+1, INT_MIN);
+
+        // Base case: If the length is 0, no cuts are needed
+        dp[0] = 0;
+
+        // Iterate through all lengths from 1 to n
+        for(int i = 1; i <= n; i++){
+            // If a cut of length x is possible, update dp[i]
+            if(i-x >= 0) dp[i] = max(dp[i], dp[i-x] + 1);
+
+            // If a cut of length y is possible, update dp[i]
+            if(i-y >= 0) dp[i] = max(dp[i], dp[i-y] + 1);
+
+            // If a cut of length z is possible, update dp[i]
+            if(i-z >= 0) dp[i] = max(dp[i], dp[i-z] + 1);
+        }
+
+        // Return the maximum number of cuts for length n
+        // If no valid cuts are possible, dp[n] will still be INT_MIN
+        return dp[n];
+    }
+    
+    // Function to calculate the maximum number of cuts that can be made
+    int maximizeTheCuts(int n, int x, int y, int z)
+    {
+        // Call the solve function to calculate the maximum cuts
+        int maxCuts = solve(n, x, y, z);
+
+        // If the result is negative, it means no valid cuts are possible
+        if(maxCuts < 0) return 0;
+
+        // Return the maximum number of valid cuts
+        return maxCuts;
+    }
+};