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

Author: JawadSher

24 - Dynamic Programming Problems/08 - Maximum The Cut Segments/02 - Top-Down | DP | Approach.cpp

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+class Solution
+{
+    public:
+    // Recursive function with memoization to calculate the maximum number of cuts
+    int solve(int n, int x, int y, int z, vector<int>& dp){
+        // Base case: If the length is exactly 0, no further cuts are needed
+        if(n == 0) return 0;
+        // Base case: If the length becomes negative, this is an invalid case
+        if(n < 0) return INT_MIN;
+
+        // If the result for the current length `n` is already computed, return it
+        if(dp[n] != -1) return dp[n];
+
+        // Try cutting a segment of length x and recursively calculate the cuts
+        int a = solve(n-x, x, y, z, dp) + 1;
+        // Try cutting a segment of length y and recursively calculate the cuts
+        int b = solve(n-y, x, y, z, dp) + 1;
+        // Try cutting a segment of length z and recursively calculate the cuts
+        int c = solve(n-z, x, y, z, dp) + 1;
+
+        // Store the result in the dp array for future use
+        dp[n] = max(a, max(b, c));
+
+        // Return the computed result for the current length `n`
+        return dp[n];
+    }
+
+    // Function to find the maximum number of cuts that can be made
+    int maximizeTheCuts(int n, int x, int y, int z)
+    {
+        // Create a dp array initialized with -1 to indicate uncomputed states
+        vector<int> dp(n+1, -1);
+
+        // Call the recursive function with memoization to calculate the maximum cuts
+        int maxCuts = solve(n, x, y, z, dp);
+
+        // 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;
+    }
+};